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1Chapter 1: Mathematical Modeling And Engineering Problem Solvin g

UNIT 1

1

MATHEMATICAL MODELING AND

ENGINEERING PROBLEM SOLVING

Unit Structure

1.0 ObMectives

1.1 Introduction

1.2 Mathematical Modeling

1.3 Conservation Law and Engineering Problems

1.4 Summary

1.5 Exercises

1.0 Objectives

The obMective of this chapter is to introduce the reader to mat hematical modeling

and its application in engineering problem solving. This chapte r also discuss the

conservation laws and how numerical methods are useful in engin eering problem

solving.

1.1 Introduction

Mathematics is widely used in solving real-world problems espec ially due to the

increasing computational power of digital computers and computi ng methods,

which have facilitated the easy handling of lengthy and complic ated problems.

Numerical methods are technique s by which mathematical problems are

formulated so that they can be solved with arithmetic operation s. Although there

are many kinds of numerical methods, they have one common chara cteristic: they

invariably involve large numbers of tedious arithmetic calculat ions and with the

development of fast, efficient digital computers, the role of n umerical methods in

solving engineering problems has increased.

Translating a real-life problem into a mathematical form can gi ve a better

representation of certain problems and helps to find a solution for the problem . munotes.in

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2NUMERICAL AND STATISTICAL METHODS

There are many advantages of translating a real-life problem in to a mathematical

form like

- Mathematics helps us to formulate a real-world problem.

- Computers can be used to perform lengthy numerical calculations .

- Mathematics has well defined rules for manipulations.

In the coming sections we will discuss the concepts of mathemat ical modelling and

how to model a given problem. We will also discuss how mathemat ical modeling

can be used in engineering problem solving.

1.2 Mathematical Modeling

The method of translating a real-life problem into a mathematic al form is called

mathematical modeling. Mathematical modeling is a tool widely u sed in science

and engineering. They provide a rigorous description of various real-world

phenomena. Mathematical modeling helps to understand and analys e these real-

world phenomena.

In mathematical modelling, we consider a real-world problem and write it as a

mathematical problem. We then solve this mathematical problem, and interpret the

solution in terms of the real-world problem which was considere d. We will see how

to formulate a given problem using a very simple example.

Example 1.2.1: To travel 500 kms if 50 litres of petrol is required. How much

petrol is needed to go to a place which is 200 kms away"

Solution. Step I : We will first mathematically formulate this given problem:

Let ݔ be the litres of petrol needed and ݕ be the kilometres travelled. Then we know

that more the distance travelled, more is the petrol required. That is ݔ is directly

proportional to ݕ .

Thus, we can write

ݔൌݕ݇

where ݇ is a positive constant.

Step II : Solving the model using the given data:

Now, since it is given that to travel 500 kms if 50 litres of p etrol are required, we

get

ͷͲ ൌ ݇ ൈͷͲͲ munotes.in

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3Chapter 1: Mathematical Modeling And Engineering Problem Solvin g

That is

݇ൌͳ

ͳͲ

Step III : Using the solution obtained in Step II to interpret the given problem:

To find the litres of petrol required to travel 200kms, we subs titute the values in the

mathematical model. That is

ݔൌͳ

ͳͲൈʹͲͲ

which gives

ݔൌʹ Ͳ

That is, we need 20 litres of petrol to travel 200kms away.

Now, we will see one more example to get a better understanding of mathematical

modeling.

Example 1.2.2: A motorboat goes upstream on a river and covers the distance

between two towns on the riverbank in 12 hours. It covers this distance downstream

in 10 hours. If the speed of the stream is 3 km/hr, find the sp eed of the boat in still

water.

Solution. Step I : We will first mathematically formulate this given problem:

Let ݔ be the speed of the boat, ݐ be the time taken and ݕ be the distance travelled.

Then using the formula for speed, distance and time we get

ݕൌݔݐ

Step II : Solving the model using the given data:

We know that while going upstream, the actual speed of the boat is equal to

ݐܾ݄ܽ݁ݐ݂݀݁݁ݏȂݎ݁ݒ݅ݎ݄݁ݐ݂݀݁݁ݏ

Hence, in upstream, the actual speed of the boat is ݔെ͵.

Similarly, while going downstream, the actual speed of the boat is equal to

ݐܾ݄ܽ݁ݐ݂݀݁݁ݏݎ݁ݒ݅ݎ݄݁ݐ݂݀݁݁ݏ

Hence, in downstream, the actual speed of the boat is ݔ͵.

Since, the time taken to travel upstream is 12 hours, we have

ݕൌͳ ʹ ሺ ݔെ͵ ሻ munotes.in

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4NUMERICAL AND STATISTICAL METHODS

and as the time taken to travel downstream is 12 hours, we have

ݕൌͳ Ͳ ሺ ݔ͵ ሻ

Thus,

ͳʹሺݔെ͵ሻൌͳ Ͳ ሺ ݔ͵ ሻ

which gives ݔൌ͵ ͵ Ǥ

Step III : Using the solution obtained in Step II to interpret the given problem:

From Step II it is clear that the speed of the boat in still ri ver is 33 km/hr.

Our next example, will show how mathematical modeling can be us ed to study

population of a country.

Example 1.2.: Suppose the current population is 200,000,000 and the birth rat e

and death rates are 0.04 and 0.02 respectively. What will be th e population in 5

years"

Solution. Step I : We will first mathematically formulate this given problem:

We know that the population of a country increases with birth a nd decreases with

death. Let ݐ denote time in years where ݐൌͲ implies the present time. Let ሺݐሻ

denote the population at time ݐǤ

If ܤሺݐሻ denote the number of births and ܦሺݐሻ denote the number of deaths in year

ݐ then

ሺݐͳሻൌሺݐሻܤሺݐሻെܦሺݐሻ

Let ܾൌሺ௧ሻ

ሺ௧ሻ be the birth rate for the interval ݐǡݐͳ and ݀ൌሺ௧ሻ

ሺ௧ሻ be the death rate

for the interval ݐǡݐͳ.

Then

ሺݐͳሻൌሺݐሻܾሺݐሻെ݀ሺݐሻ

ൌሺͳܾെ݀ ሻሺݐሻ

Then, when ݐൌͲ we get

ሺͳሻൌሺͳܾെ݀ ሻሺͲሻ

Similarly, taking ݐൌͳ we get

ሺʹሻൌሺͳܾെ݀ ሻሺͳሻ

ൌሺͳܾെ݀ ሻଶሺͲሻ munotes.in

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5Chapter 1: Mathematical Modeling And Engineering Problem Solvin g

Continuing in this manner we get

ሺݐሻൌሺͳܾെ݀ ሻ௧ሺͲሻ

Taking ሺͳെܾെ݀ ሻൌܿ we get

ሺݐሻൌܿ௧ሺͲሻ

where ܿ is called the growth rate.

Step II : Solving the model using the given data:

From the given data, we have ܾൌͲ.04 and ݀ൌͲ Ǥ Ͳ ʹ Ǥ Thus

ܿൌͳܾെ݀

ൌͳ Ǥ Ͳ ʹ

Step III : Using the solution obtained in Step II to interpret the given problem:

The population in 5 years can be estimated as

ሺͷሻൌሺͳǤͲʹሻହൈʹͲͲǡͲͲͲǡͲͲͲ

ൌʹ ʹ Ͳ ͺ ͳ ͳ Ͳ Ǥ

Since population cannot be in decimal, we round it to 220816161 .

1. Conservation Law and Engineering Problems

Conservation laws of science and engineering basically deals wi th

݄݁݃݊ܽܥ ൌ ݁ݏܽ݁ݎܿ݊ܫ െ݁ݏܽ݁ݎܿ݁ܦ

This equation incorporates the fundamental way in which conserv ation laws are

used in engineering that is t o determine the change with respec t to time.

Another way to use conservation laws is the case in which the c hange is 0, that is

݁ݏܽ݁ݎܿ݊ܫ ൌ ݁ݏܽ݁ݎܿ݁ܦ

This case is also known as ݕݐ݀ܽ݁ݐܵെ݁ݐܽݐܵ computation.

Let ܣ be the quantity of interest defined on a domain ȳǤ Then the rate of change of

ܳ is equal to the total amount of ܳ produced or destroyed in ك߱ȳ and the flux of

ܳ across the boundary ߲߱ that is the amount of ܳ that either goes in or comes out

of ߱Ǥ

This can be mathematically expressed as

݀

ݐ݀නܳ

ఠݔ݀ ൌ නݔ݀ܵ

ఠെන ݒܨ

డఠߪ݀ሺݔሻ munotes.in

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6NUMERICAL AND STATISTICAL METHODS

where ݒ is the unit outward normal, ߪ݀ሺݔሻ is the surface measure and ܨǡܵ are flux

and the quantity produced or destroyed respectively.

On simplifying, using integration by parts rule, we get

ܷ௧ݒ݅݀ሺܨሻൌܵ

where ݒ݅݀ሺܨሻ is the divergence.

When ܵ is taken to be zero, we get

ܷ௧ݒ݅݀ሺܨሻൌͲ

which is called the conservation law, as the only change in ܷ comes from the

quantity entering or leaving the domain of interest.

As an example, consider the following:

Heat Equation:

Assume that a hot rod is heated at one end and is left to cool, without providing any

further source of heat. The heat spreads uniformly (that is dif fuses out) and the

temperature of the rod becomes uniform after some time.

Here, let ܷ be the temperature of the material. The diffusion of the heat is given by

Fourier’s la w as

ܨሺܷሻൌെ ܷ݇

where ݇ is the conductivity of the medium.

Thus, the heat equation is obtained as

ܷ௧െݒ݅݀ሺܷ݇ሻൌͲ

Conservation laws arise in many models in science and numerical methods play a

very important role in approximating or simulating the solution s of conservation

laws. Most of the engineering probl ems deals with conservation laws.

x Chemical engineering focus on mass balance reactors.

x In civil engineering force balances are utilized to analyse str uctures.

x In mechanical engineering the same principles are used to analy se the

transient up-and down motion or vibrations of an automobile.

x Electrical engineering uses conservation of energy voltage bala nce. munotes.in

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7Chapter 1: Mathematical Modeling And Engineering Problem Solvin g

1.4 Summary

This chapter

x gives an introduction to mathematical modeling.

x discusses the application of mathematical modelling in engineer ing problem

solving.

x discusses the conservation laws and how numerical methods are u seful in

engineering problem solving.

1.5 Exercises

1. An investor invested ͳͲǡͲͲͲȀെ at ͳͲΨ simple interest per year. With the

return from the investment, he wants to buy a T.V. that costs ʹͲǡͲͲͲȀെǤ

After how many years will he be able to buy the T.V."

2. A car starts from a place ܣ and travels at a speed of ͵Ͳ km/hr towards another

place ܤǤ At the same time another car starts from ܤ and travels towards ܣ at

a speed of ʹͲ km/hr. If the distance between ܣ and ܤ is ͳʹͲ kms, after how

much time will the cars meet"

3. A farmhouse uses atleast 1000 kg of special food daily. The spe cial food is a

mixture of corn and bean with the composition Material Nutrient Present Per Kg Cost Per Kg Protein Fibre Corn 0.09 0.02 10 Bean 0.6 0.06 20

The dietary requirements of the special food are atleast 30 pr otein and atmost 5

fibre. Formulate this problem to minimise the cost of the food.

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8NUMERICAL AND STATISTICAL METHODSUNIT 1

2

APPRO;IMATION ROUND-OFF ERRORS

Unit Structure

2.0 ObMectives

2.1 Introduction

2.2 Significant Figures

2.3 Accuracy Precision

2.4 Error Definitions

2.4.1 Calculation of Errors

2.4.2 Error Estimates for Iterative Methods

2.5 Round-Off Errors

2.6 Problems on Errors Significant numbers

2.7 Summary

2.8 References

2.9 Exercises

2.0 Objectives

After reading this chapter, you should be able to:

1. Know the concept of significant digits.

2. Know the concept Accuracy precision.

3. find the true and relative true error,

4. find the approximate and relative approximate error,

5. relate the absolute relative approximate error to the number of s i g n i f i c a n t

digits

6. Know other different errors including round-off error.

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9Chapter 2: Approximation Round-o൵ Errors2.1 Introduction

Approximate numbers : There are two types of numbers exact and approximate.

Exact numbers are 2, 4, 9,

ଶ , 6.45, .... etc. but there are numbers such that

ସ

ଷ (=1.333…..) , √ 2 ( 1.414213 ...) and S ( 3.141592. …) which cannot be

expressed by a finite number of digits. These may be approximat ed by numbers

1.3333,1.4141, and 3.1416, respectively. Such numbers, which re present the given

numbers to a certain degree of accuracy, are called approximate numbers.

Rounding-off : There are numbers with many digits, e.g., ଶଶ

3.142857143.

In practice, it is desirable to limit such numbers to a managea ble number of digits,

such as 3.14 or 3.143. This process of dropping unwanted digits is called rounding-

off.

In this chapter information concerned with the quantification o f error is discussed

in the first sections. This is followed by a section on one of the two maMor forms of

numerical error: round-off error. Round-off error is due to the fact that computers

can represent only quantities with a finite number of digits.

2.2 Significant Figures

Whenever we use a number in a computation, we must have assuran ce that it can

be used with confidence. For example, Fig. 2.1 depicts a speedo meter and odometer

from an automobile.

Fig. 2.1: An automobile speedometer and odometer illustrating t he concept

of a significant figure.

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10NUMERICAL AND STATISTICAL METHODSVisual inspection of the speedometer indicates the speed betwee n 180 and 200

MPH. Because the indicator is lesser than the midpoint between the markers on the

gauge, it can be said with assurance that the car is traveling at approximately 190

MPH. However, let us say that we insist that the speed be estim ated to one decimal

place. For this case, one person might say 188.8, whereas anoth er might say 188.9

MPH. Therefore, because of the l imits of this instrument, only the first three digits

can be used with confidence. Estimates of the fourth digit (or higher) must be

viewed as approximations. It would be nonsensical to claim, on the basis of this

speedometer, that the automobile is traveling at 188.8642138 MP H. In contrast, the

odometer provides up to six certai n digits. From Fig. 2.1, we c an conclude that the

car has traveled slightly less than 248.5 km during its lifetim e. In this case, the fifth

digit (and higher) is uncertain.

Definition: The number of significant figures or significant digits in the

representation of a number is the number of digits that can be used with confidence .

In particular, for our purposes, the number of significant digi ts is equal to the

number of digits that are known (or assumed) to be correct plus one estimated digit.

o The concept of a significant figure, or digit, has been develop ed to formally

designate the reliability of a numerical value.

o The significant digits of a numbe r are those that can be used w ith

confidence.

o They correspond to the number of certain digits plus one estima ted digit.

Significant digits . The digits used to express a number are called significant di gits.

The digits 1, 2, 3, 4, 5, 6, 7, 8, 9 are significant digits. ‘0’ is also a si gnificant digit

except when it is used to fix the decimal point or to fill the places of unknown or

discarded digits.

For example , each of the numbers 7845, 3.589, and 0.4758 contains 4 signif icant

figures while the numbers 0.00386, 0.000587, 0.0000296 contain only three

significant figures (since zer os only help to fix the position of the decimal point).

Similarly, in the number 0.0003090, the first four ‘0’ s’ are n ot significant digits

since they serve only to fix t he position of the decimal point and indicate the place

values of the other digits. The other two ‘0’ s’ are significan t.

To be more clear, the number 3.0686 contains five significant d igits.

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11Chapter 2: Approximation Round-o൵ ErrorsThe significant figure in a numbe r in positional notation consi sts of

1) All non-zero digits

2) =ero digits which

lie between significant digits

lie to the right of decimal point and at the same time to the r ight of a

non-zero digit

are specifically indicated to be significant.

The significant figure in a number written in scientific notati on

(e.g., M î 10k) consists of all the digits explicitly in M.

Example 1

Give some examples of showing t he number of significant digits.

Solution

a) 0.0459 has three significant digits

b) 4.590 has four significant digits

c) 4008 has four significant digits

d) 4008.0 has five significant digits

e) 310 079.1 u has four significant digits

f) 310 0790.1 u has five significant digits

g) 310 07900.1 u has six significant digits

Significant digits are counted from left to right starting with the non- zero digit on

the left.

A list is provided to help students understand how to calculate significant digits in

a given number: Number Significant digits Number of significant digits 3969 3, 9, 6, 9 04 3060 3, 0, 6 03 3900 3, 9 02 39.69 3, 9, 6, 9 04 0.3969 3, 9, 6, 9 04 munotes.in

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12NUMERICAL AND STATISTICAL METHODS Number Significant digits Number of significant digits 39.00 3, 9, 0, 0 04 0.00039 3, 9 02 0.00390 3, 9, 0 03 3.0069 3, 0, 0, 6, 9 05 3.9 î 10 6 3, 9 02 3.909 î 10 5 3, 9, 0, 9 04 6 î 10 –2 6 01 The concept of significant figures has two important implicatio ns for study of

numerical methods:

1. Numerical methods yield approximate results . We must, therefore, develop

criteria to specify how confident we are in our approximate res ult. One way

to do this is in terms of significant figures. For example, we might decide that

our approximation is acceptable if it is correct to four signif icant figures.

2. Although quantities such as π, e, or √7 represent specific qu antities, they

cannot be expressed exactly by a limited number of digits.

For example,

π = 3.141592653589793238462643. . .

ad infinitum. Because computers retain only a finite number of significant figures,

such numbers can never be represented exactly. The omission of the remaining

significant figures is called round-off error.

The concept of significant figures will have relevance to our d efinition of accuracy

and precision in the next section.

Applications:

The output from a physical measuring device or sensor is genera lly known to be

correct up to a fixed number of digits.

For example, if the temperature is measured with a thermometer is calibrated

between 85.6 and 85.7 degrees. the first three digits of the te mperature are known

i.e. 8, 5, and 6, but do not know the value of any subsequent d igits.

The first unknown digit is sometimes estimated at half of the v alue of the

calibration size, or 0.05 degrees in our case. If we did this, the temperature would

be reported to four significant digits as 85.65 degrees. munotes.in

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13Chapter 2: Approximation Round-o൵ ErrorsAlternatively, we can choose to report the temperature to only three significant

digits.

In this case, we could truncate or chop off the unknown digits to give a result of

48.6 degrees, or round off the result to the nearest tenth of a degree to give either

48.6 or 48.7 degrees depending on whether the actual reading wa s more or less than

half-way between the two calibrations.

Round-off is generally the preferred procedure in this example, but without

knowing which technique was adopted, we would really only be co nfident in the

first two digits of the temperature.

Hence, if the temperature is reported as 48.6 degrees without a ny further

explanation, we do not know whether the 6 is a correct digit or an estimated digit.

2. Accuracy and Precision

The errors associated with both calculations and measurements c an be

characterized with regard to their accuracy and precision.

Accuracy refers to how closely a computed or measured value agrees with the true

value .

Accuracy tells us how close a measured value is to the actual v alue.

It is associated with the quality of data and numbers of errors present in the data

set. Accuracy can be calculated using a single factor or measur ement.

Precision tells us how close measured values are to each other .

It often results in round off errors. To calculate precision, m ultiple measurements

are required.

Example :- The distance between point A and B is 7.15. On measuring wit h

different devices the distance appears as:

Data set 1: 6.34, 6.31, 6.32

Data set 2 : 7.11, 7.19, 7.9

Data set 1 is more precise since they are close to each other a nd data set 2 is more

accurate since they are close to the actual value.

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14NUMERICAL AND STATISTICAL METHODSTo round-off a number to n significant digits to get accuracy, discard all

digits to the right of the nth digit and if this discarded number is

o less than 5 in ( n 1)th place, leave the nth digit unaltered. e.g., 7.893

to 7.89.

o greater than 5 in ( n + 1)th place, increase the nth digit by unity, e.g.,

6.3456 to 6.346.

o exactly 5 in ( n 1)th place, increase the nth digit by unity i f it is odd,

otherwise leave it unchanged.

e.g., 12.675 a 12.68 12.685 a 12.68 The number thus rounded-off is said to be correct to n significant figures. A

list is provided for explanatory proposes: Number Rounded-off to Three digits Four digits Five digits 00.543241 00.543 00.5432 00.54324 39.5255 39.5 39.52 39.526 69.4155 69.4 69.42 69.416 00.667676 00.668 00.6677 00.66768

Graphical Illustration of Precision Accuracy

Precision Accuracy is illustrated graphically using an analog y from target

practice.

The bullet holes on each target in Fig. 2.2 can be thought of a s the predictions of a

numerical technique, whereas the bull’s -eye represents the truth.

o Inaccuracy (also called bias) is defined as systematic deviation from the

truth. Is illustrated on circles on right side

o Imprecision (also called uncertainty ), on the other hand, refers to the

magnitude of the scatter is represented on left side circles wh ere the shots are

tightly grouped. munotes.in

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15Chapter 2: Approximation Round-o൵ Errors

Fig 2.2 Graphical Illustration of Precision & Accuracy

2.4 Error Definitions

Machine epsilon

We know that a computer has a finite word length, so only a fix ed number of

digits is stored and used during computation. Hence, even in st oring an exact

decimal number in its converted form in the computer memory, an e r r o r i s

introduced. This error is machine dependant and is called machi ne epsilon.

In any numerical computation, we come across the following type s of errors:

Inherent errors. Errors which are already present in the statement of a

problem before its solution are called inherent errors. Such er rors arise either

due to the fact that the given data is approximate or due to limitations of

mathematical tables, calculators, or the digital computer. Error True value – Approximate

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16NUMERICAL AND STATISTICAL METHODSInherent errors can be minimized by taking better data or by us ing high

precision computing aids. Accuracy refers to the number of sig nificant digits

in a value, for example, 53.965 is accurate to 5 significant di gits.

Precision refers to the number of decimal positions or order of magnitude of the

last digit in the value. For example, in 53.965, precision is 1 0–3.

Example. Which of the following numbers has the greatest precision?

4.3201, 4.32, 4.320106.

Sol. In 4.3201, precision is 10 –4

In 4.32, precision is 10 –2

In 4.320106, precision is 10 –6.

Truncation errors

Truncation errors are caused by using approximate results or by replacing an

infinite process with a finite one.

If we are using a decimal computer having a fixed word length o f 4 digits,

rounding-off of 13.658 gives 13.66, whereas truncation gives 13 .65.

Therefore this difference in last significant digits causes err or as result of difference

in processing of rounding-off truncating a number.

Numerical errors arise from the use of approximations to repres ent exact

mathematical operations and quantities. These include truncatio n errors, which

result when approximations are used to represent exact mathemat ical procedures,

and round-off errors, which result when numbers having limited significant figures

are used to represent exact numbers.

For both types, the relationship between the exact, or true, re sult and the

approximation can be formulated as

True value approximation error (2.1)

By rearranging Eq. (2.1), we find that the numerical error is e qual to the

discrepancy between the truth and the approximation, as in

ܧ௧ = true value − approximation (2.2) munotes.in

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17Chapter 2: Approximation Round-o൵ Errorswhere ܧ௧ is used to designate the exact value of the error. The subscri pt t is included

to designate that this is the “true” error. This is in contrast to other cas es, as

described shortly, where an “approximate” estimate of the error must be employed.

A shortcoming of this definition is that it takes no account of the order of magnitude

of the value under examination. For example, an error of a cent imeter is much more

significant if we are measuring a rivet rather than a bridge. O ne way to account for

the magnitudes of the quantities being evaluated is to normaliz e the error to the true

value, as in

ൌݎݎݎ݁݁ݑݎݐ

݁ݑ݈ܽݒ݁ݑݎݐ

where, as specified by Eq. (2.2), error true value − approximation.

The relative error can also be multiplied by 100 percent to exp ress it as

ߝ௧ൌ௧௨

௧௨௩௨ܺͳͲͲ (2.3)

where ߝ௧ designates the true percent relative error.

2.4.1 Calculation of Errors

Example 2.4.1

Problem Statement . Suppose that you have the tas k of measuring the lengths of a

bridge and a rivet and come up with 9999 and 9 cm, respectively . If the true values

are 10,000 and 10 cm, respectively, compute (a) the true error and (b) the true

percent relative error for each case.

Solution.

(a) The error for measuring the bridge is >Eq. (2.2)@

ܧ௧ = 10,000 − 9999 = 1 cm

and for the rivet it is

ܧ௧ = 10 − 9 = 1 cm

(b) The percent relative error for the bridge is >Eq. (2.3)@

ߝ௧ൌͳ

ͳͲǡͲͲͲܺͳͲͲ ൌ ͲǤͲͳΨ

and for the rivet it is

ߝ௧ൌͳ

ͳͲܺͳͲͲ ൌ ͳͲΨ munotes.in

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18NUMERICAL AND STATISTICAL METHODSThus, although both measurements have an error of 1 cm, the rel ative error for the

rivet is much greater. We would conclude that we have done an a dequate Mob of

measuring the bridge, whereas our estimate for the rivet leaves something to be

desired.

Notice that for Eqs. (2.2) and (2 .3), E and ε are subscripted with a t to signify that

the error is normalized to the true value.

In real-world applications, we will obviously not know the true answer a priori. For

these situations, an alternative is to normalize the error usin g the best available

estimate of the true value, that is, to the approximation itsel f, as

ߝൌୟ୮୮୰୭୶୧୫ୟ୲ୣୣ୰୰୭୰

ୟ୮୮୰୭୶୧୫ୟ୲୧୭୬ܺͳͲͲ (2.4)

where the subscript a signifies that the error is normalized to an approximate

value.

Note also that for real-world applications, Eq. (2.2) cannot be used to calculate the

error term for Eq. (2.4). One of the challenges of numerical me thods is to determine

error estimates in the absence of knowledge regarding the true value.

For example, certain numerical methods use an iterative approac h to compute

answers. In such an approach, a present approximation is made o n the basis of a

previous approximation. This process is performed repeatedly, o r iteratively, to

successively compute (we hope) better and better approximations .

For such cases, the error is often estimated as the difference between previous and

current approximations. Thus, percent relative error is determi ned according to

ߝൌୡ୳୰୰ୣ୬୲ୟ୮୮୰୭୶୧୫ୟ୲୧୭୬ି୮୰ୣ୴୧୭୳ୱୟ୮୮୰୭୶୧୫ୟ୲୧୭୬

ୡ୳୰୰ୣ୬୲ୟ୮୮୰୭୶୧୫ୟ୲୧୭୬ܺͳͲͲ (2.5)

The signs of Eqs. (2.2) through (2.5) may be either positive or negative. If the

approximation is greater than the true value (or the previous a pproximation is

greater than the current approximati on), the error is negative if the approximation

is less than the true value, the error is positive. Also, for E qs. (2.3) to (2.5), the

denominator may be less than zer o, which can also lead to a neg ative error. Often,

when performing computations, we may not be concerned with the sign of the error,

but we are interested in whether the percent absolute value is lower than a

prespecified percent tolerance ߝ௦ . Therefore, it is often useful to employ the

absolute value of Eqs. (2.2) through (2.5). For such cases, the c o m p u t a t i o n i s

repeated until

_ ߝ_ ߝ௦ (2.6) munotes.in

## Page 19

19Chapter 2: Approximation Round-o൵ ErrorsIf this relationship holds, our result is assumed to be within the prespecified

acceptable level ߝ௦ . Note that for the remainder of this text, we will almost

exclusively employ absolute valu es when we use relative errors.

It is also convenient to relate these errors to the number of s ignificant figures in the

approximation.

It can be shown (Scarborough, 1966) that if the following crite rion is met, we can

be assured that the result is cor rect to at least n significant figures.

ߝ௦ = (0.5 × 102−n)% (2.7)

2.4.2 Error Estimates for Iterative Methods

Problem Statement . In mathematics, functions can often be represented by infinit e

series. For example, the exponential function can be computed u sing

݁௫ൌͳݔ௫మ

ଶǨ௫య

ଷǨڮ௫

Ǩ (2.8)

Thus, as more terms are added in sequence, the approximation be comes a better

and better estimate of the true value of ݁௫. Equation (Eq. 2.8) is called a Maclaurin

series expansion.

Starting with the simplest version, ݁௫ 1, add terms one at a time to estimate ݁Ǥହ.

After each new term is added, compute the true and approximate percent relative

errors with Eqs. (2.3) and (2.5), respectively. Note that the t rue value is

݁Ǥହ 1.648721 . . . Add terms until the absolute value of the app roximate error

estimate ߝ falls below a prespecified error criterion ߝ௦ conforming to three

significant figures.

Solution. First, Eq. (2.7) can be employed to determine the error crite rion that

ensures a result is correct to at least three significant figur es:

ߝ௦ = (0.5 × 102−3)% = 0.05%

Thus, we will add terms to the series until ߝ falls below this level.

The first estimate is simply equal to Eq. (2.8) with a single t erm. Thus, the first

estimate is equal to 1. The sec ond estimate is then generated b y adding the second

term, as in

ex 1 x

or for x 0.5,

e0.5 1 0.5 1.5 munotes.in

## Page 20

20NUMERICAL AND STATISTICAL METHODSThis represents a true per cent relative error of >Eq. (2.3)@

ߝ௧ൌͳǤͶͺʹͳെͳǤͷ

ͳǤͶͺʹͳܺͳͲͲ ൌ ͻǤͲʹΨ

Equation (2.5) can be used to determine an approximate estimate of the error, as in

ߝൌͳǤͷെͳ

ͳǤͷܺͳͲͲ ൌ ͵͵Ψ

Because ߝ is not less than the required value of ߝ௦, we would continue the

computation by adding another term, x2/2, and repeating the error calculations.

The process is continued until ߝ ߝ௦ .

The entire computation can be summarized as Terms Result ࢿ )(ࢿ )(1 1 39.3 2 1.5 9.02 33.3 3 1.625 1.44 7.69 4 1.645833333 0.175 1.27 5 1.648437500 0.0172 0.158 6 1.648697917 0.00142 0.0158 Thus, after six terms are included, the approximate error falls below ߝ௦ 0.05

and the computation is terminated. However, notice that, rather t h a n t h r e e

significant figures, the result is accurate to five This is be cause, for this case, both

Eqs. (2.5) and (2.7) are conservative.

That is, they ensure that the result is at least as good as the y specify.

2.5 Round-Off errors

Rounding errors. Rounding errors arise from the process of rounding- off

numbers during the computation. They are also called procedual errors or

numerical errors. Such errors are unavoidable in most of the ca lculations due

to limitations of computing aids.

These errors can be reduced, however, by

I. changing the calculation procedure so as to avoid subtractio n of nearly

equal numbers or division by a small number

II. retaining at least one more significant digit at each step and rounding-off

at the last step. Rounding-off may be executed in two ways:

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## Page 21

21Chapter 2: Approximation Round-o൵ Errors

Chopping. In chopping, extra digits are dropped by truncation of number.

Suppose we are using a computer with a fixed word length of fou r digits, then

a number like 12.92364 will be stored as 12.92.

We can express the number 12.92364 in the floating print form as

True x 12.92364

0.1292364 î 10 2 (0.1292 0.0000364) î 10 2

0.1292 î 10 2 0.364 î 10 –4 2

fx . 10E gx . 10E – d

Approximate x Error

Error gx . 10E – d, 0 ≤ gx ≤ d

Here, gx is the mantissa, d is the length of mantissa and E is exponent

Since 0 ≤ gx 1

Absolute error ≤ 10E – d

Case I. If gx 0.5 then approximate x fx . 10E

Case II. If gx ≥ .5 then approximate x fx . 10E 10 E – d

Error True value – Approximate value

fx . 10E gx . 10E – d – fx .10E – 10E – d

( gx – 1) . 10 E – d

absolute error ≤ 0.5.(10) E – d.

Symmetric round-off. In symmetric round-off, the l ast retained significant

digit is rounded up by unity if the first discarded digit is ≥ 5, otherwise the

last retained digit is unchanged.

2.6 Problems on Error Definitions, Significant numbers

accuracy

Problem 1. Suppose 1.414 is used as an approximation to √2 . Find the

absolute and relative errors.

Sol. True value 1.41421356

Approximate value 1.414

Error True value – Approximate value

– 1.414 1.41421356 – 1.414

0.00021356

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## Page 22

22NUMERICAL AND STATISTICAL METHODSAbsolute error ea _ Error _

_ 0.00021356 _ 0.21356 î 10 –3

Relative error er ea True value

0.151 î 10 –3.

Problem 2. If 0.333 is the approximate value of ଵ

ଷ , f i n d t h e a b s o l u t e ,

relative, and percentage errors.

Solution. True value (;) ଵ

ଷ

Approximate value (;

) 0.333

A b s o l u t e e r r o r ea _ ; – ;’_

_ ଵ

ଷെͲǤ͵͵͵_ _ 0.333333 – 0.333 _ . 000333

Relative error ݁ൌ ೌ

௫ Ǥଷଷଷ

Ǥଷଷଷଷଷଷ 0.000999

Percentage error ep er î 100 .000999 î 100 .099

Problem . An approximate value of π i s given by 3.1428571 and its true

value is 3.1415926. Find the absolute and relative errors.

Sol. True value 3.1415926

Approximate value 3.1428571

Error True value – Approximate value

3.1415926 – 3.1428571

– 0.0012645

Absolute error ea _ Error _ 0.0012645

Relative error ݁ൌ ೌ

்௨௨ Ǥଵଶସହ

ଷǤଵସଵହଽଶ

0.000402502.

Problem 4. Three approximate values of the number ଵ

ଷ are given as 0.30,

0.33,and 0.34. Which of these three is the best approximation"

Sol. The best approximation will be the one which has the least abso lute error.

True value ଵ

ଷ 0.33333

0.21356 u 103

2 munotes.in

## Page 23

23Chapter 2: Approximation Round-o൵ ErrorsCase I. Approximate value 0.30

Absolute error _ True value – Approximate value _

_ 0.33333 – 0.30 _

0.03333

Case II. Approximate value 0.33

Absolute error _ True value – Approximate value _

_ 0.33333 – 0.33 _

0.00333.

Case III. Approximate value 0.34

Absolute error _ True value – Approximate value _

_ 0.33333 – 0.34 _

_ – 0.00667 _ 0.00667

Since the absolute error is least in case II, 0.33 is the best approximation.

Problem 5. Find the relative error of the number 8.6 if both of its digits

are correct.

Sol. Here, ea 0.5 ݁ ൌଵ

ଶܺͳͲିଵ

݁ൌ Ǥହ

଼Ǥ ൌǤͲͲͷͺ

Problem 6. Find the relative error if ଶ

ଷ is approximated to 0.667.

Sol. True value ଶ

ଷൌ 0.666666

Approximate value 0.667

Absolute error ea _ True value – approximate value _

_ .666666 – .667 _ .000334

Relative error ݁ൌ Ǥଷଷସ

Ǥ .0005

Problem 7 . Find the percentage error if 625.483 is approximated to

three significant figures.

Sol. ea _ 625.483 – 625 _ 0.483

݁ൌ ೌ

ଶହǤସ଼ଷ Ǥସ଼ଷ

ଶହǤସ଼ଷ 0.000772

e p er î 100

ep 0.000772 î 100 0.77 munotes.in

## Page 24

24NUMERICAL AND STATISTICAL METHODSProblem . Round-off the numbers 865250 and 37.46235 to four

significant figures and compute e a, er, ep in each case.

Sol. (i) Number rounded-off to four significant digits 865200

; 865250

;

865200

Error ; – ;

865250 – 865200 50

Absolute error ea _ error _ 50

R e l a t i v e e r r o r ݁ൌ ೌ

ଡ଼ ହ

଼ହଶହ 5.77 î 10 –5

Percentage error ep er î 100 5.77 î 10 –3

( ii) Number rounded-off to four significant digits 37.46

; 37.46235

;

37.46

Error ; – ;

0.00235

Absolute error ea _ error _ 0.00235

Relative error ݁ൌ ೌ

ଡ଼ Ǥଶଷହ

ଷǤସଶଷହ 6.2729î 10 –5

Percentage error ep er î 100 6.2729 î 10 –3.

Problem . Round-off the number 75462 to four significant digits and

then calculate the absolute error and percentage error .

Sol. Number rounded-off to four significant digits 75460

Absolute error ea _ 75462 – 75460 _ 2

Relative error ݁ൌ ೌ

ହସଶ ଶ

ହସଶ 0.0000265

Percentage error ep er î 100 0.00265

Problem 10. Find the absolute, relative, a nd percentage errors if x is

rounded- off to three decimal digits. Given x 0.005998.

Sol. Number rounded-off to three decimal digits .006

Error .005998 – .006 – .000002

Absolute error ea _ error _ .000002

Relative error ݁ൌ ೌ

Ǥହଽଽ଼ Ǥଶ

Ǥହଽଽ଼ 0.0033344

Percentage error ep er î 100 0. 33344 munotes.in

## Page 25

25Chapter 2: Approximation Round-o൵ Errors2.7 Summary

The second chapter of this book introduces the learner with the c o n c e p t s o f

approximations, rounding, significant digits and error which pl ays a very crucial

role in solving a problem using numerical methods. Rules to rou nd-off a given data

and rules to determine significant digits are discussed which i s useful in achieving

the desired accuracy in the given problem.

2. References

Following books are recommended for further reading :-

1) Numerical Methods for Engineers Steven C Chapra Raymond P Can ale –

6th Edition

2) Introductory Methods of Numerical Methods by S S Shastri

3) Computer based Numerical Statistical Techniques – M. Goyal

4) Numerical Statistical Methods by Bhupendra T Kesria, Himalaya

Publishing House

2. Exercises

1. Round-off the following numbers correct to four significant dig its:

3.26425, 35.46735, 4985561, 0.70035, 0.00032217, 1.6583, 30.05 67,

0.859378, 3.14159.

2. The height of an observation tower was estimated to be 47 m.

whereas its actual height was 45 m. Calculate the percentage of

relative error in the measurement.

3. If true value ଵ

ଷ , approximate value 3.33, find the absolute and

relative errors.

4. Round-off the following numbe rs to two decimal

4 8 . 2 1 4 1 6 , 2 . 3 7 4 2 , 5 2 . 2 7 5 , 2 . 3 7 5 , 2 . 3 8 5 , 8 1 . 2 5 5 .

5. If ; 2.536, find the absolute error and relative error when

ȋȌ ; is rounded-off

ȋȌ ; is truncated to two decimal digits. munotes.in

## Page 26

26NUMERICAL AND STATISTICAL METHODS6. If ߨൌଶଶ

is approximated as 3.14, find the absolute error, relative

error, and percentage of relative error.

7. Given the solution of a problem as ;

35.25 with the relative error

in the solution atmost 2, find, to four decimal digits, the ra nge of

values within which t he exact value of the solution must lie.

8. Given that:

a 10.00 0.05, b 0.0356 0.0002

c 1 5 3 0 0 1 0 0 , d 6 2 0 0 0 5 0 0

Find the maximum value o f the absolute error in

ȋȌ a b c d (ii) a 5c – d (iii) d3.

9. What do you understand by machine epsilon of a computer" Explai n.

10. What do you mean by truncation error" Explain with examples.

munotes.in

## Page 27

27Chapter 3: Truncation Errors the Taylor SeriesUNIT 1

TRUNCATION ERRORS

THE TAYLOR SERIES

Unit Structure

3.0 ObMectives

3.1 Introduction

3.2 The Taylor ’s Series

3.2.1 Taylor’s Theorem

3.2.2 Taylor Series Approximation of a Polynomial

3.2.3 The Remainder for the Taylor Series Expansion

3.2.4 Error in Taylor Series

3.3 Error Propagation

3.3.1 Functions of a Single Variable

3.3.2 Propagation in a Function of a Single Variable

3.4 Total Numerical Errors

3.5 Formulation Errors and Data Uncertainty

3.6 Summary

3.7 References

3.8 Exercise

.0 Objectives

After reading this chapter, you should be able to

1. Understand the basics of Taylor’s theorem,

2. Write trigonometric functions as Taylor’s polynomial,

3. Use Taylor’s theorem to find the values of a function at any poi nt, given the

values of the function and all its derivatives at a particular point,

4. Calculate errors and error bounds of approximating a function b y Taylor

series

5. Revisit whenever Taylor’s theorem is used to derive or explain numerical

methods for various mathematical procedures. munotes.in

## Page 28

28NUMERICAL AND STATISTICAL METHODS.1 Introduction

Truncation errors are those that result from using an approxima tion in place of an

exact mathematical procedure. A truncation error was introduced i n t o t h e

numerical solution because the difference equation only approxi mates the true

value of the derivative. In order to gain insight into the prop erties of such errors,

we now turn to a mathematical formulation that is used widely i n numerical

methods to express functions in an approximate manner —the Taylor series.

In this chapter Understanding the basics of Taylor’s theorem, Taylors series

Truncation errors, remainder in taylor series, error propagatio n, Total numerical

errors is covered

.2 The Taylor ’s Series

Taylor’s theorem and its associated formula, the Taylor series, is of great val ue in

the study of numerical methods. In essence, the Taylor series provides a means to

predict a function value at one point in terms of the function value and its

derivatives at another point . In particular, the theorem states that any smooth

function can be approximated as a polynomial.

A useful way to gain insight into the Taylor series is to build it term by term. For

example, the first term in the series is

f(xi+1) ؆ f(x i) (3.1)

This relationship, called the zero-order approximation , indicates that the value of

f at the new point is the same as its value at the old point. T his result makes

intuitive sense because if x i and x i1 are close to each other, it is likely that the

new value is probably similar to the old value.

Equation (3.1) provides a perfect estimate if the function bein g approximated is,

in fact, a constant. However, if the function changes at all ov er the interval,

additional terms of the Taylor series are required to provide a better estimate.

For example, the first-order approximation is developed by adding another term

to yield

f(xi+1) ؆ f(x i) + f '(x i) (x i+1 − xi) (3.2)

The additional first-order term consists of a slope f(xi) multiplied by the distance

between xi and xi+1. Thus, the expression is now in the form of a straight line an d

is capable of predicting an increase or decrease of the functio n between xi

and xi+1. munotes.in

## Page 29

29Chapter 3: Truncation Errors the Taylor SeriesAlthough Eq. (3.2) can predict a change, it is exact only for a s t r a i g h t - l i n e , o r

linear,

trend. Therefore, a second-order term is added to the series to capture some of the

curvature that the function might exhibit:

݂ሺݔାଵሻ؆݂ሺݔሻ݂Ԣሺݔ ሻሺݔାଵെݔሻᇲᇲሺ௫ሻ

ଶǨሺݔାଵെݔሻଶ (3.3)

In a similar manner, additional terms can be included to develo p the complete

Taylor series expansion:

݂ሺݔାଵሻ؆݂ሺݔሻ݂Ԣሺݔ ሻሺݔାଵെݔሻᇲᇲሺ௫ሻ

ଶǨሺݔାଵെݔሻଶ

ᇲᇲᇲሺ௫ሻ

ଷǨሺݔାଵെݔሻଷڮሺ௫ሻ

Ǩሺݔାଵെݔሻܴ (3.4)

Note that because Eq. (3.4) is an infinite series, an equal sig n replaces the

approximate sign that was used in Eqs. (3.1) through (3.3). A r emainder term is

included to account for all terms from n 1 to infinity:

ܴൌశభሺஞሻ

ሺାଵሻǨሺݔାଵെݔሻାଵ (3.5)

where the subscript n connotes that this is the remainder for the nth-order

approximation and ξ is a value of x that lies somewhere between ݔ and ݔାଵ.

ξ is a value that provides an exa ct determination of the error.

It is often convenient to simplify the Taylor series by definin g a step size

h ݔାଵ – ݔ and expressing Eq. (3.4) as

݂ሺݔାଵሻ؆݂ሺݔሻ݂Ԣሺݔ ሻ݄ᇲᇲሺ௫ሻ

ଶǨ݄ଶᇲᇲᇲሺ௫ሻ

ଷǨ݄ଷڮሺ௫ሻ

Ǩ݄ܴ (3.6)

where the remainder term is now

ܴൌశభሺஞሻ

ሺାଵሻǨ݄ାଵ (3.7)

.2.1 Taylor’s Theorem

If the function f and its first n 1 derivatives are continuous on an interval

containing a and x, then the value of the function at x is given by

݂ሺݔሻൌ݂ሺܽሻ݂Ԣሺܽሻሺݔ െܽሻᇲᇲሺሻ

ଶǨሺݔെܽሻଶᇲᇲᇲሺሻ

ଷǨሺݔ െܽሻଷڮ

ሺሻ

Ǩሺݔ െܽሻܴ (3.8) munotes.in

## Page 30

30NUMERICAL AND STATISTICAL METHODSwhere the remainder Rn is defined as

ܴൌሺ௫ି௧ሻ

Ǩ݂ሺାଵሻሺݐሻݐ݀௫

(3.9)

where t a dummy variable. Equation (3.8) is called the Taylor series or Taylor’s

formula.

If the remainder is omitted, the right side of Eq. (3.8) is th e Taylor polynomial

approximation to f (x). In essence, the theorem sta tes that any smooth function

can be approximated as a polynomial.

Equation (3.9) is but one way, called the integral form, by whi ch the remainder

can be expressed. An alternative formulation can be derived on the basis of the

integral mean-value theorem.

.2.2.Taylor Series Approximation of a Polynomial

Problem Statement. Use zero- t hrough fourth-order Taylor series e x p a n s i o n s t o

approximate the function

f(x) = −0.1x4 − 0.15x3 − 0.5x2 − 0.25x + 1.2

from ݔ 0 with h = 1. That is, predict the function’s value at ݔାଵ 1.

Solution. Because we are dealing with a known function, we can compute values

for

f (x) between 0 and 1.

Therefore f(x) by substituting x 0 becomes

f(0) = −0.1(0)4 − 0.15(0)3 − 0.5(0)2 − 0.25(0) + 1.2

= −0 − 0 − 0 −0 + 1.2

f (0) 1.2

and

f(x) by substituting x 1 becomes

f(1 ) = −0.1(1)4 − 0.15(1)3 − 0.5(1)2 − 0.25(1) + 1.2

= −0.1 – 0.15 – 0.5 −0.25 + 1.2

f (1) 0.2

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## Page 31

31Chapter 3: Truncation Errors the Taylor SeriesThe results (Fig. 3.1) indicate that the function starts at f(0)=1.2 and then curves

downward to f (1) 0.2

Thus, the true value that we are trying to predict is 0.2.

The Taylor series approximation with n 0 is (Eq. 3.1)

݂ሺݔାଵሻͳǤʹ

݂ሺݔାଵሻൌ݂ሺݔሻ݂Ԣሺݔሻ݄ᇲᇲሺ௫ሻ

ଶǨ݄ଶ

Fig .1

The approximation of f(x) = −0.1x4 − 0.15x3 − 0.5x2 − 0.25x + 1.2 at x 1 by

zero-order, first-order, and second -order Taylor series expansi ons.

Thus, as in Fig. 3.1, the zero-order approximation is a constan t. Using this

formulation results in a truncation error ܧ௧

ܧ௧ true value – approximation >Refer Eq. (2.2) in prev. chapter@

Et 0.2 − 1.2 −1.0

at x 1.

For n 1, the first derivative must be determined and evaluated at x 0:

f' (x) −0.1 x 4 x3 − 0.15 x 3 x2 − 0.5 x 2 x − 0.25 x 1 −0.25

f' (0) −0.4(0.0)3 − 0.45(0.0)2 − 1.0(0.0) − 0.25 −0.25 Second order first order

1.0

i + 1 i i

0.5

xi = 0

xi + 1 = 1

x

h True munotes.in

## Page 32

32NUMERICAL AND STATISTICAL METHODS=Therefore, the first-order approximation is >Eq. (3.2)@

f(x i1) = 1.2 − 0.25h

which can be used to compute f (1) 0.95. Consequently, the approximation

begins to capture the downward traMectory of the function in th e form of a

sloping straight line (Fig. 3.1). This results in a reduction o f the truncation error to

Et 0.2 − 0.95 −0.75

For n 2, the second derivative is evaluated at x 0:

f

(0) −1.2(0.0)2 − 0.9(0.0) − 1.0 −1.0

Therefore, according to Eq. (3.3),

f(xi1) 1.2 − 0.25h − 0.5h2

and substituting h = 1, f (1) 0.45. The inclusion of the second derivative now adds

some downward curvature resulting in an improved estimate, as seen in Fig. 3.1.

The truncation error is reduced further to 0.2 − 0.45 −0.25.

Additional terms would improve the approximation even more. In fact, the

inclusion of the third and the fourth derivatives results in ex actly the same

equation we started with:

f(x) 1.2 − 0.25h − 0.5h2 − 0.15h3 − 0.1h4

where the remainder term is

ܴସൌ݂ሺହሻሺɌሻ

ͷǨ݄ହൌͲ

because the fifth derivative of a fourth-order polynomial is ze ro. Consequently, the

Taylor series expansion to the fourth derivative yields an exac t estimate at

xi1 1:

f(1) 1.2 − 0.25(1) − 0.5(1)2 − 0.15(1)3 − 0.1(1)4 0.2

Eq. (3.7) is useful for gaining insight into truncation errors .

This is because we do have control over the term h in the equation. In other

words, we can choose how far away from x we want to evaluate f (x), and we can

control the number of terms we include in the expansion. Conseq uently, Eq. (3.9)

is usually expressed as

Rn O(݄ାଵ)

where the nomenclature O(݄ାଵ)means that the truncation error is of the order of

݄ାଵ. munotes.in

## Page 33

33Chapter 3: Truncation Errors the Taylor Series.2. The Remainder for the Taylor Series Expansion

Before demonstrating how the Taylor series is actually used to estimate numerical

errors, we must explain why we included the argument ξ in Eq. ( 3.7). A

mathematical derivation is presented in point 3.2.1. We will no w develop an

alternative exposition based on a somewhat more visual interpre tation. Then we

can extend this specific case t o the more general formulation.

Suppose that we truncated the Taylor series expansion >Eq. (3.6 )@ after the zero-

order term to yield

f(xi1) f(xi )

A visual depiction of this zero-order prediction is shown in Fi g. 3.2. The

remainder, or error, of this pre diction, which is also shown in the illustration,

consists of the infinite series of terms that were truncated:

ܴൌ݂Ԣሺݔሻ݄݂ᇱᇱሺݔሻ

ʹǨ݄ଶ݂ᇱᇱᇱሺݔሻ

͵Ǩ݄ଷڮ

It is obviously inconvenient to deal with the remainder in this i n f i n i t e s e r i e s

format.

One simplification might be to truncate the remainder itself, a s in

ܴ؆݂Ԣሺݔሻ݄

Fig .2 Graphical depiction of a zero-order Taylor series prediction a nd

remainder.

Exact prediction

R0

xi

xi + 1

x

h munotes.in

## Page 34

34NUMERICAL AND STATISTICAL METHODSThe use of Taylor series exists in so many aspects of numerical methods. For

example, you must have come across expressions such as

(1) 6421) cos(6 4 2x x xx

(2) 753) sin(7 5 3x x xx x

(3) 3213 2x xx ex

All the above expressions are actually a special case of Taylor series called the

Maclaurin series. Why are these applications of Taylor’s theor em important for

numerical methods" Expressions such as given in Equations (1), (2) and (3) give

you a way to find the approximate values of these functions by using the basic

arithmetic operations of additi on, subtraction, division, and m ultiplication.

Example 1

Find the value of 25.0e using the first five terms of the Maclaurin series.

Solution

The first five terms of the Maclaurin series for xeis 43214 3 2x x xx ex| 425.0

325.0

225.025.014 3 2

25.0|e

2840.1

The exact value of 25.0e up to 5 significant digits is also 1.2840.

But the above discussion and example do not answer our question of what a

Taylor series is.

Here it is, for a function xf cccccc 3 2

3 2hxfhxfhxfxf hxf ( 4 )

Provided all derivatives of xf exist and are continuous between x and hx. munotes.in

## Page 35

35Chapter 3: Truncation Errors the Taylor SeriesAs Archimedes would have said ( without the fine print ), “Give me the value of the

function at a single point, and the value of all (first, second, and so on) its

derivatives, and I can give you the valu e of the function at any other point ”.

It is very important to note that the Taylor series is not aski ng for the expression

of the function and its derivatives, Must the value of the func tion and its derivatives

at a single point.

Example 2

Take x xf sin , we all know the value of 12sin ¸

¹·¨

©§S. We also know the x xf cos c and 02cos ¸

¹·¨

©§S. Similarly ) sin(x xf cc and 12sin ¸

¹·¨

©§S. In

a way, we know the value of xsin and all its derivatives at 2S x. We do not

need to use any calculators, Must plain differential calculus a nd trigonometry

would do. Can you use Taylor series and this information to fi nd the value of 2sin"

Solution 2S x 2 hx x h 2 22S 42920.0

So cccccccccc 4)(3 24 3 2hx fhxfhxfhxf xf hxf 2S x 42920.0 h x xf sin , ¸

¹·¨

©§ ¸

¹·¨

©§

2sin2SSf 1 munotes.in

## Page 36

36NUMERICAL AND STATISTICAL METHODS x xf cos c, 02 ¸

¹·¨

©§cSf x xf sin cc, 12 ¸¹·¨©§ccSf ) cos(x xf ccc, 02 ¸

¹·¨

©§cccSf ) sin(x xf cccc,12 ¸

¹·¨

©§ccccSf

Hence ¸

¹·¨

©§cccc¸

¹·¨

©§ccc¸

¹·¨

©§cc¸

¹·¨

©§c¸

¹·¨

©§ ¸

¹·¨

©§42 32 22 2 2 24 3 2hfhfhfh f f h fS S S SS S ¸

¹·¨

©§442920.01342920.00242920.01 42920.001 42920.024 3 2Sf

00141393.00 092106.001

90931.0#

The value of 2sin I get from my calculator is 90930.0which is very close to the

value I Must obtained. Now you can get a better value by using more terms of the

series. In addition, you can now use the value calculated for 2sin coupled with

the value of 2cos (which can be calculated by Taylor series Must like this

example or by using the 1 cos sin2 2{x x i d e n t i t y ) t o f i n d v a l u e o f xsin a t

some other point. In this way, we can find the value of xsin for any value from 0 x to S2 and then can use the periodicity of xsin, that is ,2,1, 2 sin sin nnx x S t o c a l c u l a t e t h e v a l u e o f xsin a t a n y o t h e r

point.

Example

Derive the Maclaurin series of 753sin7 5 3x x xx x

Solution

In the previous example, we wrote the Taylor series for xsin around the point 2S x. Maclaurin series is simply a Taylor series for the point 0 x. munotes.in

## Page 37

37Chapter 3: Truncation Errors the Taylor Series x xf sin , 00 f x xf cos c,10 cf x xf sin cc,00 ccf x xf cos ccc,1 0 cccf x xf sin cccc,00 ccccf ) cos(x x f ccccc,10 cccccf

Using the Taylor series now, ccccccccccccccc 50403020 0 0 05 4 3 2hfhfhfhfh f f h f ccccccccccccccc 50403020 0 05 4 3 2hfhfhfhfh f f hf

ccccccccccccccc 5 4 3 25 4 3 2hx fhxfhxfhxfhxfxf hxf 51403120 105 4 3 2h h h hh

535 3h hh

So 535 3x xx xf 53sin5 3x xx x

Example 4

Find the value of 6f given that1254 f, 744 cf, 304 ccf, 64 cccf

and all other higher derivatives of xf at 4 x are zero.

Solution cccccc 3 23 2hxfhxfhxf xf hxf 4 x 46 h 2

Since fourth and higher derivatives of xf are zero at 4 x. munotes.in

## Page 38

38NUMERICAL AND STATISTICAL METHODS 324224 24 4 243 2

f f f f f cccccc ¸¸

¹·

¨¨

©§¸¸

¹·

¨¨

©§ 3262230274 12563 2

f

860 148 125

341

Note that to find 6f exactly, we only needed the value of the function and all

its derivatives at some other point, in this case, 4 x. We did not need the

expression for the function and all its derivatives. Taylor se ries application would

be redundant if we needed to know the expression for the functi on, as we could

Must substitute 6 x in it to get the value of6f.

Actually the problem posed above was obtained from a known fun ction 5 2 32 3 x x xxf w h e r e1254 f, 744 cf, 304 ccf, 64 cccf,

and all other higher derivatives are zero.

.2.4 Error in Taylor Series

As you have noticed, the Taylor series has infinite terms. Onl y in special cases

such as a finite polynomial does it have a finite number of ter ms. So whenever

you are using a Taylor series to calculate the value of a funct ion, it is being

calculated approximately.

The Taylor polynomial of order n o f a f u n c t i o n )(xf w i t h )1(n continuous

derivatives in the domain @ ,> hxx is given by hxRnhxfhxfhxf xf hxfnn

n c 2

2

where the remainder is given by c fnhhxRnn

n11

)1(

.

where hxcx

that is,c is some point in the domain hxx,.

munotes.in

## Page 39

39Chapter 3: Truncation Errors the Taylor SeriesExample 5

The Taylor series for xeat point 0 x is given by 543215 4 3 2x x x xx ex

a) What is the truncation (true) error in the representation of 1e if only four terms

of the series are used"

b) Use the remainder theorem to find the bounds of the truncati on error.

Solution

a) If only four terms of the series are used, then 3213 2x xx ex| 31

21113 2

1|e

66667.2

The truncation (true) error would be the unused terms of the Ta ylor series, which

then are 545 4x xEt

51

415 4

0516152.0#

b) But is there any way to know the bounds of this error other tha n

calculating it directly" n c

where

c fnhhxRnn

n11

1

, hxcx , and c is some point in the domain hxx,. munotes.in

## Page 40

40NUMERICAL AND STATISTICAL METHODSSo in this case, if we are using four terms of the Taylor serie s, the remainder is

given by 3,0 n x

c f R1313

313110

cf4

41

24ce

Since hxcx 10 0 c 1 0 c

The error is bound between 241241

30eRe 241241

3eR 113261.01 041667.03R

So the bound of the error is less than 113261.0 w h i c h d o e s c o n c u r w i t h t h e

calculated error of 0516152.0.

Example 6

The Taylor series for xeat point 0 x is given by 543215 4 3 2x x x xx ex

As you can see in the previous example that by taking more term s, the error

bounds decrease and hence you have a better estimate of 1e. How many terms it

would require to get an approximation of 1e within a magnitude of true error of

less than610" munotes.in

## Page 41

41Chapter 3: Truncation Errors the Taylor SeriesSolution

Using 1n terms of the Taylor series gives an error bound of

c fnhhxRnn

n11

1

xexf h x )(,1,0

c fnRnn

n11

111

cn

en 111

Since hxcx 10 0 c 1 0 c )1(1)1(1

neRnn

So if we want to find out how many terms it would require to ge t an

approximation of 1e within a magnitude of true error of less than610, 610)1(ne e n610)1( ! 3 10)1(6u!n (as we do not know the value of ebut it is less than 3). 9tn

So 9 terms or more will get 1e within an error of 610 in its value.

We can do calculations such as the ones given above only for si mple functions.

To do a similar analysis of how many terms of the series are ne eded for a

specified accuracy for any general function, we can do that bas ed on the concept

of absolute relative approximate errors discussed in Chapter 01 .02 as follows. munotes.in

## Page 42

42NUMERICAL AND STATISTICAL METHODSWe use the concept of absolute relative approximate error (see Chapter 02 for

details), which is calculated after each term in the series is added. The maximum

value of m, for which the absolute relative approximate error is less tha n mu2105.0 is the least number of significant digits correct in the answ er. It

establishes the accuracy of the approximate value of a function w i t h o u t t h e

knowledge of remainder of Taylor series or the true error.

. Error Propagation

The purpose of this section is to study how errors in numbers c an propagate

through mathematical functions. For example, if we multiply two n u m b e r s t h a t

have errors, we would like to estimate the error in the product .

Numerical solutions involve a series of computational steps. Th erefore, it is

necessary to understand the way the error propagates with progr essive

computations.

Propagation of error (or propagation of uncer tainty) is the effect of variables,

uncertainties (or errors) on the uncertainty of a function based on them.

..1 Functions of a Single Variable

Suppose that we have a function f (x) that is dependent on a single independent

variable x. Assume that x is an approximation of x. We, therefore, would like to

assess the effect of the discrepancy between x and x on the value of the function.

That is, we would like to estimate

∆f(ݔҧ) _ f(x) − f(ݔҧ)_

The problem with evaluating ∆ f(ݔҧሻis that f (x) is unknown because x is

unknown. We can overcome this difficulty if x˜ is close to x and f(ݔҧሻis continuous

and differentiable. If these conditions hold, a Taylor series c an be employed to

compute f (x) near f(ݔҧሻas in

݂ሺݔሻൌ݂ሺݔതሻ݂Ԣሺݔҧሻሺݔ െݔҧሻ݂ᇱᇱሺݔҧሻ

ʹǨሺݔ െݔҧሻଶڮ

Dropping the second- and higher-order terms and rearranging yie lds

f(x) − f(ݔҧ) ؆݂Ԣሺݔതሻ(x − ݔҧሻ

or

∆݂ሺݔതሻ _ ݂Ԣሺݔതሻ_∆ݔҧ (3.3.1) munotes.in

## Page 43

43Chapter 3: Truncation Errors the Taylor SeriesWhere ∆݂ሺݔതሻ _݂ሺݔሻ - ݂ሺݔതሻ_

represents an estimate of the error of the function and ∆ݔҧ _ݔ- ݔത_

represents an estimate of the error of x. Equation (3.3.1) provides the capability to

approximate the error in f (x) given the derivative of a function and an estimate of

the error in the independent variable. Figure 3.3.1 is a graphi cal illustration of the

operation.

FIGURE ..1 Graphical depiction of fir st order error propagation.

..2 Error Propagation in a Function of a Single Variable

Problem Statement

Given a value of xҧ = 2.5 with an error of ∆ xҧ 0.01, estimate the resulting error in

the function, f ( x) x3.

Solution. Using Eq. (4.25),

∆f(xҧ) ؆ 3(2.5)2(0.01) 0.1875

Because f (2.5) 15.625, we predict that

f(2.5) 15.625 0.1875

munotes.in

## Page 44

44NUMERICAL AND STATISTICAL METHODSor that the true value lies between 15.4375 and 15.8125. In fac t, if x were actually

2.49, the function could be evaluated as 15.4382, and if x were 2.51, it would be

15.8132. For this case, the first-order error analysis provides a fairly close

estimate of the true error.

.4 Total Numerical Error

The total numerical error is the summation of the truncation an d round-off errors.

FIGURE .4.1

A graphical depiction of the trade-off between round-off and tr uncation error that

sometimes comes into play in the course of a numerical method. The point of

diminishing returns is shown, where round-off error begins to n egate the benefits

of step-size reduction.

In general, the only way to minimize round-off errors is to inc rease the number of

significant figures of the computer. Further, we have noted tha t round-off error

will increase due to subtractive cancellation or due to an incr ease in the number of

computations in an analysis. The truncation error can be reduce d by decreasing

the step size. Because a decrease in step size can lead to subt ractive cancellation

or to an increase in computations , the truncation errors are de creased as the

round-off errors are increased. Therefore, we are faced by the following dilemma:

The strategy for decreasing one component of the total error le ads to an increase

of the other component. In a computation, we could conceivably decrease the step

size to minimize truncation errors only to discover that in doi ng so, the round-off

error begins to dominate the solution and the total error grows Thus, our remedy

becomes our problem (Fig. 3.4.1). One challenge that we face is to determine an

munotes.in

## Page 45

45Chapter 3: Truncation Errors the Taylor Seriesappropriate step size for a particular computation. We would li ke to choose a

large step size in order to decrease the amount of calculations a n d r o u n d - o f f

errors without incurring the penalty of a large truncation erro r. If the total error is

as shown in Fig. 3.4.1, the challe nge is to identify the point of diminishing returns

where round-off error begins to negate the benefits of step-siz e reduction.

In actual cases, however, such situations are relatively uncomm on because most

computers carry enough significa nt figures that round-off error s do not

predominate. Nevertheless, they sometimes do occur and suggest a sort of

“numerical uncertainty principle” that places an absolute limit on the ac curacy

that may be obtained using certain computerized numerical metho ds.

.5 Formulation Error and Data Uncertainty

Formulation Errors

Formulation, or model, are the errors resulting from incomplete m a t h e m a t i c a l

models when some latent effects are not taken into account or i gnored.

Data Uncertainty

Errors resulting from the acc uracy and/or precision of the data

o When with biased (underestimation/overestimation) or imprecise

instruments

o We can use descriptive statistics (viz. mean and variance) to p rovide a

measure of the bias and imprecision.

.6 Summary

The third chapter of this book discusses the truncation error a nd Taylor series

expansion which is one of the mo st widely used example to illus trate truncation

error and truncation error bound. Formulation error and data un certainty is also

discussed in this chapter.

.7 References

Following books are recommended for further reading:-

1) Introductory Methods of Numerical Methods by S S Shastri munotes.in

## Page 46

46NUMERICAL AND STATISTICAL METHODS2) Numerical Methods for Engineers Steven C Chapra Raymond P Can ale –

6th Edition

3) Numerical Statistical Methods by Bhupendra T Kesria, Himalaya

Publishing House

4) Computer based Numerical Statistical Techniques – M. Goyal

. Exercise

1. Finding Taylor polynomials

Find the Taylor polynomials of orders 0,1 and 2 genera ted by f at a

a) f (x) ଵ

௫ , a 2

b) f (x) ଵ

ሺ௫ାଶሻ , a 0

c) f (x) sinx , a గ

ସ

d) f (x) cosx , a గ

ସ

2. Find Maclaurin series for the following functions

a) ݁ି௫

b) ݁ೣ

మ

c) ଵ

ሺଵା௫ሻ

d) ଵ

ሺଵି௫ሻ

e) ݊݅ݏ͵ݔ

f) sin௫

ଶ

. Finding Taylor Series

Find the Taylor series generated by f at x a.

a) f (x) ݔଷ-2x4, a 2

b) f (x) ʹݔଷݔଶ3x-8, a 1

c) f (x) ݔସݔଶ1, a -2 munotes.in

## Page 47

47Chapter 3: Truncation Errors the Taylor Seriesd) f (x) ͵ݔହ-ݔସ2ݔଷݔଶ-2 , a -1

e) f (x) ଵ

௫మ, a 1

f) f (x) ଵ

ሺଵି௫ሻ, a 0

g) f (x) ݁௫, a 2

h) f (x) ʹ௫, a 1

4. Use the taylor series generated by at x a to show that

݁௫ൌ݁ ቂͳሺݔെܽሻሺ௫ିሻమ

ଶǨڮቃ

5. ሺ ሻ Find the Taylor series generated

by ݁௫ at x 1 compare your answer with the formula in example 4

munotes.in

## Page 48

48NUMERICAL AND STATISTICAL METHODSUNIT 2

4

SOLUTIONS OF ALGEBRAIC AND

TRANSCENDENTAL E4UATIONS

Unit Structure

4.0 ObMectives

4.1 Introduction

4.1.1 Simple and Multiple roots

4.1.2 Algebraic and Transcendental Equations

4 . 1 . 3 D i r e c t m e t h o d s a n d I t e r a t i v e m e t h o d s

4 . 1 . 4 I n t e r m e d i a t e V a l u e T h e o r e m

4.1.5 Rate of Convergence

4.2 Bisection Method

4.3 Newton-Raphson Method

4 . 3 . 1 G e o m e t r i c a l I n t e r p r e t a t i o n

4.3.2 Rate of Convergence

4.4 Regula-Falsi Method

4 . 4 . 2 R a t e o f C o n v e r g e n c e

4.5 Secant Method

4 . 5 . 2 R a t e o f C o n v e r g e n c e

4.6 Geometrical Interpretation of Secant and Regula Falsi Metho d

4.7 Summary

4.8 Exercises

4.0 Objectives

This chapter will enable the learner to:

x understand the concepts of simple root, multiple roots, algebra ic equations,

transcendental equations, direct methods, iterative methods. munotes.in

## Page 49

49Chapter 4: Solutions of Algebraic And Transcendental Equationsx find roots of an equation using Bisection method, Newton-Raphso n method,

Regula-Falsi method, Secant method.

x understand the geometrical interpretation of these methods and derive the rate

of convergence.

4.1 Introduction

The solution of an equation of the form f(x) 0 is the set of all values which when

substituted for unknowns, make an equation true. Finding roots of an equation is a

problem of great importance in the fields of mathematics and en gineering. In this

chapter we see different methods to solve a given equation.

4.1.1 Simple and Multiple Roots

Definition 4.1.1.1 (Root of an Equation). A number ζ is said to be a root or a zero

of an equation f(x) 0 if f(ζ) 0.

Definition 4.1.1.2 (Simple Root). A number ζ is said to be a simple root of an

equation f(x) 0 if f(ζ) 0 and f

(ζ) ≠ 0. In this case we can write f(x) as

f(x) (x − ζ)g(x), where g(ζ) ≠ 0.

Definition 4.1.1. (Multiple Root). A number ζ is said to be a multiple root of

multiplicity m of an equation f(x) 0 if f(ζ) 0, f

(ζ) 0, ..., f(m−1)(ζ) 0 and f(m)(ζ)

≠ 0. In this case we can write f(x) as

f(x) (x − ζ)mg(x), where g(ζ) ≠ 0.

4.1.2 Algebraic and Transcendental Equations

Definition 4.1.2.1 (Algebraic Equation). A polynomial equation of the form f(x)

a0 a1x a2x2 āāā an−1xn−1 anxn 0, an ≠ 0 where ai אԧ for 0 ≤ i ≤ n is called

an algebraic equation of degree n.

Definition 4.1.2.2 (Transcendental Equation). A n e q u a t i o n w h i c h c o n t a i n s

exponential functions, logarithmic functions, trigonometric fun ctions etc. is called

a transcendental function.

4.1. Direct Methods and Iterative Methods

Definition 4.1.2.1 (Direct Methods). A method which gives an exact root in a

finite number of steps is called direct method. munotes.in

## Page 50

50NUMERICAL AND STATISTICAL METHODSDefinition 4.1.2.2 (Iterative Methods). A method based on successive

approximations, that is starting with one or more initial appro ximations to the root,

to obtain a sequence of approximations or iterates which conver ge to the root is

called an iterative method.

4.1.4 Intermediate Value Theorem

Iterative methods are based on successive approximations to the root starting with

one or more initial approximations. Choosing an initial approxi mation for an

iterative method plays an important role in solving the given e quation in a smaller

number of iterates. Initial approximation to the root can be ta ken from the physical

considerations of the given problem or by graphical methods.

As an example of finding initial approximation using physical c onsiderations of the

given problem, consider f(x) x3 − 28. Then to find the root of f(x) 0, one of the

best initial approximation is x 3 as cube of x 3 is close to the given value 28 To find initial approximation graphically, consider an example of

Figure 1: f(x) x3 − 3x

f(x) x3 − 3x. The value of x at which the graph of f(x) intersects x- axis gives the

root of f(x) 0. From Figure 1, it is clear that the roots of f(x) 0 lies close to 2 and

−2. Hence the best initial appr oximations will be 2 and −2.

Intermediate Value Theorem is another commonly used method to o btain the initial

approximations to the root of a given equation.

Theorem 4.1.4.1 (Intermediate Value Theorem). If f(x) is a continuous function

on some interval > a,b@ and f(a)f(b) < 0 then the equation f(x) 0 has at least one

real root or an odd number of real roots in the interval ( a,b).

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## Page 51

51Chapter 4: Solutions of Algebraic And Transcendental Equations4.1.5 Rate of Convergence

An iterative method is said to have a rate of convergence α, if α is the largest

positive real number for which there exists a finite constant C ≠ 0 such that

where is the error in the it iterate for i אN^0`. The constant C is called

the asymptotic error constant.

4.2 Bisection Method

Bisection method is based on the repeated application of interm ediate value

theorem. Let I0 (a0,b0) contain the root of the equation f(x) 0. We find

by bisecting the interval I0. Let I1 be the interval ( a0,m1),

if f(a0)f(m1) < 0 or the interval ( m1,b0), if f(m1)f(b0) < 0. Thus interval I1 also

contains the root of f(x) 0. We bisect the interval I1 and take I2 as the subinterval

at whose end points the function f(x) takes the values of opposite signs and hence

I2 also contains the root.

Repeating the process of bisecting intervals, we get a sequence o f n e s t e d

subintervals I0 ـ I1 ـ I2 ـāāā such that each subinterval contains the root. The

midpoint of the last sub interval is taken as the desired approx imate root.

Example 4.2.1. F i n d t h e s m a l l e s t p o s i t i v e r o o t o f f(x) x3 − 5x 1 0 by

performing three iterations of Bisection Method.

Solution: Here, f(0) 1 and f(1) = −3. That is f(0)f(1) < 0 and hence the smallest

positive root lies in the interval (0 ,1). Taking a0 0 and b0 1, we get

(First Iteration)

Since, f(m1) = −1.375 and f(a0)f(m1) < 0, the root lies in the interval (0 ,0.5).

Taking a1 0 and b1 0.5, we get

(Second Iteration)

Since, f(m2) = −0.234375 and f(a1)f(m2) < 0, the root lies in the interval (0 ,0.25).

munotes.in

## Page 52

52NUMERICAL AND STATISTICAL METHODSTaking a2 0 and b2 0.25, we get

(Third Iteration)

Since, f(m3) 0 .37695 and f(m3)f(b2) < 0, the approximate root lies in the interval

(0.125,0.25).

Since we have to perform three iterations, we take the approxim ate root as midpoint

of the interval obtained in the third iteration, that is (0 .125,0.25). Hence the

approximate root is 0 .1875.

4. Newton-Raphson Method

Newton-Raphson method is based on approximating the given equat ion f(x) 0

with a first degree equation in x. Thus, we write

f(x) a0x a1 0

whose root is given by such that the parameters a1 and a0 are to be

determined. Let xk be the kth approximation to the root. Then

f(xk) a0xk a1 (4.3.1)

and

f

(xk) a0 (4.3.2)

Substituting the value of a0 in 4.3.1 we get a1 f(xk) − f

(xk)xk. Hence,

.

Representing the approximate value of x by xk1 we get

(4.3.3)

This method is called the Newto n-Raphson method to find the roo ts of f(x) 0.

Alternative:

Let xk be the kth approximation to the root of the equation f(x) 0 and h be an

increment such that xk h is an exact root. Then

f(xk h) 0 .

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53Chapter 4: Solutions of Algebraic And Transcendental EquationsUsing Taylor series expansion on f(xk h) we get,

Neglecting the second and higher powers of h we get,

f(xk) h f

(xk) 0 .

Thus,

We put xk1 xk h and obtain the iteration method as

Newton-Raphson method requires two function evaluations fk and fk’ for each

iteration.

4..1 Geometrical Interpretation

We approximate f(x) by a line taken as a tangent to the curve at xk which gives the

next approximation xk1 as the x-intercept as in Figure 2.

Example 4.3.1. Find the approximate root correct upto two decim al places for f(x)

x4 − x − 10 using Newton-Raphson Method with initial approximation x0 1.

Figure 2: Newton-Raphson Method

Solution: Here f(x) x4 − x − 10 0 implies f

(x) 4 x3 − 1. For x0 1, f(x0) = −10

and f

(x0) 3. By Newton-Raphson iteration Formula,

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54NUMERICAL AND STATISTICAL METHODSThus,

.

Similarly, x3 2.5562 , x4 2.0982 , x5 1.8956 , x6 1.8568 , x7 1.8555.

4..2 Rate of Convergence

Let f(x) be a continuous function. Then the Newton-Raphson method to f ind the

approximate root of f(x) 0 is given by equation 4.3.3.

Let ζ be the exact root of f(x). Then we write xk1, xk in terms of ζ as

and , where I is the error in the ith iteration.

Let

.

Thus, equation 4.3.3 becomes

.

That is

. (4.3.4)

Using Taylor series expansion on and , we get

and

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55Chapter 4: Solutions of Algebraic And Transcendental EquationsThus, equation 4.3.4 becomes

Neglecting the third and higher powers of ߳, we get

where . Thus, Newton-Raphson method has a second order rate of

convergence.

4.4 Regula-Falsi Method

Given a continuous function f(x), we approximate it by a first-degree equation of

the form a0x a1, such that a0 ≠ 0 in the neighbourhood of the root. Thus, we write f(x) a0x a1 0

Then (4.4.1) f (xk) a0xk a1 and (4.4.2) f(ݔିଵ) ܽݔିଵܽଵ (4.4.3) On solving equation 4.4.2 and 4.4.3, we get

and

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56NUMERICAL AND STATISTICAL METHODSFrom 4.4.1, we have . Hence substituting the values of a0 and a1 and

writing x as xk1 we get

That is (4.4.4)

which can be expressed as

( 4 . 4 . 5 )

Here, we take the approximations xk and xk−1 such that f(xk)f(xk−1) < 0. This method

is known as Regula Falsi Method or False Position Method. This method requires

only one function evaluation per iteration.

Example 4.4.1. P e r f o r m f o u r i t e r a t i o n s o f R e g u l a F a l s i m e t h o d f o r

f(x) x3 − 5x 1 such that the root lies in the interval (0 ,1).

Solution: Since the root lies in the interval (0 ,1), we take x0 0 and x1 1. Then

f(x0) f(0) 1 and f(x1) f(1) = −3. By Regula Falsi Method,

and

As f(x0)f(x2) < 0 and f(x1)f(x2) > 0, by Intermediate Value property, the root lies in

the interval ( x0,x2). Hence

and f(x3) = −0.004352297.

Similarly, we get x4 0.201654 and x5 0.20164. Hence, we get the approximate

root as 0 .20164.

4.4.2 Rate of Convergence

Let f(x) be a continuous function. Then the Regula Falsi method to fin d the

approximate root of f(x) 0 is given by equation 4.4.5. Let ζ be the exact root of

f(x). Then we write xk1, xk, xk−1 in terms of ζ as

where is the error in the ith iteration.

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57Chapter 4: Solutions of Algebraic And Transcendental EquationsHence, .

Thus, equation 4.4.5 becomes

Applying Taylor expansion on and we get

Since ζ is an exact root of f(x) 0 we get

Neglecting the higher powers of ߳ and , we get

Hence

(4.4.6)

where .

Since in Regula Falsi method, one of the x0 or x1 is fixed, equation 4.4.6 becomes

, if x0 is fixed and , if x1 is fixed.

,,,,,,,,,

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58NUMERICAL AND STATISTICAL METHODSIn both the cases, the rate of convergence is 1. Thus, Regula F alsi method has a

linear rate of convergence.

4.5 Secant Method

Given a continuous function f(x), we approximate it by a first-degree equation of

the form a0x a1, such that a0 ≠ 0 in the neighbourhood of the root. Thus, we write

f(x) a0x a1 0 (4.5.1)

Then

f(xk) a0xk a1 (4.5.2)

and

f(xk−1) a0xk−1 a1 (4.5.3)

On solving equation 4.5.2 and 4.5.3, we get

and

From 4.5.1, we have . Hence substituting the values of a0 and a1 and

writing x as xk1 we get

That is

(4.5.4)

which can be expressed as

(4.5.5)

This method is known as Secant Method or Chord Method.

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59Chapter 4: Solutions of Algebraic And Transcendental EquationsExample 4.5.1. Using Secant Method, find the root of f(x) cosx−xex 0 taking

the initial approximations as 0 and 1.

Solution: Here x0 0, f(x0) 1 and x1 1, f(x1) = −2.17798. Using Secant formula

we get

and f(x2) 0 .519872. Then

and f(x3) 0 .2036. Similarly, we get x4 0.531706 and x5 0.51691. Hence the

approximate root is 0 .51691.

4.5.2 Rate of Convergence

Let f(x) be a continuous function. Then t he Secant method to find the approximate

root of f(x) 0 is given by equation 4.5.5. Let ζ be the exact root of f(x). Then we

write xk1, xk, xk−1 in terms of ζ as

where is the error in the ith iteration.

Hence, . Thus, equation

4.5.5 becomes

Applying Taylor expansion on and we get

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60NUMERICAL AND STATISTICAL METHODSSince ζ is an exact root of f(x) 0 we get

Neglecting the higher powers of k and , we get

Hence (4.5.6)

where .

Considering the general equation of rate of convergence, we hav e

( 4 . 5 . 7 )

which implies

.

Then

Substituting the value of in 4.5.6 we get

d

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61Chapter 4: Solutions of Algebraic And Transcendental EquationsFrom equation 4.5.7

(4.5.8)

Comparing the powers of ߳ we get

which implies

Neglecting the negative value of p, we get the rate of convergence of Secant method

as . On comparing the constants of 4.5.8 we get,

.

Thus, where and .

4.6 Geometrical Interpretation of Secant and Regula Falsi Metho d

Geometrically, in Secant and Regula Falsi method we replace the function f(x) by

a chord passing through ( xk,fk) and ( xk−1,fk−1). We take the next root approximation

as the point of intersection of the chord with the x- axis.

Figure 3: Secant Method Figure 4: Re gula Falsi Method

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62NUMERICAL AND STATISTICAL METHODS4.7 Summary

In this chapter, iteration methods to find the approximate root s of an equation is

discussed.

The concepts of simple roots, multiple roots, algebraic and tra nscendental

equations are discussed.

Intermediate value property to find the initial approximations to the root is

discussed. Bisection method, Newton-Raphson method, Regula Fals i method and

Secant method to find the approximate root of an equation is di scussed.

Geometrical interpretation and r ate of convergence of each meth od are discussed.

4. Exercise

1. Perform four iterations of bisection method to obtain a root of

f(x) cosx − xex.

2. Determine the initial approximation to find the smallest positi ve root for f(x)

x4 − 3x2 x − 10 and find the root correct to five decimal places by Newton

Raphson Method.

3. Perform four iterations of Newton Raphson method to obtain the approximate

value of ͳభ

య starting with the initial approximation x0 2.

4. Using Regula Falsi Method, find the root of f(x) cosx − xex 0 taking the

initial approximations as 0 and 1.

5. Perform three iterations of Secant Method for f(x) x3 −5x1 such that the

root lies in the interval (0 ,1).

6. For f(x) x4 −x−10 determine the initial approximations to find the smallest

positive root correct up to three decimal places using Newton R aphson

method, Secant Method and Regula Falsi Method.

7. Show that the Newton-Raphson method leads to the recurrence

to find the square-root of a.

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63Chapter 4: Solutions of Algebraic And Transcendental Equations8. For f(x) x−e−x 0 determine the initial approximation to find the smallest

positive root. Find the root correct to three decimal places u sing Regula Falsi

and Secant method.

9. Find the approximate root correct upto three decimal places for f(x) cosx −

xex using Newton- Raphson Method with initial approximation x0 1.

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64NUMERICAL AND STATISTICAL METHODSUNIT 2

5

INTERPOLATION

Unit Structure

5.0 ObMectives

5.1 Introduction

5.1.1 Existence and Uniqueness of Interpolating Polynomial

5.2 Lagrange Interpolation

5 . 2 . 1 L i n e a r I n t e r p o l a t i o n

5.2.2 4uadratic Interpolation

5 . 2 . 3 H i g h e r O r d e r I n t e r p o l a t i o n

5.3 Newton Divided Difference Interpolation

5 . 3 . 1 L i n e a r I n t e r p o l a t i o n

5.3.2 Higher Order Interpolation

5.4 Finite Difference Operators

5.5 Interpolating polynomials using finite difference operators

5 . 5 . 1 N e w t o n F o r w a r d D i f f e r e n c e I n t e r p o l a t i o n

5 . 5 . 2 N e w t o n B a c k w a r d D i f f e r e n c e I n t e r p o l a t i o n

5.6 Summary

5.7 Exercises

5.0 Objectives

This chapter will enable the learner to:

x Understand the concepts of interpolation and interpolating poly nomial. Prove

the existence and uniqueness of an interpolating polynomial.

x To find interpolating polynomial u sing Lagrange method and Newt on

Divided Difference method. munotes.in

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65Chapter 5: Interpolationx To understand the concepts of finite difference operators and t o relate

between difference operators and divided differences.

x To find interpolating polynomial using finite differences using N e w t o n

Forward Difference and Backwa rd Difference Interpolation.

5.1 Introduction

If we have a set of values of a function y f(x) as follows: x x0 x1 x2 āāā xn y y0 y1 y2 āāā yn then the process of finding the value of f(x) corresponding to any value of x xi

between x0 and xn is called interpolation. If the function f(x) is explicitly known,

then the value of f(x) corresponding to any value of x xi can be found easily. But,

if the explicit form of the function is not known then finding the value of f(x) is not

easy. In this case, we approximate the function by simpler func tions like

polynomials which assumes the same values as those of f(x) at the given points x0,

x1, x2,āāā ,xn.

Definition 5.1.1 (Interpolating Polynomial). A polynomial P(x) is said to be an

interpolating polynomial of a function f(x) if the values of P(x) and/or its certain

order derivatives coincides with those of f(x) for given values of x.

If we know the values of f(x) at n 1 distinct points say x0 < x 1 < x 2 < āāā < x n, then

interpolation is the process of finding a polynomial P(x) such that

(a) P(xi) f(xi), i 0,1,2,āāā ,n

or

(a) P(xi) f(xi), i 0,1,2,āāā ,n

(b) P’(xi) f ’(xi), i 0,1,2,āāā ,n munotes.in

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66NUMERICAL AND STATISTICAL METHODS5.1.1 Existence and Uniqueness of Interpolating Polynomial

Theorem 5.1.1.1. L e t f(x) be a continuous function on > a,b@ and let

a x0 < x 1 < x 2 < āāā < x n b be the n 1 distinct points such that the value of

f(x) is known at these points. Then there exist a unique polynomia l P(x) such that

P(xi) f(xi), for i 0,1,2,āāā ,n.

Proof. We intend to find a polynomial P(x) a0 a1x a2x2 āāā anxn such that

P(xi) f(xi), for i 0,1,2,āāā ,n. That is

(5.1.1.1)

Then the polynomial P(x) exists only if the system of equations 5.1.1.1 has a unique

solution. That is, the polynomial P(x) exists only if the Vandermonde

s determinant

is non-zero (in other words, P(x) exists only if the determinant of the co-efficient

matrix is non-zero). Let

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67Chapter 5: Interpolation

5.2 Lagrange Interpolation

5.2.1 Linear Interpolation

For linear interpolation, n 1 and we find a 1-degree polynomial

P1(x) a1x a0

such that

f(x0) P1(x0) a1x0 a0

and

f(x1) P1(x1) a1x1 a0.

We eliminate a0 and a1 to obtain P1(x) as follows:

On simplifying, we get

P1(x)(x0 − x1) − f(x0)(x − x1) f(x1)(x − x0) 0 and hence

(5.2.1.2)

where and are called the Lagrange

Fundamental Polynomials which satisfies the condition l0(x) l1(x) 1.

Example 5.2.1. Find the unique polynomial P(x) such that P(2) 1 .5, P(5) 4

using Lagrange interpolation.

Solution: By Lagrange interpolation formula,

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68NUMERICAL AND STATISTICAL METHODS5.2.2 4uadratic Interpolation

For quadratic interpolation, n 2 and we find a 2-degree polynomial

P2(x) a2x2 a1x a0

such that

݂ሺݔሻൌܲଶሺݔሻൌܽଶݔଶܽଵݔܽ

݂ሺݔଵሻൌܲʹሺݔ ଵሻൌܽ ଶݔଵଶܽଵݔଵ a0.

f(x2) P2(x2) ܽଶݔଶଶܽଵݔଶ a0

We eliminate a0, a1 and a2 to obtain P2(x) as follows:

On simplifying, we get

where l0(x) l1(x) l2(x) 1.

Example 5.2.2. Find the unique polynomial P(x) such that P(3) 1, P(4) 2 and

P(5) 4 using Lagrange interpolation.

Solution: By Lagrange interpolation formula,

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69Chapter 5: Interpolation5.2. Higher Order Interpolation

The Lagrange Interpolating polynomial P(x) of degree n for given n 1 distinct

points a x0 < x 1 < x 2 < āāā < x n b is given by

(5.2.3.1)

where,

for i 0,1,āāā ,n.

5. Newton Divided Difference Interpolation

5..1 Linear Interpolation

For linear interpolation, n 1 and we find a 1-degree polynomial

P1(x) a1x a0

such that

f(x0) P1(x0) a1x0 a0

and

f(x1) P1(x1) a1x1 a0.

We eliminate a0 and a1 to obtain P1(x) as follows:

On simplifying, we get

Let , then

P1(x) f(x0) (x − x0) f>x0, x1@ (5.3.1.1)

The ratio f>x0, x1@ is called the first divided difference of f(x) relative to x0 and x1.

The equation 5.3.1.1 is called the linear Newton interpolating polynomial with

divided differences.

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70NUMERICAL AND STATISTICAL METHODSExample 5..1. Find the unique polynomial P(x) such that P(2) 1 .5, P(5) 4

using Newton divided difference interpolation.

Solution: Here

Hence, by Newton divided difference interpolation,

5..2 Higher Order Interpolation

We define higher order divided differences as

Hence, in general we have

for k 3,4,āāā ,n and in terms of function values, we have

Then the Newton’s Divided Difference interpolating polynomial is given by

Pn(x) f>x0@ (x − x0)f>x0,x1@ (x − x0)(x − x1)f>x0,x1,x2@āāā

( x − x0)(x − x1)āāā(x − xn−1)f>x0,x1,āāā ,xn@ (5.3.2.1)

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71Chapter 5: InterpolationExample 5..2. Construct the divided difference table for the given data and hence

find the interpolating polynomial. x 0.5 1.5 3.0 5.0 6.5 8.0 f(x) 1.625 5.875 31.000 131.000 282.125 521.000 Solution: The divided difference table will be as follows:

x f (x) I order d.d. II order d.d. III order d.d. IV order d.d.

0.5 1.625

4.25

1.5 5.875 5

16.75 1

3.0 31.000 9.5 0

50.00 1

5.0 131.000 14.5 0

100.75 1

6.5 282.125 19.5

159.25

8.0 521.000

From the table, as the fourth divided differences are zero, the i n t e r p o l a t i n g

polynomial will be given as

P3(x) f>x0@ (x − x0)f>x0,x1@ (x − x0)(x − x1)f>x0,x1,x2@

( x − x0)(x − x1)(x − x2)f>x0,x1,x2,x3@

1 .625 ( x − 0.5)(4.25) 5( x − 0.5)(x − 1.5)

1( x − 0.5)(x − 1.5)(x − 3)

x3 x 1

5.4 Finite Difference Operators

Consider n1 equally spaced points x0,x1,āāā ,xn such that xi x0ih,i 0,1,āāā ,n.

We define the following operators:

1. Shift Operator (E): E(f(xi)) f(xi h). munotes.in

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72NUMERICAL AND STATISTICAL METHODS2. Forward Difference Operator ( ∆): ∆(f(xi)) f(xi h) − f(xi).

3. Backward Difference Operator ( :)(f(xi)) f(xi)−f(xi −h).

4. Central Difference Operator ( .

5. Average Operator ( .

Some properties of these operators:

.

Now we write the Newton’s divided differences in terms of forward and backward

difference operators. Consider

As xi x0 ih, i 0,1,āāā ,n, we get x1 − x0 h and hence

Now we consider

Thus, by induction we have

.

Similarly, for backward difference operator, consider

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73Chapter 5: Interpolation

Thus, by induction we have

.

5.5 Interpolating polynomials using finite differences operator s

5.5.1 Newton Forward Difference Interpolation

Newton’s forward difference interpolating polynomial is obtained by substituting

the divided differences in 5.3.2.1 with the forward differences . That is

(5.5.1.1)

Let , then 5.5.1.1 can be written as

(5.5.1.2)

5.5.2 Newton Backward Difference Interpolation

Newton interpolation with divided differences can be expressed in terms of

backward differences by evaluati ng the differences at the end p oint xn. Hence, we,

write

where .

t

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74NUMERICAL AND STATISTICAL METHODSExpanding (1 −)−u in binomial series, we get

Neglecting the higher order differences, we get the interpolati ng polynomial as

P n(x) Pn(xn hu)

(5.5.2.1)

Example 5.5. Obtain the Newton’s forward and backward difference interpolating

polynomial for the given data x 0.1 0.2 0.3 0.4 0.5 f(x) 1.40 1.56 1.76 2.00 2.28 Solution: The difference table will be as follows:

x f (x) I order d.d. II order d.d. III order d.d. IV order d.d.

0.1 1.40

0.16

0.2 1.56 0.04

0.20 0

0.3 1.76 0.04 0

0.24 0

0.4 2.00 0.04

0.28

0.5 2.28

From the table, as the third differences onwards are zero, the interpolating

polynomial using Newton’s forward differences will be given as

and the Newton’s backward difference interpolating polynomial will be given as

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75Chapter 5: Interpolation

5.6 Summary

In this chapter, interpolation methods to approximate a functio n by a family of

simpler functions like polynomials is discussed.

Lagrange Interpolation method and Newton Divided Difference Int erpolation

method are discussed.

Difference operators namely shift operator, forward, backward a nd central

difference operators and average operator are discussed.

The relation between divided difference and the forward and bac kward difference

operators are discussed and hence the Newton’s divided difference interpolation is

expressed in terms of forward and backward difference operators .

5.7 Exercises

1. Given f(2) 4 and f(2.5) 5 .5, find the linear interpolating polynomial using

Lagrange interpolation and Newton divided difference interpolat ion.

2. Using the data sin0 .1 0 .09983 and sin0 .2 0 .19867, find an approximate

value of sin0 .15 using Lagrange interpolation.

3. Using Newton divided difference interpolation, find the unique polynomial

of degree 2 such that f(0) 1 ,f(1) 3 and f(3) 55.

4. Calculate the nth divided difference of for points x0,x1,āāā ,xn.

5. Show that δ (1 −)−1/2 and .

6. For the given data, find the Newton’s forward and backward difference

polynomials. x 0 0.1 0.2 0.3 0.4 0.5 f(x) −1.5 −1.27 −0.98 −0.63 −0.22 0.25 7. Calculate .

8. If f(x) eax, show that ∆nf(x) (eah − 1)neax.

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76NUMERICAL AND STATISTICAL METHODS9. Using the Newton’s backward difference interpolation, construct the

interpolating polynomial that fits data. x 0.1 0.3 0.5 0.7 0.9 1.1 f(x) −1.699 −1.073 −0.375 0.443 1.429 2.631 10. Find the interpolating polynomial that fits the data as follows : x −2 −1 0 1 3 4 f(x) 9 16 17 18 44 81

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77Chapter 6: Solution of Simultaneous Algebraic Equations

UNIT

6

SOLUTION OF SIMULTANEOUS

ALGEBRAIC E4UATIONS

Unit Structure

6.0 ObMectives

6.1 Introduction

6.2 Gauss-Jordan Method

6.3 Gauss-Seidel Method

6.4 Summary

6.5 Bibliography

6.6 Unit End Exercise

6.0 Objectives

Student will be able to understand the following from the Chapt er:

Method to represent equations in Matrix Form.

Rules of Elementary Transformation.

Application of Dynamic Iteration Method.

6.1 Introduction

An equation is an expression having two equal sides, which are separated by an

equal to sign. For Example: 9 4 13

Here, a mathematical operation has been performed in which 4 an d 9 has been

added and the result 13 has been written alongside with an equal to sign.

Similarly, an Algebraic equation is a mathematical expression having two equal

sides separated by an equal to sign, in which one side the expression is formulated

by a set of variables and on the other side there is a constant value. For example

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78NUMERICAL AND STATISTICAL METHODS

Here, the set of variables x and y has been used to provide a unique pair of values

whose sum will be equal to 9.

If the powers (degrees) of the variables used is 1, then these algebraic equa tions

are known as Linear Algebraic equation. For example: 2 x 3y 5z 8.

It may happen, that there are more than equations,where there w ill be at-lest one

unique pair which will satisfy all the Algebraic equations. The procedure to find

these unique pairs are known as Solutions of Simultaneous Algeb raic Equations.

There are two types of methods:

1. Gauss-Jordan Method

2. Gauss-Seidel Method

6.2 Gauss-Jordan Method

Gauss Jordan Method is an algorithm used to find the solution o f simultaneous

equations. The algorithm uses Matrix Approach to determine the solution.

The method requires elementary transformation or elimination us ing ow

operations. Hence, it is also known as Gauss-Elimination Method .

Steps of Algorithm:

i Represent the set of equation in the following format:

A î X B

where,

A: Coefficient Matrix

X: Variable Matrix B: Constant Matrix

Example:

Convert the following equations in Matrix format:

3x 5y 12

2x y 1

The Matrix representation is:

In the above set of equations, the coefficients are the values which are written

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79Chapter 6: Solution of Simultaneous Algebraic Equations

The Constant matrix are the values which are written after the equal to sign.

Hence, the coefficient matrix is given as:

Hence, the variable matrix is given as:

And, the Constant matrix is:

ii Temporarily Combine the Coefficient A and Constant B Matrices in the

following format.

>A : B@

iii Perform row Transformation considering following Do’s and Don’t: Allowed Not Allowed Swapping of Rows

Ra ↔ Rb

where a 6 b Swapping between Row and

Column

Ra ↔ Cb Mathematical Operations between

Rows allowed:

Addition, Subtraction.

Ra ↔ Ra Rb

Mathematical Operations with

constant value allowed:

Multiplication, Division.

Mathematical Operations between

Rows not allowed:

Multiplication, Division.

Mathematical Operations with

constant value allowed:

Addition, Subtraction .

Ra ↔ Ra k Row Transformation is done to convert the Coefficient Matrix A to Unit Matrix

of same dimension as that of A.

6.2.1 Solved Examples:

i. Solve the system-

6x y z 20 x 4y − z 6 x − y 5z 7

using Gauss-Jordan Method

Sol. Given:

6x y z 20 x 4y − z 6 x − y 5z 7

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80NUMERICAL AND STATISTICAL METHODS

The matrix representation is-

Followed by the echolon form, formed by combining Coefficient M atrix and

Constant Matrix.

Perform elementary ow transformation in the above matrix to con vert matrix A to the following: ͳͲͲ

ͲͳͲ

ͲͲͳ൩

ͳ ͳǣ ʹ Ͳ

ͳͶെ ͳ ǣ

ͳെ ͳ ͷ ǣ ൩

ܴଵ՞ܴଶ

ͳͶെ ͳ ǣ

ͳ ͳǣ ʹ Ͳ

ͳെ ͳ ͷ ǣ ൩

ܴଶ՞ܴଶെܴଵ

ͳͶെ ͳ ǣ

Ͳെ ʹ ͵ ǣെ ͳ

ͳെ ͳ ͷǣ ൩

ܴଷ՞ܴଷെܴଵ

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81Chapter 6: Solution of Simultaneous Algebraic Equations

The solution of the equations are: x 3 y 1 and z 1.

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82NUMERICAL AND STATISTICAL METHODS

ii. Solve the system-

2x y z 10 3 x 2y 3z 18 x 4y 9z 16

using Gauss-Jordan Method

Sol. Given:

2x y z 10 3 x 2y 3z 18 x 4y 9z 16

The matrix representation is-

Followed by the echolon form, formed by combining Coefficient M atrix and

Constant Matrix.

Perform elementary ow transformation in the above matrix to con vert matrix A to

the following:

ͳͲͲ

ͲͳͲ

ͲͲͳ൩

ʹͳͳǣͳ Ͳ

͵ʹ͵ǣͳ ͺ

ͳͶͻǣͳ ൩

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83Chapter 6: Solution of Simultaneous Algebraic Equations

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84NUMERICAL AND STATISTICAL METHODS

The solution of the equations are: x 7 y = −9 and z 5.

6. Gauss-Seidel Method

Gauss Seidel Method uses Iterative methods to find the unique s olution of Linear

Algebraic equations. In this method, the present value of a var iable depends on the

past and present value of the other variables. This type of Ite ration is known as

Dynamic Iteration Method .

To achieve convergence of the values it is important to have Diagonal Dominance .

In Diagonal Dominance, the first equation should have the highe st coefficient

among the set of x coefficients as well as it should be the highest coefficient wi thin

the same equation. Similarly, the second equation should have h ighest coefficient

of y as well as among its own equation.

After ensuring Diagonal Dominance, the variable of each equatio n is represented

as the function of other variables.

6..1 Solved Examples:

i. Solve the equation: x 4y − z 6 6 x y z 20 x − y 5z 7

by using Gauss-Seidel Method

Sol. Given:

x 4y − z 6 6 x y z 20 x − y 5z 7

On comparing the coefficient of x among the given set of equations. The maximum

value is 6 which is present in the second equation. Considering the second equation,

the maximum coefficient present among the variable is also 6 (C oefficient of x)

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85Chapter 6: Solution of Simultaneous Algebraic Equations

Hence, the first equation is: 6xy] 20 —(i)

Similarly, among First and Third equation, the maximum value of the Coefficient

of y is 4.

Hence, the second equation is: x4y-] 6 —(ii)

And, the third equation is: x-y5] 7 —(iii)

Now represent each variable as a function of other two variable s like: Using

equation 1:

Using equation 2:

Using equation 3:

Consider the initial values of x, y and z as 0.

Now,

To implement the iteration the equations are re-written as:

Where, xn, yn and zn are the Present values of x, y and z respectively. xn−1, yn−1 and

zn−1 are the Past values of x, y and z respectively.

Means,

To calculate the Present value of x, we require Past values of y and z.

To calculate the Present value of y, we require Present value of x and Past value of

z.

To calculate the Present value of z, we require Present values of x and y.

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86NUMERICAL AND STATISTICAL METHODS

Hence the values of x, y and z will be: i xn yn zn 0 0 0 0 1 3.33 0.6675 0.8675 2 3.0775 0.9475 0.974 3 3.0131 0.97412 0.992 4 3.0056 0.99665 0.9982 5 3.0035 0.99815 0.99893 6 3.0005 0.9996 0.99982 7 3.00009 0.99993 0.999668

On Approximation:

x 3 y 1 and z 1 ii. Solve the equation:

x1 10x2 4x3 6 2 x1 10x2 4x3 = −15

9x1 2x2 4x3 20

by using Gauss-Seidel Method

Sol. Given:

x1 10x2 4x3 6

2x1 − 4x2 10x3 = −15

9x1 2x2 4x3 20

On re-arranging to achieve the Diagonal Dominance:

9x1 2x2 4x3 20 — (1) x1 10x2 4x3 6 —(2)

2x1 − 4x2 10x3 = −15 — (3)

Therefore,

(From equation (1))

(From equation (2))

(From equation (3)) Hence the values of x, y and z will

be:

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87Chapter 6: Solution of Simultaneous Algebraic Equationsi xn yn zn 0 0 0 0 1 2.2222 0.3778 -1.7933 2 2.9353 1.0238 -1.6775 3 2.7403 0.99697 -1.6493 4 2.7337 0.98635 -1.6522 5 2.7373 0.98715 -1.6526 6 2.7373 0.98731 -1.6525 7 2.7373 0.9873 -1.6525 On Approximation:

x 2.7373 y 0.9873 and z = −1.6525

6.4 Summary

Linear Algebraic equations can be solved by using two methods.

– Gauss Seidel Method

– Gauss Jordan Method

Gauss Seidel Method uses iterative approach, following Diagonal D o m i n a n c e

principle.

Gauss Jordan Method uses Matrix approach of the form: A î X B, following

Elementary Transformation principle.

6.5 References

(a) S. S. Shastry ”Introductory Met hods of Numerical Methods”. (Chp 3)

(b) Steven C. Chapra, Raymond P. Canale ”Numerical Methods for Engineers”.

6.6 Unit End Exercise

Find the solution of the following set of equation by using Gau ss Jordan method.

(a) 3x 2y 4z 7 2 x y z 7 x 3y 5z 2

(b) 10x y z 12 2 x 10y z 13 x y 3z 5

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88NUMERICAL AND STATISTICAL METHODS

(d) 2x y − z = −1 x − 2y 3z 9

3 x − y 5z 14

Find the solution of the following set of equation by using Gau ss Seidel method.

(a) 10x y z 12 2 x 10y z 13 x y 3z 5

(b) 28x 4y − z 32 2 x 17y 4z 35 x 3y 10z 24

(c) 7x1 2x2 − 3x3 = −12 2x1 5x2 − 3x3 = −20 x1 − x2 − 6x3 = −26

(d) 7x1 2x2 − 3x3 = −12 2x1 5x2 − 3x3 = −20 x1 − x2 − 6x3 = −26

(e) 10x y z 12 x 10y z 12 x y 10z 12

(Assume x(0) 0.4, y(0) 0.6, z(0) 0.8)

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## Page 89

89Chapter 7: Numerical Di൵ erentiation and Integration

UNIT

7

NUMERICAL DIFFERENTIATION AND

INTEGRATION

Unit Structure

7.0 ObMectives

7.1 Introduction

7.2 Numerical Differentiation

7.3 Numerical Integration

7.4 Summary

7.5 Bibliography

7.6 Unit End Exercise

7.0 Objectives

Student will be able to understand the following from the Chapt er:

Methods to compute value of Differentiation of a function at a particular value.

Methods to find the area covere d by the curve within the the gi ven interval.

Understand the importance of Interpolation Method.

7.1 Introduction

Differentiation is the method used to determine the slope of th e curve at a particular

point. Whereas, Integration is the method used to find the area between two values.

The solution of Differentiation and Integration at and between particular values can

be easily determined by using some Standard rules. There are so me complex

function whose differentiation and Integration solution is very d i f f i c u l t t o

determine. Hence, there are some practical approaches which can be used to find

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90NUMERICAL AND STATISTICAL METHODS

7.2 Numerical Differentiation

The value of differentiation or derivative of a function at a p articular value can be

determined by using Interpolation Methods.

(a) Newton’s Difference Method (If the step size is constant.)

(b) La-grange’s Interpolation Method (If step size is not constant.)

Newton’s Difference Method is further divi ded into two types depending on the

position of the sample present in the input data set.

Newton’s Forward Difference Method: To be used when the input value is lying

towards the start of the input data set.

Newton’s Backward Difference Method: To be used when the input value is lying

at the end of the input data set.

7.2.1 Newton Forward Difference Method

For a given set of values ( xi,yi), i 0,1,2,...n where xi are at equal intervals and h is

the interval of of the input values i.e. xi x0 i î h then,

Since the above equation is in form of “ p”, therefore, chain rule is to be used for

differentiation. Hence,

We know that,

Where, x Unknown Value x0 First Input value h Step Size

Hence,

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91Chapter 7: Numerical Di൵ erentiation and Integration

And,

Hence due to simplification,

On differentiating second time, the expression becomes:

7.2.2 Newton Backward Difference Method

For a given set of values ( xi,yi), i 0,1,2,...n where xi are at equal intervals and h is

the interval of of the input values i.e. xi x0 i î h then,

Since the above equation is in form of “ p”, therefore, chain rule is to be used for

differentiation. Hence,

We know that,

Where, x Unknown Value xn Final Input value h Step Size

Hence,

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92NUMERICAL AND STATISTICAL METHODS

And,

Hence due to simplification,

On differentiating second time, the expression becomes:

7.2.3 Solved Examples

i. From the data table given below obtain and x 1.0 1.2 1.4 1.6 1.8 2.0 2.2 y 2.7183 3.3201 4.0552 4.9530 6.0496 7.3891 9.0250 Sol. The first step is to identify which method to be used.

Since in the question the value of and 2 is to be determined.

Hence, Forward Difference to be used as the value lies at the start of the data set.

Therefore, Forward Difference Table is to be formed. x y 4y 42y 43y 44y 45y 46y 1 2.7183 0.6018 0.1333 0.0294 0.0067 0.0013 0.0001 1.2 3.3201 0.7351 0.1627 0.0361 0.0080 0.0014 1.4 4.0552 0.8978 0.1988 0.0441 0.0094 1.6 4.9530 1.0966 0.2429 0.0535 1.8 6.0496 1.3395 0.2964 2.0 7.3891 1.6359 2.2 9.0250

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93Chapter 7: Numerical Di൵ erentiation and Integration

Now select the values corresponding to the row of x 1.2. Since, value of and

is to be determined at x 1.2

Hence, the values are: 4 y0 0.7351, 42y0 0.1627, 43y0 0.0361 , 44y0 0.0080,

45y0 0.0014.

Substituting the values in the following formula:

Similarly, to find the value of , Substitute the values in the following equation:

where, 0 .0014.

Therefore,

i. From the data table given below obtain and x 1.0 1.2 1.4 1.6 1.8 2.0 2.2 y 2.7183 3.3201 4.0552 4.9530 6.0496 7.3891 9.0250 Sol. The first step is to identify which method to be used.

Since in the question the value of and 8 is to be determined.

Hence, Backward Difference to be used as the value lies at the end of the data set.

Therefore, Backward Difference Table is to be formed. x y 5y 52y 53y 54y 55y 56y 1 2.7183 1.2 3.3201 0.6018 1.4 4.0552 0.7351 0.1333

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## Page 94

94NUMERICAL AND STATISTICAL METHODSx y 5y 52y 53y 54y 55y 56y 1.6 4.9530 0.8978 0.1627 0.0294 1.8 6.0496 1.0966 0.1988 0.0067 2.0 7.3891 1.3395 0.2429 0.0441 0.0080 0.0013 2.2 9.0250 1.6359 0.2964 0.0535 0.0094 0.0014 0.0001 Now select the values corresponding to the row of x 1.2. Since,

value of and is to be determined at x 1.8

Hence, the values are:

Substituting the values in the following formula:

Similarly, to find the value of , Substitute the values in the following equation:

where,

Therefore,

If step si]e of the input value is not constant , then the derivatives can be

determined by simply different iating the expression provided by the Langrange’s

Polynomial .

7.2.4 Solved Examples

i. Tabulate the following function: y x3 −10x6 at x0 0.5, x1 1 and x2 2.

Compute its 1st and 2nd derivatives at x 1.00 using Lagrange’s interpolation

method.

Sol. Given: y x3 − 10x 6

At x0 0.5 y0 0.53 − 10 î 0 .5 6 1 .125

At x1 1 y1 13 − 10 × 1 + 6 = −3

At x2 2 y2 23 − 10 × 2 + 6 = −6

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95Chapter 7: Numerical Di൵ erentiation and Integration

Hence, the Lagange’s Formula is:

Hence, will be obtained by differentiating both the sides

So therefore, at x 1 the will be:

7. Numerical Integration

Numerical Integration provides a set of methods to compute the Definite

Integration of a Function between the given set of values.

There are three methods used to find the value of Integration:

x Trapezoidal Rule.

x Simpson’s Rule.

x Simpson’s Rule.

7..1 Trape]oidal Rule

In this method the curve is divided into small trapeziums.

These trapeziums are then added to to find the complete area of the curve between

two values. Hence,

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96NUMERICAL AND STATISTICAL METHODS

Let y f(x) then,

Where,

xn: Upper Limit x0: Lower Limit y0, y1, y2, āāā, yn are the values of of y corresponding

to x0, x1, x2, āāā, xn

7.3.2 Simpson’s Rule

Let y f(x) then,

(Sum of Odd osition Terms) 2 î

(Sum of Even osition Terms)

Where,

xn: Upper Limit

x0: Lower Limit

y0, y1, y2, āāā, yn are the values of of y corresponding to x0, x1, x2, āāā, xn

7.3.2 Simpson’s Rule

Let y f(x) then,

(Sum of Multiple of 3, position terms)

3 î (Sum of Remaining terms)

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97Chapter 7: Numerical Di൵ erentiation and Integration

Where,

݄ൌݔെݔ

݊

xn: Upper Limit

x0: Lower Limit

y0, y1, y2, āāā, yn are the values of of y corresponding to x0, x1, x2, āāā, xn

7.3.3 Solved Examples

i. A solid of revolution is formed by rotating about the x-axis, the area between the

x-axis,the lines x 0 and x 1 and the curve through the points below: x 0.00 0.25 0.50 0.75 1.00 x 1.000 0.9896 0.9589 0.9089 0.8415 Estimate the volume of the solid formed.

Sol. Let “V” be the volume of the solid formed by rotating the curve around x- axis,

then

Therefore, the tables is updated as: x 0.00 0.25 0.50 0.75 1.00 x 1.000 0.9793 0.9195 0.8261 0.7081 Rewriting the same table to find the value of n. x 0.00 x0 0.25 x1 0.50 x2 0.75 x3 1.00 x4 x 1.000 0.9793 0.9195 0.8261 0.7081 As the extreme or the last value is x4. Hence, n 4

Therefore,

(a) TRAPE=OIDAL RULE

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98NUMERICAL AND STATISTICAL METHODS

where, h 0.25, y0 1.000, y1 0.9793, y2 0.9195, y3 0.8261, y4 0.7081

On substituting the values we get:

(b) SIMPSON’S RULE

where, h 0.25, y0 1.000, y1 0.9793, y2 0.9195, y3 0.8261, y4 0.7081

On substituting the values we get:

(c) SIMPSON’S ଷ

଼th RULE

ߨൈනԝଵ

ݕଶݔ݀ ൌ ߨൈ͵݄

ͺ൫ݕݕସʹൈሺݕଷሻ͵ൈሺݕଵݕଶሻ൯

where, ݄ൌͲ Ǥ ʹ ͷ ǡݕ ൌͳ Ǥ Ͳ Ͳ Ͳ ǡݕ ଵൌͲ Ǥ ͻ ͻ ͵ ǡݕ ଶൌͲ Ǥ ͻ ͳ ͻ ͷ ǡݕ ଷൌ

ͲǤͺʹͳǡݕ ସൌͲ Ǥ Ͳ ͺ ͳ

On substituting the values we get:

ߨൈනԝଵ

ݕଶݔ݀ ൌ ʹǤͶͳ

7.4 Summary

Numerical Differentiation uses the methods to find the value of First ans Second

Order Derivatives at a par ticular value of x or the input varia ble.

Numerical Integration provides methods to find the Definite Int egration or the area

covered by the curve between two points.

7.5 References

(a) S. S. Shastry ”Introductory Methods of Numerical Methods”.

(b) Steven C. Chapra, Raymond P. Canale ”Numerical Methods for Engineers”.

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99Chapter 7: Numerical Di൵ erentiation and Integration

7.6 Unit End Exercise

(a) Find and 2 and x 0.4 from the

following table:

( 0 ,1.0) (0 .1,0.9975) (0 .2,0.9900) (0 .3,0.9776) (0 .4,0.9604)

(b) The following table gives angular displacement a at different t ime t (time).

(0,0.052) (0 .02,0.105) (0 .04,0.168) (0 .06,0.242) (0 .08,0.327)

(0.10,0.408)

Calculate angular velocity and acceleration at t 0.04, 0.06, an d 0.1

Angular Velocity : Angular Acceleration :

(c) A cubic function y f(x) satisfies the following data. x 0 1 3 4 y 1 4 40 85 D e t e r m i n e f(x) and hence find f ’(2) and f ’’(2)

(d) Use Trapezoidal Rule to evaluate integral width of sub-interva l (h)

0.5.

(e) Using Simpson’s rules. Evaluate take n 6.

(f) Using Simpson’s rules. Evaluate take n 6.

(g) Using Simpson’s rules. Evaluate take n 6.

(h) Use Trapezoidal Rule to evaluate integral width of sub-interval

(h) 0 .5.

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## Page 100

100NUMERICAL AND STATISTICAL METHODSUNIT

NUMERICAL DIFFERENTIATION E4UATION

Unit Structure

8.0 ObMectives

8.1 Introduction

8.2 Euler’s Method

8.3 Euler’s Modified Method

8.4 Range Kutta Method

8.5 Summary

8.6 Bibliography

8.7 Unit End Exercise

.0 Objectives

Student will be able to understa nd the following from the Chapt er:

Methods to compute value of Differential Equation of a function at a particular

value.

Practical or Software Implemented method to find the solution o f Differential

equation.

.1 Introduction

Differential equation is defined as an expression which contain s derivative terms.

For example: . Here the term is indicating that the occurrence of

a Differential Equation.

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101Chapter 8: Numerical Di൵ erentiation EquationThe differential equations can be solved analytically using var ious methods like:

Variable separable , Substitution , Linear Differential Equation , Solution to

Homogeneous Equation , etc. but in this chapter, various practically approachable

methods will be discussed to find the solution of a given diffe rential equation at a

particular value.

A Differential equation is defined on the basic of two terminol ogies:

Order: Number of times a variable is getting differentiated by an an other variable.

Degree: Power of the highest order deriva tive in a differential equat ion.

For Example : Consider the following equations: A.

B. C.

D.

In the above equations, eqn. A has Order 1, because y is differentiated only

1 time w.r.t. x. The degree 1 because the power of the only differential equation

is 1.

In eqn. B has Order 2, because y is differentiated 2 times w.r.t. x.

The degree 1 because the power of the only differential equation is 1.

In eqn. C has Order 1, because y is differentiated only 1 time w.r.t. x.

The degree 2 because the power of the only differential equation is 2.

In eqn. D, there are two derivatives present. Hence, the term with maxim um

differentiation present is to be selected. Therefore, the equat ion has Order 2,

because y is differentiated only 2 time w.r.t. x. The degree 1 because the power

of the derivative of maximum order is 1.

8.2 Euler’s Method

Euler’s Method is a Practically implemented method used to find the solution of

first order Differential equation.

Suppose the given differential equation is ) with initial condi tions y

(x0) y0 (The value of y at x x0 is y0.) Then by Euler’s method:

yn1 yn h î f(xn,yn) for n 0,1,2,...

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102NUMERICAL AND STATISTICAL METHODSWhere,

yn1 is Future Value of y . yn is Present Value of y .

h is Common Difference or Step Size .

.2.1 Solved Examples

Find the value of y when x 0.1, given that y(0) 1 and y0 x2 y by using Euler’s

Method.

Sol. Given: y0 x2 y with initial condition y(0) 1

and the meaning of y(0) 1 is the value of y at x 0

is 1. Hence,

x0 0 and y0 1

To find the value of y at x 0.1, value of h is required.

Hence, h 0.05.

According to Euler’s Method, yn1 yn h î f (xn,yn).

Iteration 1:

y1 y0 h î f (x0,y0) y0 1 x0 0 h 0.05 f (xn,yn) x2 y y1 1 0 .05(02 1)

y1 1.05 (The value of y at x x0 h 0 0 .05 0 .05

Iteration 2:

y2 y1 h î f (x1,y1) y2 1.05 x1 x0 h 0.05 y2 1.05 0 .05(0.052 1.05) y1

1.0265 (The value of y at x x1 h 0.05 0 .05 0 .10

8.3 Euler’s Modified Method

The values of y determined at every iteration may have some error depending on

the value of selected Common Difference or Step Size h. Hence, to find the accurate

value of y at a particular x, Euler’s Modified Method is used.

In this method the value of the corresponding iteration is ensu red by providing a

process called Iteration within Iteration method .

This method is used to minimize the error. The iterative formul a is given as

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103Chapter 8: Numerical Di൵ erentiation EquationWhere,

xm, ym are the values to be used in the basic iteration and y(m 1)n is the value

obtained while saturating the given “ y” value in the same iteration.

The steps to use Euler’s Modified Method is:

ȋȌ Apply Euler’s Method at every iteration to find the approximate value at new

value of x using Euler’s Method .

ȋȌ Apply Euler’s Modified Method in a particular iteration to saturate the value

of y.

..1 Solved Examples

Find the value of y when x 0.1, given that y(0) 1 and y0 x2 y by using Euler’s

Modified Method.

Sol. Given: y0 x2 y with initial condition y(0) 1

and the meaning of y(0) 1 is the value of y at x 0

is 1. Hence,

x0 0 and y0 1

To find the value of y at x 0.1, value of h is required.

Hence, h 0.05.

Iteration 1:

Using Euler’s Method y1 y0 h î f (x0,y0) y0 1 x0 0 h 0.05 f (xn,yn) x2

y y1 1 0 .05(02 1) y1 1.05

Iteration 1 a :

Using Euler’s Modified Method,

where, x1 x0 h 0 0 .05 0 .05 and is the initial value obtained by using

Euler’s Method.

Since, 0500 and 0513 are not equal. Hence, apply 2nd iteration.

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104NUMERICAL AND STATISTICAL METHODSIteration 1 b :

Using Euler’s Modified Method,

where, x1 x0h 00 .05 0 .05 and 0513 is the value obtained in

Iteration 1(a) .

Since, values of and are equal. Hence, the value of y at x 0.05 is 1 .0513

Iteration 2:

Using Euler’s Method y2 y1 h î f (x1,y1) y1 1.0513 x1 x0 h 0.05 h 0.05

f (xn,yn) x2 y

y2 1.0513 0 .05(0.052 1.0513) y2 1.1044

Iteration 2 a :

Using Euler’s Modified Method,

where, x2 x1h 0.050 .05 0 .10 and is the initial value obtained by using

Euler’s Method.

Since, 1044 and 1055 are not equal. Hence, apply 2nd iteration.

Iteration 1 b :

Using Euler’s Modified Method,

where, x2 x1 h 0.05 0 .05 0 .10 and is the value obtained at

d

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## Page 105

105Chapter 8: Numerical Di൵ erentiation EquationIteration 2 a.

Since, values of and are equal. Hence,

the value of y at x 0.1 is 1 .1055

.4 Range Kutta Method

Range Kutta Method is an another method to find the solution of a F irst Order

Differential Equations. It is mainly divided into two methods d epending on the

number of parameters used in a method.

Range Kutta 2nd Order Method.

Range Kutta 4th Order Method.

Range Kutta 2nd order Method uses Two parameters k1 and k2 to find the value of

yn1. The expression is given as:

where

k1 h î f (xn,yn) k2 h î f (xn h,yn k1)

Range Kutta 4th order Method uses Four parameters k1, k2, k3 and k4 to find the

value of yn1. The expression is given as:

where,

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106NUMERICAL AND STATISTICAL METHODS.4.1 Solved Examples

Given , where y0 2, Find y(0.1) and y(0.2), correct upto 4 decimal

places.

Sol. Given: y(0) 2

y0 2 and x0 0

Range Kutta 2nd Order Method:

h 0.1

Iteration 1:

So,

y1 2.2050 at x 0.1 Iteration 2:

k1 h î f (x1,y1)

k1 0.1(2.205 − 0.1) k1 0.2105 k2 h î f (x1 h,y1 k1)

k2 0.1 î ((2 .2050 0 .2105) − (0 .1 0 .1))

k2 0.22155

So,

y1 2.421025 at x 0.2 Runge Kutta 4th Order:

yyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

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## Page 107

107Chapter 8: Numerical Di൵ erentiation EquationWhere,

Let h 0.1,

Iteration 1:

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108NUMERICAL AND STATISTICAL METHODSIteration 2:

.5 Taylor Series

The methods discussed in the previous sections, are applicable only for First

order Differential Equation. But, Taylor Series method can be u sed to find the

solution for higher order differential equations.

Taylor Series is a method used to represent a Function as a sum of infinite series

represented in terms of derivatives derived at a particular poi nt. Taylor series is

mathematically expressed as:

Similarly, to get an approximate value of y at x x0 h is given by following

expression:

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109Chapter 8: Numerical Di൵ erentiation Equation.5.1 Solved Examples

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110NUMERICAL AND STATISTICAL METHODS.6 Summary

Euler’s Method, Euler’s Modified and Range Kutta Method are app licable only in

First Order Differential Equation.

The First order differential equation should be of the form

Solution of Higher Order Differential Equation can be done by u sing Taylor’s

Method.

.7 References

ȋȌ S. S. Shastry ”Introductory Methods of Numerical Methods”.

ȋȌ Steven C. Chapra, Raymond P. Canale ”Numerical Methods for Engineers”.

.7 Unit End Exercise

ȋȌ Use Euler’s method to estimate y(0.5) of the following equation with

h 0.25 and

ȋȌ Apply Euler’s method to solve with y(0) 1.

Hence, find y(1). Take h 0.2

ȋ
Ȍ Using Euler’s method, find y(2) where and y(1) 1,

take h 0.2.

ȋȌ Using Euler’s method, find y(2) where and y(1) 1,

take h 0.2.

ȋȌ Solve y0 = 1 − y, y(0) = 0 by Euler’s Modified Method and obtain y

at x 0.1 and x 0.2

ȋȌ Apply Euler’s Modified method to find y(1.2). Given

0 .1 and y(1) 2 .718. Correct upto three decimal places.

ȋȌ Solve (1) 2. Compute y for x 1.2 and x 1.4 using

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111Chapter 8: Numerical Di൵ erentiation Equation Euler’s Modified Method.

ȋȌ Using Range-Kutta Method of second order to find y(0.2).

Given x y, y(0) 2, h 0.1

ȋȌ Use Range-Kutta Method of fourth order to find y(0.1), y(0.2).

Given (0) 1. (Take h 0.1)

ȋȌ Use Range-Kutta Method of fourth order to find y(0.1).

Given (0) 1. (Take h 0.1)

ȋȌ Use Range-Kutta Method of second order to find y(1.2).

Given y2 x2, y(1) 0. (Take h 0.1)

ȋȌ Solve , where y(0) 1, to find y(0.1) using

Range-Kutta method.

ȋȌ By using Runge-Kutta method of order 4 to evaluate y(2.4) from the

following differential equation y0 f(x,y) where f(x,y) ( x1)y. Initial

condition y(2) 1, h 0.2 correct upto 4 decimal places.

ȋȌ Use Taylor series method, for the equation, and y(0) 2 to find

the value of y at x 1.

ȋȌ Use Taylor series method, for the equation, and y(0) 1

to find the value of y at x 0.1,0.2,0.3.

ȋȌ Use Taylor series method, for the equation, and y(4) 4

to find the value of y at x 4.1,4.2.

ȋȌ Use Taylor’s series method, for the equation, y0 x2 − y and y(0) 1

to find y(0.1)

xxxxxxxxxxx

yyyyyyyyyyy

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## Page 112

112NUMERICAL AND STATISTICAL METHODSUnit 4

9

LEAST-SQUARES REGRESSION

Unit Structure

9.0 Objectives

9.1 Introduction

9.2 Basic Concepts of Correlation

9.2.1 Types of Correlation

9.2.2 Scatter Diagram

9.2.3 Karl Pearson's Coefficient of correlation

9.3 Linear Regression

9.3.1 Regression Equations

9.3.2 Method of Least Squares

9.3.3 Properties of Regression Coefficients

9.4 Polynomial Regression

9.5 Non- liner Regression

9.5.1 Polynomial fit

9.6 Multiple Linear Regression

9.7 General Linear Least Squares

9.7.1 Ordinary Least Square(OLS)

9.7.2 Generalized Least Square(GLS)

9.8 Summary

9.9 References

9.10 Exercises

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113Chapter 9: Least-Squares Regression.0 Objectives

This chapter would make you understand the following concepts:

x Correlation.

x Scatter Diagram .

x Linear Regression.

x Polynomial Regression

x Non- liner Regression.

x Multiple Linear Regression.

.1 Introduction

The functional relationship between two or more variables is ca lled 'Regression'.

here in regression, the variables may be considered as independ ent or dependent.

For example imagine the person walking on the straight road fro m point A to

point B he took 10min, the distance from A to B is 100m, now to reach point C

from B, he took another 10min while the distance from B to C is 100m, then how

many minutes will he take to reach point E from C which is 200 m long from

Point C ? or if he walks continuously 20min & straight like thi s from point C

where will he reach ? also what if the walking speed is not uni form ? or road is

not straight ?

.2 Basic Concepts of Correlation

Definition:

Two variables are said to be correlated if the change in the value of one variable

causes corresponding change in the value of other variable.

9.2.1 Types of Correlation

1) Positive & Negative Correlation : If changes in the value of one variable

causes change in value of other variable in the same directi on then the

variables are said to be positively correlated.

eg. The more petrol you put in your car, the farther it can go .

If changes in the value of one variable causes change in valu e of other

variable in opposite direction then the variable are said to be negatively

correlated eg. As a biker's speed increases, his time to get t o the finish line

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## Page 114

114NUMERICAL AND STATISTICAL METHODS2) Simple, Multiple & Partial Correlation : The correlation between two

variables is simple correlation. The Correlation between three or more

variables is called multivariate correlation, For example the r elationship

between supply, price and profit of a commodity. in partial cor relation

though more than two factors are involved but correlation is st udied only

between two factors and the other factors are assumed to be con stant.

3) Linear and Non-linear Correlation : if the nature of the graph is straight

line the correlation is called linear and if the graph is not a straight line but

curve is called non-linear correlation

9.2.2 Scatter Diagram

This is a graphical method to study correlation. In this method each pair of

observations is represented by a point in a plane. The diagram formed by plotting

all points is called scatter diagram. in this following Scatter diagram each dot

coordinate is like ሺݔଵǡݕଵሻǡሺݔଶǡݕଶሻǡǥǤǤሺݔ ǡݕሻ on XY- plane

The Following are types of scatter diagram:

Merits & Demerits of Scatter Diagram

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115Chapter 9: Least-Squares RegressionMerits:

i) Scatter diagram is easy to draw.

ii) It is not influenced by extreme items.

iii) It is easy to understand and non-mathematical method of s tudying

correlation between two variables

Demerits:

It does not give the exact degree of correlation between two va riables. it gives

only a rough idea.

9.2.3 Karl Pearson's Coefficient of correlation (Product moment Coefficient

of correlation)

Karl Pearson's Coefficient of Correlation is an extensively use d mathematical

method in which the numerical representation is applied to meas ure the level of

relation between linear related variables. The coefficient of correlation is

expressed by ǲݎǳ.

Steps involved to calculate ̶̶:

Step 1: Calculate the actual mean of ݔ and actual mean of ݕ

Step 2: Take deviation from the actual mean of ݔ series, It gives column

ܺൌሺݔെݔҧሻǤ

Step 3: Take deviation from the actual mean of ݕ s e r i e s , I t g i v e s c o l u m n

ܻ ൌ ሺݕെݕ തሻǤ

Step 4: Calculate summation of deviation product of ܺ and ܻ i.e. σܻܺǤ

Step 5 : Square the deviations of ܺand ܻ and calculate the summation

i.e. σܺଶand σܻଶǤ

Step 6 : Use the following formula to calculate ݎ

ݎൌσܻܺ

ඥσܺଶσܻଶ

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116NUMERICAL AND STATISTICAL METHODSInterpretation of the values of :

Example 9.2.3.1 :

Find the Coefficient of correlation for the following data and comment on its

value. 6 13 9 10 6 4 2 15 17 13 7 6

Solution : Here ݊ൌ ǡ ഥൌσ

ൌૡ

ൌ ͺ and ഥൌσ

ൌ

ൌͳ Ͳ

Consider the following table for calculation ൌሺ െ ഥሻൌሺ െ ഥሻʹെʹെͺͳͶͶͳ͵ͳͷͷͷʹͷʹͷʹͷͻͳͳͳͶͻͳͲͳ͵ʹ͵Ͷͻെʹെ͵ͶͻͶെͶെͶͳͳͳૡૠૠ

ݎൌσ

ඥσమσమ =

ξହସൈଵଶൌͲ Ǥ ͻ

ݎൌͲ Ǥ ͻ

Hence, Strongly positive correlation between and

Similarly we can calculate ݎvalue without deviation form means (i.e. Direct

method)

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## Page 117

117Chapter 9: Least-Squares Regressionwhere ൌσିσσ

ඨσିሺσሻ

ൈටσିሺσሻ

, where ݊ൌnumber of observations.

Also we can calculate ݎ value by using assumed means of ݔand ݕ

where ൌσିσσ

ඨσିሺσሻ

ൈටσିሺσሻ

,

where ݊ൌnumber of observations,

ݑൌݔെܣ ǡ here A is assumed mean of ݔseries,

ݒൌݕെܤ ǡ here B is assumed mean of ݕseries.

. Linear Regression

Regression is the method of estimating the value of one variabl e when other

variable is known and when the variables are correlated. when t he points of

scatter diagram concentrate around straight line, the regressio n is called linear and

this straight line is known as the line of regression

9.3.1 Regression Equations

Regression equations are algebraic forms of the regression line s. Since there are

two regression lines, there are two regression equations the re gression of ݔ on

ݕ( ݕ is independent and ݔ is dependent) is used to estimate the values of ݔ for the

given changes in ݕ and the regression equation of ݕ on ݔ( ݔ is independent and

ݕ is dependent ) is used to estimate the values of ݕ for the given changes in ݔ .

a) Regression equation of ݕ on ݔ :

The regression equation of ݕ on ݔ is expressed as ൌ (--- 1)

where ݔ = independent variable, ݕൌ dependent variable,

ܽൌݕ intercept, ܾൌ slope of the said line & ݊ = number of observations.

The values of Ԣܽᇱand ԢܾԢ can be obtained by the Method of least squares. in this

method the following two algebraic Normal equations are solved to determine the

values of ܽ and ܾ :

σൌ σ (---2)

σ ൌ σσ ---(3) munotes.in

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118NUMERICAL AND STATISTICAL METHODSb) Regression equation of ݔ on ݕ :

The regression equation of ݔon ݕ is expressed as ݔൌܿݕ݀ (--- 1')

where ݕ ൌ independent variable, ݔൌ dependent variable,

ܿൌݕ intercept, ݀ൌ slope of the said line & ݊ = number of observations.

The values of ԢܿԢ and Ԣ݀Ԣ can be obtained by the Method of least squares. in this

method the following two algebraic Normal equations are solved to determine the

values of ܿ and ݀ :

σൌ σ (---2')

σ ൌ σσ ---(3')

Regression equation using regression coefficients:

Direct method:

a) Regression equation of ݕ on ݔ

When the values of ݔ and ݕ are large. In such case,

the equation ݕൌܽݔܾ is changed to

െ ഥൌሺെഥሻ ---(4)

where ݕഥand ݔҧ are arithmetic mean of ݕ andݔ respectively

Dividing both side of equation (2) by ݊ ,we get ݕതൌܽݔܾ ҧ

so that ܽൌݕതݔܾҧ

Substituting this in equation (1), we get

ݕെݕതൌܾ ሺ ݔെݔ ҧሻ

Writing ܾ with the usual subscript we get the equation (4)

again, multiplying equation (2) by σݔ and equation (3) by ݊ we have

σݔσݕൌܽ݊ σݔܾሺσݔሻଶ

݊σݕݔ ൌ ܽ݊σݔܾ݊ሺ σݔሻଶ

subtracting the first from second, we get

݊σݕݔെσݔσݕൌܽ݊ ሺσݔሻଶെܾሺσݔሻଶ munotes.in

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119Chapter 9: Least-Squares Regression ܾൌσ௫௬ିσೣσ

σ௫మିሺσೣሻమ

writing ܾ with the usual subscript we get

Regression coefficient of ݕ on ݔ is ൌσିσσ

σିሺσሻ

b) Regression equation of ݔ on ݕ

Processing the same way as before, the regression equation of ݔ on ݕ is

െഥൌሺെഥሻ

Where Regression coefficient of ݔ on ݕ is ൌσିσσ

σିሺσሻ

Example 9.3.1.1:

Find the two-regression equation for the following data: 16 18 20 23 26 27 11 12 14 15 17 16

Solution :

Calculation for regression equations 16 11 176 256 121 18 12 216 324 144 20 14 280 400 196 23 15 345 529 225 26 17 442 676 289 27 16 432 729 256 σ=130 σൌ 85 σ ൌ1891 σൌ2914 σൌ1231 ഥൌσ

ൌ

= 21.67, ഥൌ

ൌૡ

ൌ Ǥ ૠ

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120NUMERICAL AND STATISTICAL METHODSi) Regression Equation of on is െഥൌሺെഥሻ

Regression coefficient of ݕ on ݔ is ൌσିσσ

σିሺσሻ

ൌૡૢିכૡ

ૢି

= 0.50

Regression Equation of on is

ݕെݕതൌܾ௬௫ሺݔെݔҧሻ

ݕെͳͶǤͳൌͲǤͷͲ ሺݔെʹͳǤ ሻ

ൌ Ǥ Ǥ

ii) Regression Equation of on is െഥൌሺെഥሻ

Regression coefficient of ݔ on ݕ is ൌσିσσ

σିሺσሻ

ൌૡૢିૡכ

ିૡ

ൌ Ǥૡ

Regression Equation of on is

ݔെݔҧൌܾ௫௬ሺݕെݕതሻ

ݔെʹ ͳ Ǥ ൌͳ Ǥ ͺ Ͷ ሺ ݕെͳ Ͷ Ǥ ͳ ሻ

ൌ Ǥૡ െ Ǥ

Deviations taken from arithmetic mean of and

a) Regression equation of ݕ on ݔ :

Another way of expressing regression equation of ݕ on ݔ is by taking deviation

ofݔ andݕ series from their respective ac tual means. in this case also, the equation

ݕൌܽݔܾ is changed to െഥൌሺെഥሻ

The value of ܾ௬௫ can be easily obtained as follows:

ൌσ

σ

Where χൌሺݔെݔҧሻ and γൌሺ ݕെݕതሻ munotes.in

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121Chapter 9: Least-Squares RegressionThe two normal equation which we had written earlier when chang ed in terms of

χ and γ becomes

σߛൌܽ݊ ܾ σ߯( --- 5)

σߛ߯ ൌ ܽσܾ߯σ߯ଶ --- (6)

S i n c e σ߯ൌσߛൌͲ (deviation being taken from mean)

E q u a t i o n ( 5 ) r e d u c e s t o ܽ݊ ൌ Ͳ ܽൌͲ

E q u a t i o n ( 6 ) r e d u c e s t o σߛ߯ ൌ ܾσ߯ଶ

or ൌσࢽ

σ

After obtaining the value of ܾ௬௫ the regression equation can easily be written in

terms of ݔand ݕ by substituting for ߯ൌሺݔെݔҧሻ and ߛൌሺݕെݕതሻǤ

Deviations taken from arithmetic mean of and

The regression equation ݔൌܿݕ݀ is reduces to െഥൌሺെഥሻ

Where ൌσࢽ

σࢽ

Example 9.3.1.2 :

The following data related to advertising expenditure and sales Advertising expenditure

(In Lakhs) ͷͺͻSales (In Lakhs) ʹͲ͵ͲͶͲͲͷͲ

i) Find the regression equations

ii) Estimate the likely sales when advertising expenditure is 12 lakhs Rs.

iii) What would be the advertis ing expenditure if the firm wan ts to attain sales

target of 90 lakhs Rs.

Solution :

i) Let the advertising expenditure be denoted by ݔand sales by ݕ

arithmetic mean of ݔ is ഥൌσ

ൌ

ൌૠ

& arithmetic mean of ݕ is ഥൌσ

ൌ

ൌ

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122NUMERICAL AND STATISTICAL METHODSCalculation for regression equations ൌሺ െ ഥሻ ࢽൌሺ െ ഥሻ ࢽ ࢽ ͷʹͲെʹെʹͲͶͲͶͶͲͲ͵ͲെͳെͳͲͳͲͳͳͲͲͶͲͲͲͲͲͲͺͲͳʹͲʹͲͳͶͲͲͻͷͲʹͳͲʹͲͶͳͲͲ

ൌ

ൌ ൌൌࢽ

ൌૢ

ൌ ࢽ

ൌ

Regression equation of ݕ on ݔ

ܾ௬௫ൌσఞఊ

σఞమൌଽ

ଵൌͻ

Regression equation of ݕ on ݔ is

ݕെݕതൌܾ௬௫ሺݔെݔҧሻ

ݕെͶͲൌͻ ሺݔെሻ

ݕെͶͲൌͻݔെ͵

ൌૢ െ

Regression equation of ݔ on ݕ

ܾ௫௬ൌσఞఊ

σఊమൌଽ

ଵൌͲ Ǥ Ͳ ͻ

ݔെݔҧൌܾ௫௬ሺݕെݕതሻ

ݔെൌͲ Ǥ Ͳ ͻ ሺݕെͶͲሻ

ݔെൌͲ Ǥ Ͳ ͻ ݕെ͵ Ǥ

ൌ Ǥ ૢ Ǥ

ii) For estimate the likely sales when advertising expenditure is 1 2 lakhs

ൌૢ െ ൌͻ ሺͳʹሻെʹ͵ൌૡ lakhs Rs.

iii) The advertising expenditure if the firm wants to attain sales target of 90

lakhs Rs.

ൌ Ǥ ૢ ǤൌͲ Ǥ Ͳ ͻ ሺͻͲሻ͵ǤͶൌǤ lakhs Rs.

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## Page 123

123Chapter 9: Least-Squares RegressionDeviation taken from the assumed mean:

a) Regression equation of on :

The equation ݕൌܽݔܾ is changed to ݕെݕതൌܾ௬௫ሺݔെݔҧሻ

W h e n a c t u a l m e a n i s n o t a w h o l e n u m b e r , b u t a f r a c t i o n o r w h e n values of

ݔand ݕ are large

W e u s e t h e f o r m u l a : ൌσିσσ

σିሺσሻ

W h e r e ݑൌݔെܣ , ݒൌݕെܤ ǡ

ܣൌ Assumed mean for ݔǡ ܤൌ Assumed mean for ݕ

b) Regression equation of on :

T h e r e g r e s s i o n e q u a t i o n ݔൌܿݕ݀ is reduces to ݔെݔҧൌܾ௫௬ሺݕെݕതሻ

W h e r e ൌσିσσ

σିሺσሻ

Where ݑൌݔെܣ , ݒൌݕെܤ ǡ

ܣൌ Assumed mean for ݔǡ ܤൌ Assumed mean for ݕ

Also the values of correlation coefficient, means and Standard deviations of two

variables are given, in such a case we can find ܾ௫௬ and ܾ௬௫ as following:

Let ݔҧ , ݕത be the means and ߪ௫ , ߪ௬ be the standard deviation of the ݔ and ݕ

respectively and ݎ be the correlation coefficient, then ࢽൌ

ࢽ and ࢽൌࢽ

Example 9.3.1.3 :

From the following data, calculate two lines of regression 17 21 18 22 16 60 70 68 70 66 i) Estimate value of ݕ when ݔis 35

ii) Estimate value of ݔ when ݕis 75

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124NUMERICAL AND STATISTICAL METHODSSolution :

Let 21 be the assumed mean for ݔ i.e. ܣൌʹ ͳ & 90 be the assumed mean for ݕ ,

i.e. ܤൌͻ Ͳ

Calculation for regression equations ൌെ ൌെ ͳͲെͶെͳͲͶͲͳͳͲͲʹͳͲͲͲͲͲͲͳͺͺെ͵െʹͻͶʹʹͲͳͲͲͳͲͳെͷെͶʹͲʹͷͳ

ൌૢ

ൌ

ൌെ

ൌെ

ൌ

ൌ

ൌ

Regression equation of on

ൌσିσσ

σିሺσሻ

, Where ݑൌݔെܣ , ݒൌݕെܤ ǡ

ൌିሺషሻൈሺషሻ

ିሺషሻ

= ͳǤͳͶͻʹ

ݔҧൌͳ ͺ Ǥ ͺ Ƭ ݕ തൌ Ǥ ͺ

Regression equation of ݕ on ݔ is ݕെݕതൌܾ௬௫ሺݔെݔҧሻ

ݕെǤͺൌͳǤͳͶͻʹ ሺݔെͳͺǤͺሻ

ݕെǤͺൌͳǤͳͶͻʹݔെʹͳǤͲͶͻ

ൌ Ǥ ૢ Ǥ ૢ

i) When ݔis 35, ݕis

ݕൌͳ Ǥ ͳ Ͷ ͻ ʹ ሺ ͵ ͷ ሻͶ ͷ Ǥ ͳ ͻ ͷ ͳ

ൌૡ Ǥ ૠ munotes.in

## Page 125

125Chapter 9: Least-Squares RegressionRegression equation of on

ൌσିσσ

σିሺσሻ

, Where ݑൌݔെܣ , ݒൌݕെܤ ǡ

ൌିሺషሻൈሺషሻ

ିሺషሻ

ൌ Ͳ Ǥ Ͷ Ͷ

ݔҧൌͳ ͺ Ǥ ͺ Ƭ ݕ തൌ Ǥ ͺ

Regression equation of ݔ on ݕ is ݔെݔҧൌܾ௫௬ሺݕെݕതሻ

ݔെͳ ͺ Ǥ ͺ ൌͲ Ǥ Ͷ Ͷ ሺ ݕെ Ǥ ͺ ሻ

ݔെͳ ͺ Ǥ ͺൌͲ Ǥ Ͷ Ͷ ݕെʹ ͻ Ǥ ͺ ͻ ͻ

ൌ Ǥ ૠ െ Ǥ ૢ ૢ

ii) When ݕis 75, ݔis

ݔൌͲ Ǥ Ͷ Ͷ ሺ ͷ ሻെͳ ͳ Ǥ Ͳ ͻ ͻ

ൌ Ǥ ૠ

9.3.2 Method of Least Squares

The famous German mathematician Carl Friedrich Gauss had invest igated the

method of least squares as early as 1794, but unfortunately he did not publish the

method until 1809. In the meantime, the method was discovered a nd published in

1806 by the French mathematician Legendre, who quarrelled with Gauss about

who had discovered the method first. The basic idea of the meth od of least

squares is easy to understand. It may seem unusual that when se veral people

measure the same quantity, they usually do not obtain the same results. In fact, if

the same person measures the same quantity several times, the r esults will vary.

What then is the best estimate for the true measurement? The me thod of least

squares gives a way to find the best estimate, assuming that th e errors (i.e. the

differences from the true value) are random and unbiased, The m ethod of least

squares is a standard approach in regression analysis to approx imate the solution

of over determined systems by minimizing the sum of the squares of the residuals

made in the results of every single equation. The most importan t application is in

data fitting.

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## Page 126

126NUMERICAL AND STATISTICAL METHODSSteps Involved:

Step (i) Calculate the actual means of ݔ and actual mean of ݕ .

Step (ii) Calculate the summation of the ݔseries and ݕ series.

I t ' l l g i v e s σݔand σݕ respectively.

Step (iii) Square the values of the series ݔand ݕand calculate its summation.

it'll gives σݔଶ and σݕଶ respectively.

Step (iv) Multiply each value of the series ݔby the respective values of the

series ݕ and calculation its summation, it'll gives σݕݔ .

Step (v) Solve the following Normal equations to determine the values of 'ܽ '

and 'ܾ 'for regression equation ݕ on ݔ .

σݕൌܽ݊ ܾ σݔ

σݕݔ ൌ ܽσݔܾσݔଶ

after determine the value of ' ܽ 'and 'ܾ 'put these values in the following equation.

ݕൌܽݔܾ

Step (v') Solve the following Normal equations to determine the values of 'ܿ '

and '݀ 'for regression equation ݕ on ݔ .

σݔൌܿ݊ ݀ σݕ

σݔݕ ൌ ܿσݕ݀σݕଶ

after determine the value of ' ܿ 'and '݀ 'put these values in the following equation.

ݔൌܿݕ݀ .

Example 9.3.2.1 :

From the following data fit a regression line where ݕ is dependent variable and ݔ

is independent variable (i.e. ݕ on ݔ ) using the method of least square, also

estimate the value of ݕ when ݔൌ Ǥ ͺ ݔ 2 4 6 8 10 ݕ 15 14 8 7 2 munotes.in

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127Chapter 9: Least-Squares RegressionSolution : Let ݕൌܽݔܾ be the required equation of line (because ݕ on ݔሻǤ

To determine the values of ' ܽ 'and 'ܾ 'for regression equation ݕ on ݔ

The two normal equations are σݕൌܽ݊ ܾ σݔ ,

σݕݔ ൌ ܽσݔܾσݔଶ

where ݊ൌͷ (number of datasets) ʹͳͷ͵ͲͶͶͳͶͷͳͺͶͺ͵ͺͷͶͳͲʹʹͲͳͲͲݔൌ͵ͲݕൌͶݕݔൌʹͳͲݔଶൌʹʹͲSubstituting these calculated values into normal equation, we'l l get

Ͷ ൌ ͷܽ͵Ͳܾ

ʹͳͲ ൌ ͵ͲܽʹʹͲܾ

Solving the above equations, we get

ܽൌͳ ͻ Ǥ ͳ and ܾൌെ ͳ Ǥ ͷ

Hence, the required line ൌ is ൌ ૢ Ǥ ሺെǤሻ

or ൌ ૢ Ǥെ Ǥ

W h e n ൌ Ǥ ͺ , ݕis equal to ͳͻǤͳെͳǤͷሺǤͺሻ = 6.23

Hence estimate value of when ൌૠ Ǥૡ is 6.23

Example 9.3.2.2 :

From the following data fit a regression line where ݔ is dependent variable and ݕ

is independent variable (i.e. ݔ on ݕ ) using the method of least square, also

estimate the value of ݔ when ݕൌ͵ Ͳ 7 8 11 12 14 16 20 12 15 19 8 25

munotes.in

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128NUMERICAL AND STATISTICAL METHODSSolution : Let ݔൌܿݕ݀ be the required equation of line (because ݔ on ݕሻǤ

To determine the values of ' ܿ 'and '݀ 'for regression equation ݔ on ݕ

T h e t w o n o r m a l e q u a t i o n s a r e σݔൌܿ݊ ݀ σݕ

σݔݕ ൌ ܿσݕ݀σݕଶ

w h e r e ݊ൌ (number of datasets) ʹͲͳͶͲͶͲͲͺͳʹͻͳͶͶͳͳͳͷͳͷʹʹͷͳʹͳͻʹʹͺ͵ͳͳͶͺͳͳʹͶͳʹͷͶͲͲʹͷݔൌͺݕൌͻͻݔݕൌͳͳͶͳݕଶൌͳͺͳͻSubstituting this calculated values into normal equation, we'll get

ͺ ൌ ܿͻͻ݀

ͳͳͶͳ ൌ ͻͻܿͳͺͳͻ݀

Solving the above equations, we get

ܿൌͻ Ǥ Ͷ and ݀ൌ Ͳ Ǥ ͳ Ͳ

Hence, the required line ൌ is ൌૢ Ǥ Ǥ

When ൌ͵ Ͳ , ݔis equal to ͻǤͶͲǤͳͲሺ͵Ͳሻ = 12.64

Hence estimate value of when ൌ is 12.64

9.3.3 Properties of Regression Coefficients

i) Correlation coefficient is the geometric mean between the regre ssion

coefficients.

i.e. The coefficien t of regression are ݎఙ

ఙೣ and ݎఙೣ

ఙ

Geometric Mean between them ටݎఙ

ఙೣൈݎఙೣ

ఙ=ξݎଶൌേ ݎ munotes.in

## Page 129

129Chapter 9: Least-Squares RegressionBoth the regression coefficients will have the same sign, i.e., they will be wither

positive or negative. It is not possible for one to be positive and other to be

negative.

If regression coefficients are positive, then ݎis positive and if regression

coefficients are negative, ݎ is negative .if ܾ௫௬ൌͲ Ǥ ͵ and ܾ௬௫ൌͲ Ǥ

ܾ௫௬ܾ௬௫

ʹൌͲǤ͵ͲǤ

ʹൌͲ Ǥ Ͷ ͷ

then the value of ݎൌξͲǤ͵ൈͲǤൌͲ Ǥ Ͷ ʹ which is less than 0.45

The value of the correlation coefficient cannot exceed one, if one of the

regression coefficient is greater than unity, the other must be less than unity.

if ܾ௫௬ൌͳ Ǥ Ͷ and ܾ௬௫ൌͳ Ǥ ͷ the ݎൌξͳǤͶൈͳǤͷൌͳ Ǥ Ͷ ͷ which (greater than 1) is

not possible.

The point ሺݔҧǡݕതሻ satisfies both the regression equations as it lies on both the lines

so that it is the point of intersection of the two lines. This can be helpful to us

whenever the regression equations are known and the mean values of ݔand ݕ are

to be obtained, In this case, the two regression equations can be solved

simultaneously and the common solution represent ݔҧ and ݕത

Example 9.3.3.1 :

From the given data calculate equations of two lines of regress ion Mean Standard deviation 20 3 100 12 Coefficient of correlation is 0.8 i.e. ݎ = 0.8

Solution :

We have ݔҧൌʹ Ͳ ǡߪ ௫ൌ͵ ǡݕതൌͳ Ͳ Ͳ ǡߪ ௬ൌͳ ʹ and ݎൌͲ Ǥ ͺ

i) To find the regression equation of ݕ on ݔ

ܾ௬௫ൌݎൈߪ௬

ߪ௫ൌͲ Ǥ ͺൈͳʹ

͵ൌ͵ Ǥ ʹ

Regression equation of ݕ on ݔ is ሺݕെݕതሻൌܾ௬௫ሺݔെݔҧሻ

ሺݕെͳͲͲሻൌ͵ Ǥ ʹሺݔെʹ Ͳሻ

ሺݕെͳͲͲሻൌ͵ Ǥ ʹ ݔെ Ͷ

ൌ Ǥ munotes.in

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130NUMERICAL AND STATISTICAL METHODSii) To find the regression equation of ݔ on ݕ

ܾ௫௬ൌݎൈߪ௫

ߪ௬ൌͲ Ǥ ͺൈ͵

ͳʹൌͲ Ǥ ʹ

Regression equation of ݔ on ݕ is ሺݔെݔҧሻൌܾ௫௬ሺݕെݕതሻ

ሺݔെʹ ͲሻൌͲ Ǥ ʹሺݕെͳͲͲሻ

ሺݔെʹ ͲሻൌͲ Ǥ ʹ ݕെʹ Ͳ

ൌ Ǥ

Example 9.3.3.2 :

Given the two regression lines as ݔെͶ ݕൌͷ and ݔെͳ ݕൌെ Ͷ , find the

average of ݔand ݕ and correlation coefficient between ݔ and ݕ i.e. to find value

of ݔҧǡݕതand ݎ

Solution :

To find the mean of ݔ and ݕ

As per properties, ( ݔҧǡݕതሻ will lie on both the regression. thus,

ݔҧെͶݕതെͷൌͲ and ݔҧെͳݕതൌ ͶൌͲ

Solve the given two equations simultaneously.

ݔҧെͶݕതൌͷ - - - ( 1 )

ݔҧെͳݕതൌെ Ͷ ---(2)

െ

______________________

ͳʹݕതൌ ͻ

ഥൌ Ǥૠ

substituting in equation (1)

ݔҧെͶሺͷǤͷሻൌͷ

ഥൌ ૡ

ഥൌ ૡ ǡഥൌ Ǥૠ

To find correlation coefficient we have two lines of regression s. Which line is

regression line ݕ on ݔ and which is ݔ and which is ݔob ݕin not known.

Let us assume that ݔെͶ ݕൌͷ be regression line of ݕ on ݔ and ݔെͳ ݕൌ

െͶbe the regression line of ݔ on ݕ munotes.in

## Page 131

131Chapter 9: Least-Squares RegressionHence we get

ݕൌെହ

ସଵ

ସݔ and ݔൌͳ ݕെ 64

ܾ௬௫ൌଵ

ସand ܾ௫௬ൌͳ

We require to check whether the assumption is correct or not ?

Check 1: Signs: Both regression coefficients are positive.

Check 2: Product: Product of two regression coefficients.

ܾ௬௫ൈܾ௫௬ൌଵ

ସൈͳൌͶͳ Which is greater than 1

Assumption is wrong.

Let ݔെͶ ݕൌͷ be regression line of ݔ on ݕ

and ݔെͳ ݕൌെ Ͷ be the regression line of ݕ on ݔ

ܾ௫௬ൌͶ and ܾ௬௫ൌଵ

ଵ

Check 1: Signs: Both regression coefficients are positive.

Check 2: Product: Product of two regression coefficients.

ܾ௬௫ൈܾ௫௬ൌଵ

ଵൈͶൌͲǤʹͷ൏ͳ Which is less than 1

This assumption is correct.

Now, ݎൌേඥܾ௫௬ൈܾ௬௫ൌേටͶൈଵ

ଵൌേଵ

ଶ

The sing of correction coefficient is same as sign of both regr ession coefficients

ݎൌଵ

ଶorͲǤͷ

.4 Polynomial Regression

In statistics, polynomial regression is a form of regression an alysis in which the

relationship between the independent variable ݔ and the dependent variable ݕ is

modeled as an nth degree polynomial in ݔ .

Polynomial regression fits a nonlinear relationship between the value of ݔ and the munotes.in

## Page 132

132NUMERICAL AND STATISTICAL METHODScorresponding conditional mean of ݕǡand has been used to describe non linear

phenomena such as growth rate of tissues, the distribution of c arbon isotopes in

lake sediments etc.

.5 Non- liner Regression

Regression will be called non-linear if there exists a relation ship other than a

straight line between the variables under consideration.

Nonlinear regression is a form of regression analysis in which observational data

are modeled by a function which is a nonlinear combination of t he model

parameters and depends on one or more independent variables.

9.5.1 Polynomial fit :

Let ݕൌܽݔܾ ݔܿ ଶ be second degree parabolic curves of regression of ݕ on ݔ

to be fitted for the data ሺݔǡݕሻ, where ݅ൌͳ ǡ ʹ ǡ ͵ǥ݊

To fit parabolic curves for ݔ on ݕǡ

Considerݕොൌܽݔܾ ݔܿ ଶ

For each ݔǡݕොൌܽݔܾ ݔܿଶ

Error inestimation is ݕെݕపෝ and Summation of squares of error is

߮ൌ݁ଶൌ ሺݕ െݕపෝ

ୀଵ

ୀଵሻଶ

߮ൌσሺݕെܽെݔܾ െݔܿଶሻଶ

ୀଵ ǥሺሻ

To find values of ܽǡܾǡܿ such that ߮is minimum

డఝ

డൌͲ ǡడఝ

డൌͲ and డఝ

డൌͲ

Differentiating ߮w.r.t. ܽ ,we get

డఝ

డൌʹ σሺݕെܽെݔܾ െݔܿଶሻሺെͳሻ

σሺݕെܽെݔܾ െݔܿଶሻൌͲ

σݕെܽ݊െܾσݔ െܿσݔଶൌͲ

σൌ σσ ...(2) munotes.in

## Page 133

133Chapter 9: Least-Squares RegressionAgain differentiating ߮w.r.t. ܾ ,we get

డఝ

డൌʹ σሺݕെܽെݔܾ െݔܿଶሻሺെݔሻ

σሺݕെܽെݔܾ െݔܿଶሻሺݔሻൌͲ

σݔݕെܽσݔെܾσݔଶെܿσݔଷൌͲ

σ ൌσσσ ...(3)

Again differentiating ߮w.r.t. ܿ ,we get

డఝ

డൌʹ σሺݕെܽെݔܾ െݔܿଶሻሺെݔଶሻ

σሺݕെܽെݔܾ െݔܿଶሻሺݔଶሻൌͲ

σݔଶݕെܽσݔଶെܾσݔଷെܿσݔସൌͲ

σ ൌσσσ ...(4)

Equation (2) (3) and (4) are normal e quations for fitting a sec ond degree parabolic

curve.

i.e. ෝൌ

Example 9.5.1.1 :

Using method of least square fit a second degree parabola for t he following data. ͳʹ͵ͶͷͺͻʹͺͳͲͳͳͳͳͳͲͻSolution :

The second-degree polynomial equation is ൌ and Normal

Equations are

σൌ σσ

σ ൌ σσσ

σൌσσσ

munotes.in

## Page 134

134NUMERICAL AND STATISTICAL METHODSConsider the following table for calculation ͳʹͳʹͳʹͳʹͶͳʹͺʹͶͳ͵ͻʹͳʹ͵ͺͳͶͺͳ͵ʹͶͳʹͺʹͷͷͳͲʹͷͷͲͳʹͷʹͷͲʹͷͳͳ͵ʹͳ͵ͻͳʹͻͳͳͶͻ͵Ͷ͵ͷ͵ͻʹͶͲͳͺͳͲͶͺͲͷͳʹͶͲͶͲͻͻͻͺͳͺͳʹͻʹͻͷͳૠૡૠૠHere ൌૢ,

σൌ ǡσൌૠ ǡσൌ ૡ ǡσ ൌ ǡ σൌ ǡ σൌ

ૠૠǡσൌ

Substituting these values into Normal Equations we get

Ͷ ൌ ͻܽͶͷܾʹͺͷܿ (... 1 )

Ͷʹͳ ൌ ͶͷܽʹͺͷܾʹͲʹͷܿ (... 2)

ʹͳ ൌ ʹͺͷܽʹͲʹͷܾͳͷ͵͵͵ܿ (... 3 )

Solving equation (1), (2) and (3) (by simultaneously or use Cra mer's rule)

ܽൌെ Ͳ Ǥ ͻ ʹ ͺ ͷ ǡܾൌ͵ Ǥ ͷ ʹ ͵ ͳ ǡܿൌെ Ͳ Ǥ ʹ ͵

Hence, The second degree equation is

ൌെ Ǥૢ ૡ Ǥ െ Ǥ ૠ

.6 Multiple Linear Regression

Multiple linear regression is a technique that uses two or more i n d e p e n d e n t

variables to predict the outcome of a dependent variable.

The technique enables analysts to determine the variation of th e model and the

relative contribution of each independent variable in the total variance. munotes.in

## Page 135

135Chapter 9: Least-Squares RegressionMultiple regression can take two forms, i.e., linear regression and non-linear

regression.

Consider such a linear function as ݕ ൌ ܽݔܾ ݖܿ .

To fit multiple linear regression for ݔǡݕ and ݖ consider

ݕො ൌ ܽݔܾݖܿ

for each ݔ, ݕపෝൌ ܽݔܾ ݖܿ

Error in estimation is ݕെݕపෝ and summation of squares of error is

߮ൌ୧ଶൌ ሺ ୧െƸ୧୬

୧ୀଵ୬

ୀଵሻଶ

߮ൌ ሺݕെܽെݔܾ െݖܿሻଶ

ୀଵǥሺሻ

To find values of ܽǡܾǡܿsuch that ߮ is minimum

i.e. డఝ

డൌడఝ

డൌడఝ

డൌͲ

Differentiating equation (1) w.r.t. ܽ we get

డఝ

డൌʹσሺݕെܽെݔܾ െݖܿሻሺെͳሻ

σݕെܽ݊െܾ σݔെܿσݖൌͲ

ݕൌܽ݊ ܾݔ ܿݖ ǥሺሻ

Again differentiating equation (1) w.r.t ܾ we get

డఝ

డൌʹσሺݕെܽെݔܾ െݖܿሻሺെݔሻ

σݔݕെܽσݔെܾσݔଶെܿσݔݖൌͲ

ݔݕൌܽݔ ܾݔଶܿݔ ݖǥሺሻ

Again differentiating equation (1) with respect to c, we get

డఝ

డൌʹσሺݕെܽെݔܾ െݖܿሻሺെݖሻ munotes.in

## Page 136

136NUMERICAL AND STATISTICAL METHODSσݖݕെܽσݖെܾσݔݖെܿσݖଶൌͲ

ݖݕൌܽݖ ܾݔ ݖܿݖଶǥሺሻ

Here Equation (2), (3), and (4) are the normal equations for f itting a

multiple linear regression equation

ݕො ൌ ܽݔܾ ݖܿ

Example 9.6.1 :

Obtain a regression plane by using multiple regression to fit t he following data Ͳͳʹ͵Ͷͳ͵ͳͳͻʹͳʹࢠͳʹ͵Ͷͷ

Solution :

The multiple regression equation is ݕ ൌ ܽݔܾ ݖܿ

and normal equations are

σݕൌܽ݊ ܾ σݔܿσݖ

σݕݔ ൌ ܽσݔܾσݔଶܿσݖݔ

σݖݕ ൌ ܽσݖܾσݖݔܿσݖଶ

Consider the following table for calculation ࢠࢠࢠࢠͲͳ͵ͳͲͲͲͳͳ͵ͳͳʹͳͳʹͶ͵Ͷʹͳͻ͵͵ͺͶͻͷ͵ʹͳͶ͵ͻͳʹͳͺͶͶʹͷͳͲͶͳʹͲʹͷͳ͵ͲૢૡHere ݊ ൌ ͷ,

σൌ ǡσൌૢ ǡσࢠൌ ǡσ ൌ ǡ σൌ ǡσࢠ ൌ ǡ σࢠൌ

ǡσࢠ ൌ ૡ munotes.in

## Page 137

137Chapter 9: Least-Squares RegressionSubstituting these values i nto Normal Equations we get,

ͻ ൌ ͷܽͳͲܾͳͷܿ (... 1)

ʹʹʹ ൌ ͳͲܽ͵ͲܾͶͲܿ (... 2 )

͵ͳͺ ൌ ͳͷܽͶͲܾͳͷܿ (... 3 )

Solving equation (1), (2) and (3) (by simultaneously or use Cra mer's rule)

ܽൌͳ ͵ Ǥ ͺ ǡ ܾൌ͵ ǡ ܿൌͲ

ǡThe required regression plane equation is

ൌ Ǥ

.7 General Linear Least Squares

The linear least square is the least squares approximation of l inear function to

data. It is a set of formulations for solving statistical probl ems involved in linear

regression including variants for ordinary and generalized leas t squares.

9.7.1 Ordinary Least Square(OLS)

Ordinary least squares (OLS) is a type of linear least squares method for

estimating the unknown parameters in a linear regression model. OLS chooses the

parameters of a linear function of a set of explanatory variabl es by the principle of

least squares: minimizing the sum of the squares of the differe nces between the

observed dependent variable (valu es of the variable being obser ved) in the given

dataset and those predicted by the linear function of the indep endent variable.

9.7.2 Generalized Least Square (GLS)

Generalized least squares (GLS) is a technique for estimating t he unknown

parameters in a linear regression model when there is a certain degree of

correlation between the residuals in a regression model.

. Summary

x A correlation or simple linear regression analysis can determin e if two

numeric variables are significantly linearly related. A correla tion analysis

provides information on the strength and direction of the linea r relationship

between two variables, while a simple linear regression analysi s estimates

parameters in a linear equation that can be used to predict val ues of one

variable based on the other. munotes.in

## Page 138

138NUMERICAL AND STATISTICAL METHODSx Both correlation and regression can be used to examine the pres ence of a

linear relationship between two variables providing certain ass umptions

about the data are satisfied. The results of the analysis, howe ver, need to be

interpreted with care, particularly when looking for a causal r elationship or

when using the regression equation for prediction. Multiple and l o g i s t i c

regression will be the subMect of future reviews.

. References

Following books are recommended for further reading :-

x Introductory Methods of Numerical Methods - S. S. Shastri PHI.

x Numerical Methods for Engineers - Steven C. Chapra, Raymond P. Canale

Tata Mc Graw Hill.

.10 Exercises

Q.1) Calculate the product moment coefficient of correlation for the given data: 13109111214815971012134(Ans. ݎൌͲ Ǥ ͻ Ͷ ͺ ሻ

Q.2) Calculate coefficient of corre lation between the values of ݔ and ݕgiven

below. 66707068747072756874666972636965( Ans. ݎൌെ Ͳ Ǥ Ͳ ͵ ሻ

Q.3) Find the coefficient of correlation for the following data: 808490757270788286110115118105104100108112116( Ans. ݎൌͲ Ǥ ͻ ͺ ሻ

Q.4) Find the coefficient of correlation for the following data by u sing assumed

mean method ʹͳʹʹͳͶʹͲͷʹʹͲʹʹͷʹͳͶʹͳͺͷͲͲͷͳͷͷͳͳͷ͵Ͳͷʹʹͷͳͷʹͷ( Ans. ݎൎͲ Ǥ ͺ ሻ munotes.in

## Page 139

139Chapter 9: Least-Squares RegressionQ.5) Fit the straight line by using least square method for the foll owing data: ͳ2345670.52.52.04.03.56.05.2

Q.6) Find the best fit values of ܽ andܾso that ݕൌܽݔܾ fits the data given in

the table: ͲͳʹͶͳͳǤͺ͵Ǥ͵ͶǤͷ

Q.7) Consider the data below: ͳʹ͵ͶͳͳͳʹͳUse linear least-Square regression to determine function of the from ݕൌܾ݁ ௫

the given data by specifying ܾ and ݉

(Hint: Take natural log o n both side of the function)

Q.8) Fit the second order polynomial to the given below: 1234567892678101111109Q.9) Fit the second order polynomial to the given below: 01234 -4-141120

Q.10) Use the multiple regression t o fit the following data 022.5147012362ࢠ51090327

munotes.in

## Page 140

140NUMERICAL AND STATISTICAL METHODSUnit 4

10

LINEAR PROGRAMMING

Unit Structure

10.0 ObMectives

10.1 Introduction

10.2 Common terminology for LPP

10.3 Mathematical Formulation of L.P.P

10.4 Graphical Method

10.5 Summary

10.6 References

10.7 Exercise

10.0 Objectives

This chapter would make you understand the following concepts:

x Sketching the graph for linear equations.

x Formulate the LPP

x Conceptualize the feasible region.

x Solve the LPP with two variables using graphical method.

10.1 Introduction

Linear programming (LP, also called linear optimization) is a m ethod to achieve

the best outcome (such as maximum profit or lowest cost) in a m athematical

model whose requirements are represented by linear relationship s. Linear

programming is a special case of mathematical programming (also known as

mathematical optimization).

10.2 Common terminology for LPP

Linear programming is a mathematical concept used to determine the solution to

a linear problem. Typically, the goal of linear programming is to maximize or munotes.in

## Page 141

141Chapter 10: Linear Programmingminimize specified obMectives, such as profit or cost. This pro cess is known as

optimization. It relies upon three different concepts: variable s, obMectives, and

constraints.

"Linear programming is the analysis of pr oblems in which a linear function of a

number of variables is to be maximized (or minimized) when those variables are

subject to a number of restrains in the form of linear inequalities."

Objective Function : The linear function, which is to be optimized is called

obMective function The obMective function in linear programming problems is the

real-valued function whose value is to be either minimized or m aximized subMect

to the constraints defined on the given LPP over the set of fea sible solutions. The

obMective function of a LPP is a linear function of the form ܼൌܽ ଵݔଵܽଶݔଶ

ܽଷݔଷǥǤܽݔ

Decision Variables : The variables involved in LPP are called decision variables

denoted them as ሺݔǡݕሻor ሺݔଵǡݔଶሻ etc. here its refer to some quantity like units,

item production on sold quantity, Time etc.

Constraints : The constraints are limitations or restrictions on decision var iables.

they are expressed in linear equalities or inequalities i.e. ൌǡǡ

Non-negative Constraints : This is a condition in LPP specifies that the value of

variable being considered in the linear programming problem wil l never be

negative. It will be either ze ro or greater than zero but can n ever be less than zero,

Thus it is expressed in the form of ݔͲ and ݕͲ.

Feasible Solution : A feasible solution is a set of values for the decision variabl es

that satisfies all of the constraints in an optimization proble m. The set of all

feasible solutions defines the feasible region of the problem. in graph the

overlapping region is called feasible region

Optimum solution : An optimal solution to a linear program is the solution

which satisfies all constraints with maximum or minimum obMecti ve function

value.

10. Mathematical Formulation of L.P.P

To write mathematical formul ation of L.P.P following steps to b e remembered.

Step 1 : Identify the variables involved in LPP (i.e. Decision variables) and

denote them as ሺݔǡݕሻor ሺݔଵǡݔଶሻ. munotes.in

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142NUMERICAL AND STATISTICAL METHODSStep 2 : Identify the obMective function and write it as a math ematically in terms

of decision variables.

Step 3 : Identify the different constraints or restrictions an d express them

mathematically

Example 10..1 :

A bakery produces two type of cakes I and II using raw material s ܴଵ and ܴଶ. One

cake of type I is produced by using 4 units of raw material ܴଵand 6 units of raw

material ܴଶ and one cake of type II is produced by using 5 units of raw ma terial

ܴଵand 9 units of raw material ܴଶ. There are 320 units of ܴଵ and 540 units ܴଶ in

the stock. The profit per cake of type I and type II is Rs. 200 and Rs. 250

respectively. How many cakes of type I and type II be produced so as to

maximize the profit" formulate the L.P.P

Solution:

Let ݔ be the number of cakes of type I and ݕ be the number of cakes of type II to

be produce to get maximum profit.

since the production value is never negative

ǡ

This is non- negative constraints.

The profit earned by selling 1 cake of type I is Rs. 200. Henc e the profit earned

by selling ݔ cakes is Rs. 20 Ͳݔ .

Similarly, the profit earned by selling 1 cake of type II is Rs . 250 and hence profit

earned by selling ݕ cake is Rs. 250 ݕ .

The Profit earned is

ൌ

This is obMective function.

Now, after reading the given data carefully we can construct th e following table Cake Type Raw Material required Profit (units)/Cake (Units)/Cake I 4 6 200 II 5 9 250 Availability (units) 20 540 - munotes.in

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143Chapter 10: Linear Programming According to the table

1 Cake of type I consumes 4 units of ܴଵ hence ݔ cakes of type I will consume

Ͷݔ units of ܴଵ and one cake of type II consume 5 units of ܴଵ hence ݕ cakes of

type II will consume ͷݕ units of ܴଵ. But maximum number of units available of

ܴଵ is 320. Hence, the constraint is

Ͷݔ ͷݕ ͵ʹͲ .

Similarly, 1 Cake of type I consumes 6 units of ܴଶ hence ݔ cakes of type I will

consume ݔ units of ܴଶ and one cake of type II consume 9 units of ܴଶ hence ݕ

cakes of type II will consume ͻݕ units of ܴଶ. But maximum number of units

available of ܴଶ is 540. Hence, the constraint is

ݔ ͻݕ ͷͶͲ

Hence the mathematical formulation of the given L.P.P is to

Maximi]e ൌ

Subject to,

ૢ

ǡ

Example 10..2 :

A manufacture produce Ball pen and Ink pen each of which must b e processed

through two machines A and B. Machine A has maximum 220 hours a vailable

and machine B has maximum of 280 hours available . Manufacturin g a Ink pen

requires 6 hours on machine A and 3 hours on machine B. Manufac turing a Ball

pen requires 4 hours on machine A and 10 hours on machine B. If the profit are

Rs. 55 for Ink pen and Rs. 75 for Ball pen. Formulate the LPP t o have maximum

profit.

Solution :

Let Rs. ܼ be the profit, Which can be made by manufacturing and selling say

ݔ

number of Ink pens and

ݕ

number of Ball pens.

Here variable ݔ and ݕ are decision variables.

Since profit per Ink pen and ball pen is Rs. 55 and Rs. 75 resp ectively and we

want to maximize the ܼ ,Hence the obMective function is munotes.in

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144NUMERICAL AND STATISTICAL METHODSMaxܼ ൌ ͷͷݔͷݕ

We have to find ݔ and ݕ that maximize ܼ

We can construct the followi ng tabulation form of given data: Machine Time in hours required for Maximum available time in hours 1 Ink pen 1 Ball pen A ͶʹʹͲB ͵ͳͲʹͺͲA Ink pen requires 6 hr on machine A and Ball pen requires 4 hr on machine A

and maximum available time of machine A is 220 hr.

1st Constraint is

ݔͶݕ ʹʹͲ

Similarly, A Ink pen requires 3 hr on machine B and Ball pen re quires 10 hr on

machine B and maximum available time of machine B is 280 hr.

2nd Constraint is

͵ݔͳͲݕ ʹͺͲ

Here the production of Ball pen and Ink pen can not be negativ e:

we have Non-Negative constraints as ݔͲ ǡݕͲ

Hence the required formulation of LPP is as follows:

Max ൌ ૠ

S u b j e c t t o ,

ૡ

ǡ

Example 10.. :

In a workshop 2 models of agricu lture tools are manufactured ܣଵ and ܣଶ. Each ܣଵ

requires 6 hours for I processing and 3 hours for II processing . Model ܣଶ requires

2 hours for I processing and 4 hours for II processing. The wor kshop has 2 first

processing machines and 4 second processing machine each machin e of I

processing units works for 50 hrs a week. Each machine in II pr ocessing units munotes.in

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145Chapter 10: Linear Programmingworks for 40 hrs a week. The workshop gets Rs. 10/- profit on ܣଵ and Rs. 14/- on

ܣଶ on sale of each tool. Determine the maximum profit that the wo rk shop get by

allocating production capacity on production of two types of to ol A and B

Solution:

Decision Variables :

1. Let the number of units of type ܣଵ model tools be ݔ .

2. Let the number of units of type ܣଶ model tools be ݕ .

Objective function :

The obMective of the workshop is to obtain maximum profit by al locating his

production capacity between ܣଵand ܣଶ and 10 Rs. per unit profit on model ܣଵ and

ͳͶ Rs. on model ܣଶ

ܼൌͳ Ͳ ݔͳ Ͷ ݕ

Constraints :

1. For processing of ܣଵ tool and ܣଶ tools require ʹൌͺ hrs in I Processing

unit ൌ ݔʹ ݕ

2. For processing of ܣଵtype of tools and ܣଶ type requires ͵Ͷൌ hrs in II

processing unit ൌ͵ ݔͶ ݕ

Total machine hours available in I processing unit ʹൈͷͲ ൌ ͳͲͲ hrs per week.

Total machine hours availab le in II processing unit ͶൈͶͲ ൌ ͳͲ hrs per week.

Considering the time constraint , the constrain function can be written in the

following way:

ݔ ʹݕ ͳͲͲ

͵ݔ Ͷݕ ͳͲ

Non-negative constraint :

There is no possibility of negative production in the workshop

The non-negative function will be

ݔͲ ǡݕͲ

Mathematical form of the production of 2 types of tools in the work shop to

maximize profits under given constraints will be in the followi ng way. munotes.in

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146NUMERICAL AND STATISTICAL METHODS Maximi]e ൌ

S u b j e c t t o ,

, ǡ

Example 10..4 :

Diet for a sick person must contain at least 400 units of vitam ins, 500 units of

minerals and 300 calories. Two foods ܨଵ and ܨଶ c o s t R s . 2 a n d R s . 4 p e r u n i t

respectively. Each unit of food ܨଵ c o n t a i n s 1 0 u n i t s o f v i t a m i n s , 2 0 u n i t o f

minerals and 15 calories, whereas each unit of food ܨଶ contains 25 units of

vitamins, 10 units of minerals and 20 calories. Formulate the L .P.P. to satisfy sick

person’s requirement at minimum cost.

Solution :

After reading this carefully, Tabulation form of give data: Micronutrients Minimum units requirement vitamins ͳͲʹͷͶͲͲminerals ʹͲͳͲͷͲͲcalories ͳͷʹͲ͵ͲͲCost ʹͶDecision Variables :

Let ݔ for ܨଵand ݕ for ܨଶ

Objective function :

We have to find minimum the cost for a diet hence the obMective function in terms

of decision variables is

Minimize ܼൌʹ ݔͶ ݕ

Constraints :

First constraints : ͳͲݔ ʹͷݕ ͶͲͲ

S e c o n d C o n s t r a i n t s : ʹͲݔ ͳͲݕ ͷͲͲ

T h i r d c o n s t r a i n t s : ͳͷݔ ʹͲݕ ͵ͲͲ

Non-negative constraint

T h e r e i s n o p o s s i b i l i t y o f n e g a t i v e f o o d q u a n t i t y o f t h e d i e t

hence, ݔͲ ǡݕͲ munotes.in

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147Chapter 10: Linear ProgrammingMathematical formulation of LPP can be written as

Minimi]e ൌ

S u b j e c t t o ,

, ǡ

Example 10..5 :

A garden shop wishes to prepare a supply of special fertilizer at a minimal cost by

mixing two fertilizer, A and B. The mixture is contains : at le ast 45 units of

phosphate, at least 36 units of nitrate at least 40 units of am monium. Fertilizer A

cost the shop Rs 0.97 per Kg. fertilizer B cost the shop Rs.1.8 9 per Kg. Fertilizer

A contains 5 units of phosphate and 2 units of nitrate and 2 un its of ammonium,

fertilizer B contains 3 units of phosphate and 3 units of nitra te and 5 units of

ammonium. How many pounds of each fertilizer should the shop us e in order to

minimum their cost "

Solution:

After reading this carefully, Tabulation form of give data: Contains Fertili]er type Minimum units requirement A B Phosphate ͷ͵ͶͷNitrate ʹ͵͵Ammonium ʹͷͶͲCost ͲǤͻͳǤͺͻെDecision Variables :

Let ݔ for A and ݕ for B

Objective function :

We have to find minimum the cost , Hence the obMective function in terms of

decision variables is

Minimize ܼൌͲ Ǥ ͻ ݔͳ Ǥ ͺ ͻ ݕ

munotes.in

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148NUMERICAL AND STATISTICAL METHODSConstraints :

First constraints : ͷݔ ͵ݕ Ͷͷ

Second Constraints : ʹݔ͵ݕ ͵

T h i r d c o n s t r a i n t s : ʹݔͷݕ ͶͲ

Non-negative constraint

T h e r e i s n o p o s s i b i l i t y o f n e g a t i v e s u p p l y o f f e r t i l i z e r

hence, ݔͲ ǡݕͲ

Mathematical formulation of LPP can be written as

Minimi]e ൌ Ǥૢ ૠ Ǥૡ ૢ

S u b j e c t t o ,

,

ǡ

Example 10..6 :

A printing company prints two types of magazines A and B the co mpany earns

Rs. 25 and Rs. 35 on each copy of magazines A and B respectivel y. The

magazines are processed on three machines. Magazine A requires 2 hours on

machine I, 4 hours on machine II and 2 hours on machine III. Ma gazine B

requires 3 hours on machine I, 5 hours on machine II and 3 hour s on machine III.

Machines I, II and III are available for 35, 50 and 70 hours pe r week respectively

formulate the L.P.P. so as to maximize the total profit of the company.

Solution :

Tabulation form of give data: Machine Maga]ine Maximum availability A B I ʹ͵͵ͷII ͶͷͷͲIII ʹ͵Ͳ

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149Chapter 10: Linear ProgrammingDecision Variables :

Let ݔ number of copies of magazine A and ݕ number of copies of magazine B to

be printed to get maximum profit.

Objective function :

We have to Maximize the Profit, Hence the obMective function in terms of

decision variables is

M a x i m i z e ܼൌʹ ͷ ݔ͵ ͷ ݕ

Constraints :

First constraints : ʹݔ ͵ݕ ͵ͷ

Second Constraints : Ͷݔͷݕ ͷͲ

T h i r d c o n s t r a i n t s : ʹݔ͵ݕ Ͳ

Non-negative constraint

There is no possibility of negative production of magazines

The non-negative function will be

ݔͲ ǡݕͲ

Mathematical formulation of LPP can be written as

Maximi]e ൌ

S u b j e c t t o ,

ૠ

ǡ

Example 10..7 :

A food processing and distributing units has 3 production units A , B , C i n t h r e e

different parts of a city. They have five retails out lest in t he city P, 4, R, S and T

to which the food products are transported regularly. Total sto ck available at the

production units is 500 units which is in the following ways : A 200 units

B 120 units and C 180 units. Requirement at the retails outlet s of the industry

are: A 125, B 150, C 100, D 50, E 75. Cost of transportation of products from

different production centers t o different retail outlets is in the following way: munotes.in

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150NUMERICAL AND STATISTICAL METHODS P 4 R S T A 2 12 8 5 6 B 6 10 10 2 5 C 12 18 20 8 9 How the industry can minimize the cost on transportation of pro ducts. Formulate

the linear programming problem.

Solution :

The obMective of the industry is to minimize the possible cost on transportation

Let ܼ be the obMective function.

Let ݔଵǡݔଶǡݔଷǡݔସ and ݔହ are decision variables

Tabulation form of the given data Production

centers Decision

variables Retails out lets No. of units can

be supplied A ݔଵ2 12 8 5 6 200 B ݔଶ6 10 10 2 5 120 C ݔଷ12 18 20 8 9 180 Units of demand 125 150 100 50 75 500

Objective function :

The obMective is to minimize the cost

Minimize cost ൌʹ ݔଵଵͳʹݔଵଶͺݔଵଷͷݔଵସݔଵହݔଶଵͳͲݔଶଶ

ͳͲݔଶଷʹݔଶସͷݔଶହͳʹݔଷଵͳͺݔଷଶʹͲݔଷଷͺݔଷସͻݔଷହ

Constraint Function :

Supply constraint,

In the problem requirement at the retail out lets and the suppl y at the production

units is the same. therefore supply constraint will be

ܣൌݔଵଵݔଵଶݔଵଷݔଵସݔଵହൌʹ Ͳ Ͳ

ܤൌݔଶଵݔଶଶݔଶଷݔଶସݔଶହൌͳ ʹ Ͳ

ܥൌݔଷଵݔଷଶݔଷଷݔଷସݔଷହൌͳ ͺ Ͳ munotes.in

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151Chapter 10: Linear ProgrammingDemand Constraints :

P ݔଵଵݔଵଶݔଵଷൌͳ ʹ ͷ

4 ݔଶଵݔଶଶݔଶଷൌͳ ͷ Ͳ

R ݔଷଵݔଷଶݔଷଷൌͳ Ͳ Ͳ

S ݔସଵݔସଶݔସଷൌͷ Ͳ

T ݔହଵݔହଶݔହଷൌ ͷ

N o n - N e g a t i v e f u n c t i o n :

ݔଵଵǡݔଵଶǡݔଵଷǡݔଵସǡݔଵହͲ

10.4 Graphical Method

Graphical method, the inequality constraints are taken as equal ities. Each equality

constraints is drawn on the graph paper which forms a straight line. Lines are

drawn equal to the constrains. Then the region which satisfies all inequality is

located. this region is known as feasible region. Solution dete rmine with regard to

this region is called the feasible solution. Accordingly to Had ley if an optimum

(maximum or minimum) value o f a linear programming problem exit s then it

must correspond to one of the corner points of the feasible reg ion Feasible region

corresponding to a linear program ming problem can be located by constructing

graph as given below.

Steps for solving L.P.P. graphically

1. Formulate the mathematical linear programming problem. there will be 2

variables ݔ and ݕ .

2. Since both ݔ and ݕ are non negative, graphic solution will be restricted to

the first quadrant.

. Choose an appropriate scale for ݔ and ݕ axis.

4. Each inequality in the constrain ts equation can be written as e quality.

Example. ݔݕ Ͳ them make it ݔݕൌ Ͳ

5. Given any arbitrary value to one variable and get the value of other variable

by solving the equation. Similarly given another arbitrary valu e to the

variable and find the corresponding value of the other variable . Example

ݔݕൌ at any point on ݔ axis ݕwill be Ͳthe ݔݕൌ and ݔൌ then

at any point on ݕ axis ݔ will be zero ݕൌͷ, thus ሺͲǡͷሻ is a point on ݔ axis

and ሺͲǡͷሻ is a point on ݕ axis that satisfies the equation. munotes.in

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152NUMERICAL AND STATISTICAL METHODS6. Now plot these two sets of values connect these points by strai ght line. That

divides the first quadrant into two parts. Since the constraint s is an

inequality, one of the two sides satisfies inequality

7. Repeat these steps for every constraints stated in the linear p rogramming

problem.

. There forms a common area called feasible area.

. For greater than or greater than equal to constraints, the feas ible region will

be the area which lies above the constrains.

10. For less than or less than equal, the area is below these lines .

Example 10.4.1 :

Solve the following linea r programming by graphical method.

M a x i m i z e ܼൌͷ ݔ ݕ

S u b M e c t t o , ʹݔͶݕ ͳ

͵ݔݕ ͳʹ

͵ݔ͵ݕ ʹͶ , ݔͲ ǡݕͲ

Solution :

By converting the inequality equation to equality equation we g et:

ʹݔͶݕ ൌ ͳ ...(equation 1)

͵ݔݕ ൌ ͳʹ . . . ( e q u a t i o n 2 )

͵ݔ͵ݕ ൌ ʹͶ . . . ( e q u a t i o n 3 )

Equation (1)

ʹݔ Ͷݕ ൌ ͳ , When ݔൌͲ, then ݕൌଵ

ସൌͶ, When ݕൌͲ then ݔൌଵ

ଶൌͺ, ݔ08ݕ40Equation (2)

͵ݔ ݕ ൌ ͳʹ ,When ݔൌͲ then ݕൌͳ ʹ, When ݕൌͲ then ݔൌଵଶ

ଷൌͶ, ݔ04ݕ120

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153Chapter 10: Linear ProgrammingEquation (3)

͵ݔ ͵ݕ ൌ ʹͶ ,When ݔൌͲ then ݕൌଶସ

ଷൌͺ, When ݕൌͲ then ݔൌଶସ

ଷൌͺ, ݔ08ݕ80By above equation coordinates we can draw the straight lines to obtain feasible

area

Thus the feasible region is OABC

Where ܣሺͲǡͶሻ, ܤሺ͵ǤʹǡʹǤͶሻ , and ܥሺͶǡͲሻ

Consider, ܼൌͷ ݔ ݕ

at ܣሺͲǡͶሻ, ܼൌͷሺͲሻሺͶሻൌʹ Ͷ

at ܤሺ͵ǤʹǡʹǤͶሻǡܼൌͷሺ͵ǤʹሻሺʹǤͶሻൌ͵ Ͳ Ǥ Ͷ

at ܥሺͶǡͲሻ, ܼ ൌ ͷሺͶሻሺͲሻൌʹ Ͳ

Thus ܼ is maximum at point ܤ ,Thus the solution is ݔൌ͵ Ǥ ʹ and ݕൌʹ Ǥ Ͷ

Example 10.4.2 :

Solve LPP graphically, Maximum ܼൌͳ Ͳ ݔͷ ݕ

S u b M e c t t o , ݔݕͷ ,

ʹݔݕ ,

ݔͲ ǡݕͲ

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154NUMERICAL AND STATISTICAL METHODSSolution : C o n v e r t i n g t h e g i v e n c o n s t r a i n t s i n t o e q u a t i o n s ( o r i g n o r e t h e

inequality to find the co-ordinates for sketching the graph) we get,

ݔݕൌͷ --- Equation (1)

ʹݔ ݕ ൌ - - - E q u a t i o n ( 2 )

Consider the equation (1)

ݔݕൌͷ , When ݔൌͲ then ݕൌͷ and ݕൌͲ then ݔൌͷ ݔͲͷݕͷͲ

Plot ሺͲǡͷሻ and ሺͷǡͲሻ on the graph. Join the points by a straight line and shade the

region represented by, ݔݕͷ

Then, Consider the equation (2)

ʹݔ ݕ ൌ , When ݔൌͲ then ݕൌ and ݕൌͲ then ݔൌ͵ ݔͲ͵ݕͲPlot ሺͲǡሻ and ሺ͵ǡͲሻ on the graph. Join the points by a straight line and shade the

region represented by, ʹݔݕ

From the graph we get the feasible region ABCD is a feasible re gion where

ܦሺͲǡͲሻǡܣሺͲǡͷሻǡܤሺͳǡͶሻǡܥሺ͵ǡͲሻ .

Consider the value of ܼ at different values of corner points of feasible region.

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155Chapter 10: Linear ProgrammingCorner point ܼൌ ͳ Ͳ ݔͷ ݕ ܦͲܣʹͷܤ͵Ͳܥ͵Ͳ

Here ܼ attained maximum value at two points which is ܤ and ܥ

in this case any point which lies on set ܥܤ is also another optimal solution

There are infinitely many optimal solution.

Example 10.4. :

Solve the following LPP graphically

Min ܼൌʹ ݔ͵ ݕ

SubMect to, ݔݕ͵

ʹݔ ʹݕ ͳͲ

ݔͲ ǡݕͲ

Solution :

Converting the given constraints in to equations (or ignore the inequality to find

the co-ordinates for ske tching the graph) we get,

ݔݕൌ͵ --- Equation (1)

ʹݔ ʹݕ ൌ ͳͲ - - - E q u a t i o n ( 2 )

Consider the equation (1) ݔݕൌ͵ , When ݔൌͲ then ݕൌ͵ and ݕൌͲ

then ݔൌ͵ ݔͲ͵ݕ͵Ͳ

Plot ሺͲǡ͵ሻ and ሺ͵ǡͲሻ on the graph. Join the points by a straight line and shade the

region represented by, ݔݕ͵

Then, Consider the equation (2)

ʹݔ ʹݕ ൌ ͳͲ , When ݔൌͲ then ݕൌͷ and ݕൌͲ then ݔൌͷ ݔͲͷݕͷͲmunotes.in

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156NUMERICAL AND STATISTICAL METHODSPlot ሺͲǡͷሻ and ሺͷǡͲሻ on the graph. Join the points by a straight line and shade the

region represented by, ʹݔʹݕ ͳͲ

This LPP does not have common feasible region and hence no opt imum

solution.

Example 10.4.4 :

Consider a calculator company wh ich produce a scientific calcul ator and graphing

calculator. long-term proMection indicate an expected demand of a t l e a s t 1 0 0 0

scientific and 800 graphing calculators each month. Because of limitation on

production capacity, no more than 2000 scientific and 1700 grap hing calculators

can be made monthly. To satisfy a supplying contract a total of a t l e a s t 2 0 0 0

calculator must be supplied each month. if each scientific calc ulator sold result in

Rs. 120 profit and each graphing calculator sold produce Rs. 15 0 profit, how

many of each type of calculator should be made monthly to maxim ize the net

profit "

Solution :

Decision variables

Let ݔ is the number of scientific calculators produced and ݕ is the number of

graphing calculators produced.

Objective function

We have to find maximum profit. Hence obMective function in a t erms of decision

variable is,

Maximize ܼ ൌ ͳʹͲݔͳͷͲݕ

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157Chapter 10: Linear ProgrammingConstraints

at least 1000 scientif ic calculators : ݔͳ Ͳ Ͳ Ͳ

a t l e a s t 8 0 0 g r a p h i n g c a l c u l a t o r s : ݕͺ Ͳ Ͳ

n o m o r e t h a n 2 0 0 0 s c i e n t i f i c c a l c u l a t o r s : ݔʹ Ͳ Ͳ Ͳ

no more than 1700 graphing calculators : ݕͳ Ͳ Ͳ

a t o t a l o f a t l e a s t 2 0 0 0 c a l c u l a t o r s : ݔݕʹ Ͳ Ͳ Ͳ

and non-negative constraints : ݔ Ͳǡݕ Ͳ

Hence the formulation of LPP can be written as

M a x i m i z e ܼൌͳ ʹ Ͳ ݔͳ ͷ Ͳ ݕ

S u b M e c t t o , ͳͲͲͲ ݔ ʹͲͲͲ

ͺͲͲ ݕ ͳͲͲ

ݔݕʹͲͲͲ

ݔͲ ǡݕͲ

Consider, ݔݕʹ Ͳ Ͳ Ͳ

if ݔൌͲ then ݕൌʹ Ͳ Ͳ Ͳ and when ݕൌͲ then ݔൌʹ Ͳ Ͳ Ͳ

The point are ܣሺͲǡʹͲͲͲሻ and ܤሺʹͲͲͲǡͲሻ

From the graph CDEFG is feasible region

To find the value of point G, Consider ݔൌͳ Ͳ Ͳ Ͳ and ݔݕൌʹ Ͳ Ͳ Ͳ ǡ

ݕ֜ൌͳ Ͳ Ͳ Ͳ , Point G is (1000,1000)

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158NUMERICAL AND STATISTICAL METHODSTo find the value of point F, Consider ݕൌͺ Ͳ Ͳ in ݔݕൌʹ Ͳ Ͳ Ͳ ǡ

ݕ֜ൌͳ ʹ Ͳ Ͳ , Point G is (1200,800)

Here CDEFG is feasible region.

Where ܥሺͳͲͲͲǡͳͲͲ ሻǡܦሺʹͲͲͲǡͳͲͲ ሻǡܧሺʹͲͲͲǡͺͲͲ ሻǡܨሺͳʹͲͲǡͺͲͲ ሻ and

ܩሺͳͲͲͲǡͳͲͲͲሻ

Now, ܼൌͳ ʹ Ͳ ݔͳ ͷ Ͳ ݕ

at ܥሺͳͲͲͲǡͳͲͲ ሻǡ ܼൌ͵ ͷ Ͳ Ͳ Ͳ

at ܦሺʹͲͲͲǡͳͲͲ ሻǡ ܼ ൌ ͶͻͷͲͲͲ

at ܧሺʹͲͲͲǡͺͲͲ ሻǡܼ ൌ ͵ͲͲͲͲ

at ܨሺͳʹͲͲǡͺͲͲ ሻ,ܼ ൌ 264000

at ܩሺͳͲͲͲǡͳͲͲͲ ሻǡܼ ൌ ʹͲͲͲͲ

ܼis maximum at point ܦሺʹͲͲͲǡͳͲͲሻ

The maximum value of ͳʹͲݔ ͳͷͲݕ is 495000 at ሺʹͲͲͲǡͳͲͲሻ .

2000 scientific and ͳͲͲ graphing calculators should be made monthly to

maximize the net profit.

Example 10.4.5 :

Solve the following LPP graphically,

M i n i m i z e ܼൌʹ ͷ ݔͳ Ͳ ݕ

S u b M e c t t o , ͳͲݔʹݕ ʹͲ

ݔʹݕ , ݔͲ ǡݕͲ

Solution :

Consider the equation ͳͲݔ ʹݕ ൌ ʹͲ , when ݔൌͲ then ݕൌͳ Ͳ and when ݕൌͲ

then ݔൌʹ Ǥ

Consider the for 2nd equation ݔʹ ݕൌ , when ݔൌͲ then ݕൌ͵ and when ݕൌ

Ͳ then ݔൌ Ǥ

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159Chapter 10: Linear ProgrammingPlotting the graph

From the graph we get the feasible region where ܣሺͲǡͳͲሻǡܤሺͳǤͷǡʹǤʹʹሻǡܥሺǡͲሻǡ .

For Minimize ܼ ʹͷݔͳͲݕ point ܼൌ ʹ ͷ ݔͳ Ͳ ݕ ܣͳͲͲܤͳǤʹܥͳͷͲ Minimum ܼൌ ͳ Ǥ ʹ at point ܤ i.e. ݔൌͳ Ǥ ͷ and ݕൌʹ Ǥ ʹ ʹ

10.5 Summary

x The obMective of this course is to present Linear Programming. Activities

allow to train modeling practical problems in Linear programmin g. Various

tests help understanding how to solve a linear program

x Linear programming is a mathematical technique for solving cons trained

maximization and minimization problems when there are many cons traints

and the obMective function to be optimized as well as the const raints faced

are linear (i.e., can be represented by straight lines). Linear p r o g r a m m i n g

has been applied to a wide variety of constrained optimization problems.

Some of these are: the selection of the optimal production proc ess to use to

produce a product, the optimal product mix to produce, the leas t-cost input

combination to satisfy some minimum product requirement, the ma rginal

contribution to profits of the various inputs, and many others.

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160NUMERICAL AND STATISTICAL METHODS10.6 References

Following books are recommended for further reading :-

• N u m e r i c a l M e t h o d s f o r E n g i n e e r s - Steven C. Chapra, Raymond P. Canale

Tata Mc Graw Hill.

x 4uantitative Techniques - Dr. C. Satyadevi S. Chand.

10.7 Exercise

4. (1) A firm manufactures headache pills in two size A and B. size A contains 3

grains of aspirin, 6 grains of bicarbonate and 2 grain of codei ne. Size B contains 2

grains of aspirin, 8 grains of bicarbonate and 10 grain of code ine. It is found by

users that it requires at least 15 grains of aspirin, 86 grains of bicarbonate and 28

grain of codeine of providing immediate effect. it is required to determine the

least number of pills a patient should take to get immediate re lief. Formulate the

problem as a standard LPP.

4. (2) An animal feed company must produce 200lbs of mixture containin g the

ingredients A and B. A cost Rs. 5 per Lb. And B cost Rs. 10 per lb. Not more

than 100 lbs. of A can used and minimum quantity to be used for B is 80lbs. Find

how much of each ingredient should be used if the company wants to minimize

the cost. Formulate the problem as a standard LPP.

4. () A painter make two painting A and B. He spends 1 hour for drawi ng and 3

hours for coloring the painting A and he spends 3 hours for dra wing and 1 hour

for coloring the painting B. he can spend at most 8 hours and 9 hours for drawing

and coloring respectively. The profit per painting of Type A is Rs. 4000 and that

of type B is Rs. 5000. formulate as LPP to maximize the profit

4. (4) A gardener wanted to prepare a pesticide using two solutions A and B. the

cost of 1 liter of solution A is Rs. 2 and the cost of 1 liter of solutions B is Rs. 3.

He wanted to prepare at least 20 liter of pesticide. The qualit y of solution A

available in a shop is 12 liter and solution B is 15 liter. How m a n y l i t e r o f

pesticide the gardener should prepare so as to minimize the cos t " formulate the

LPP

4. (5) A bakery produce two kinds of biscuits A and B, using same ingr edients

L1 and L2. The ingredients L1 and L2 in biscuits of type A are in the ratio 4:1

and in biscuits of type B are in the ratio 9:1. The Profit per Kg for biscuits of type

A and B is Rs. 8 per Kg and Rs. 10 per Kg respectively. The bak ery had 90 Kg of

L1 and 20 Kg of L2 in stock. How many Kg of biscuits A and B be produced to

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161Chapter 10: Linear Programming4. (6) Solve graphically th e following LPP

Maximize ܼൌʹ ݔ͵ ݕ

subMect to, ݔʹ ݕ

ʹݔ െͷݕ ͳ , ݔͲ ǡݕͲ

4. (7) Solve graphically th e following LPP

Maximize ܼൌʹ ݔͷ ݕ

subMect to, Ͷݔ ʹݕ ͺͲ

ʹݔ ͷݕ ͳͺͲ , ݔͲ ǡݕͲ

4. () Solve graphically th e following LPP

Maximize ܼൌʹ ݔݕ

subMect to, Ͷݔ െݕ ͵

ʹݔ ͷݕ , ݔͲ ǡݕͲ

4. () Solve graphically th e following LPP

Maximize ܼൌͶ Ͳ Ͳ ݔͷ Ͳ Ͳ ݕ

subMect to, ݔ͵ ݕͺ

͵ݔ ݕ ͻ , ݔͲ ǡݕͲ

4. (10) Solve graphically the following LPP

Maximize ܼൌͷ Ͳ Ͳ Ͳ ݔͶ Ͳ Ͳ Ͳ ݕ

subMect to, ݔ Ͷݕ ʹͶ

ݔʹ ݕ

െݔ ݕ ͳ

ݕʹ, ݔͲ ǡݕͲ

4. (11) Solve graphically the following LPP

Minimize ܼൌ͵ Ͳ ݔͷ Ͳ ݕ

subMect to, ͵ݔ Ͷݕ ͵ͲͲ

ݔ͵ ݕʹ ͳ Ͳ , ݔͲ ǡݕͲ

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162NUMERICAL AND STATISTICAL METHODSUNIT 5

11

RANDOM VARIABLES

Unit Structure

11.0 ObMectives

11.1 Introduction

11.2 Random Variable (R.V.) – Discrete and Continuous

1 1 . 2 . 1 D i s c r e t e R a n d o m V a r i a b l e :

11.2.2 Continuous Random Variable:

11.2.3 Distinction between continuous random variable and disc rete random

variable:

11.3 Probability Distributions of Discrete Random Variable

1 1 . 3 . 1 P r o b a b i l i t y M a s s F u n c t i o n ( p . m . f . )

11.4 Probability Distributions of Continuous Random Variable

1 1 . 4 . 1 P r o b a b i l i t y D e n s i t y F u n c t i o n ( p . d . f . )

11.5 Properties of Random variable and their probability distri butions

11.6 Cumulative Distribution Function (c.d.f.)

1 1 . 5 . 1 C u m u l a t i v e D i s t r i b u t i o n F u n c t i o n ( c . d . f . ) f o r D i s c r e t e Random

Variable

1 1 . 5 . 2 C u m u l a t i v e D i s t r i b u t i o n F u n c t i o n ( c . d . f . ) f o r C o n t i n u o u s Random

Variable

11.7 Properties of Cumulative Distribution Function (c.d.f.)

11.8 Expectation or Expected Value of Random Variable

11.8.1 Expected Value of a Discrete Random Variable :

11.8.2 Expected Value of a Continuous Random Variable :

1 1 . 8 . 3 P r o p e r t i e s o f E x p e c t a t i o n

11.9 Expectation of a Function

11.10 Variance of a Random Variable

1 1 . 1 0 . 1 P r o p e r t i e s o f V a r i a n c e

11.11 Summary

11.12 Reference

11.13 Unit End Exercise munotes.in

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163Chapter 11: Random Variables11.0 Objectives

After going through this unit, you will be able to:

x Understand the concept of ra ndom variables as a function from sample

space to real line

x Understand the concept of probability distribution function of a discrete

random variable.

x Calculate the probabilities for a discrete random variable.

x Understand the Probability Distribution of Random Variable

x Understand the probability mass function and probability densit y function

x Understand the properties of random variables.

x Make familiar with the concept of random variable

x Understand the concept of cumul ative distribution function.

x Understand the expected value and variance of a random variabl e with it s

properties Andrey Nikolaevich Kolmogorov (Russian:

25 April 1903 – 2 0 O c t o b e r 1 9 8 7 ) w h o

advances many scientific fields.

In probability theory and related fields,

a stochastic or random process is

a mathematical obMect usually defined as

a family of random variables. A stochastic

process may involve several related random

variables. A quotation attributed to

Kolmogorov is “ Every mathematician

believes that he is ahead over all others .

The reason why they don’t say this in

public, is because they are intelligent

people.” 11.1 Introduction

Many problems are concerned with values associated with outcom es of random

experiments. For example we select five items randomly from the basket with

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164NUMERICAL AND STATISTICAL METHODSknown proportion of the defectives and put them in a packet. Now we want to

know the probability that the packet contain more than one defe ctive. The study of

such problems requires a concept of a random variable. Random v ariable is a

function that associates numerical values to the outcomes of ex periments.

In this chapter, we considered discrete random variable that is random variable

which could take either finite or countably infinite values. W hen a random variable

; is discrete, we can assign a positive probability to each val ue that ; can take and

determine the probability distribution for ;. The sum of all th e probabilities

associate with the different values of ; is one. However not al l experiments result

in random variables that ate discrete. There also exist random variables such as

height, weights, length of life an electric component, time tha t a bus arrives at a

specified stop or experimental laboratory error. Such random va riables can assume

infinitely many values in some interval on the real line. Such variables are called

continuous random variables. If we try to assign a positive pro bability to each of

these uncountable values, the probabilities will no longer sum to one as was the

case with discrete random variable.

11.2 Random Variable (R.V.) – Discrete and Continuous

11.2.1 Discrete Random Variable:

Sample space S or Ω contains non -numeric elements. For example, in an

experiment of tossing a coin, Ω = { H, T } . However , in practice it is easier to deal

with numerical outcomes. In turn, we associate real number with each outcome.

For instance, we may call H as 1 and T as 0. Whenever we do th is, we are dealing

with a function who se domain is the sample space Ω and whose range is the set of

real numbers . Such a function is called a random variable.

Definition 1 : Random variable : Let S / Ω be the sample space corresponding to

the outcomes of a random experiment . A function X: S → R (where R is a set of

real numbers) is called as a random variable.

Random variable is a real valued mapping. A function can eith er be one-to-one or

many-to-one correspondence. A random variable assigns a real nu mber to each

possible outcome of an experiment. A random variable is a funct ion from the

sample space of a random experiment (domain) to the set of real numbers (co-

domain).

Note: Random variable are denoted by capital letters ;,<, = et c.., where as the

values taken by them are denoted by corresponding small letters x, y, z etc. munotes.in

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165Chapter 11: Random VariablesDefinition 2: Discrete random variable : A random variable ; is said to be di screte

if it takes finite or countably infinite number of possible val ues. Thus discrete

random variable takes only isolated values.

Example1:

What are the values of a random variable ; would take if it wer e defined as number

of tails when two coins are tossed simultaneously"

Solutions:

Sample Space of the experiment (Tossing of two coins simultaneo usly) is,

S / Ω = { TT, TH, HT, HH }

Let ; be the number of tails obtained tossing two coins.

X : Ω → R

;(TT) 2, ;(TH) 1, ;(HT) 1, ;(HH) 0

Since the random variable ; is a number of tails, it takes thre e distinct values

^ 0, 1, 2 `

Remark: Several random variables can be defined on the same sample spac e Ω.

For example in the Example1, one can define < Number of heads or =

Difference between number of heads and number of tails.

Following are some of the examples of discrete variable.

a) Number of days of rainfalls in Mumbai.

b) Number of patients cured by using certain drug during pandemic.

c) Number of attempts required to pass the exam.

d) Number of accidents on a sea link road.

e) Number of customers arriving at shop.

f) Number of students attending class.

Definition : Let ; be a discrete random variable defined on a sample space Ω.

Since Ω contains either finite of countable infinite elements, and X is a function on

Ω, X can take either finite or countably infinite values. Suppose ; takes values x 1,

x2 x3,…. then the set ^ x 1, x2 x3,…} is called the range set of ;.

In Example1, the range set of ; Number of tails is ^ 0,1,2`

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166NUMERICAL AND STATISTICAL METHODS11.2.2 Continuous Random Variable:

A sample space which is finite or countably infinite is called as denumerable of

countable. If the sample space is not countable then it is call ed continuous. For a

continuous sample space Ω we can not have one to one correspond ence between Ω

and set of n atural numbers { 1,2,…}

Random variable could also be such that their set of possible v alues is uncountable.

Examples of such random variables are time taken between arriva ls of two vehicles

at petrol pump or life in hours of an electrical component.

In general if we define a random variable ;( ω) as a real valued function on domain

Ω. If the range set of X (ω) is continuous, the random var iable is continuous. The

range set will be a subset of real line.

Following are some of the examp les of continuous random variabl e

a) Daily rainfall in mm at a particular place.

b) Time taken for an angiography operation at a hospital

c) Weight of a person in kg.

d) Height of a person in cm.

e) Instrumental error (measured in suitable units) in the measurem ent.

11.2. Distinction between continuous random variable and discr ete random

variable:

1) A continuous random variable takes all possible values in a ran ge set. The ser

is in the form of interval. On the other hand discrete random v ariable takes

only specific or isolated values.

2) Since a continuous random variable takes uncountably infinite v alues, no

probability mass can be attached to a particular value of rando m variable ;.

Therefore, P(; x) 0, for all x. However in case of a discre te random

variable, probability mass is attached to individual values tak en by random

variable. In case of continuous random variable probability is attached to an

interval which is a subset of R.

11. Probability Distributions of Discrete Random Variable

Each outcome i o f a n e x p e r i m e n t h a v e a p r o b a b i l i t y P ( i) associated with it.

Similarly every value of random variable ; xi is related to the outcome i of an

experiment. Hence, for every value of random variable xi, we have a unique real munotes.in

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167Chapter 11: Random Variablesvalue P( i) associated . Thus, every random variable ; has probability P associated

with it. This function P (; xi) from the ser of all events of the sample space Ω is

called a probability distribution of the random variable.

The probability distribution (or simply distribution) of a rand om variable ; on a

sample space Ω is set of pairs (; xi, P (; xi)) for all xi : א x(Ω), where P (; xi)

is the probability that ; takes the value xi .

Consider the experiment of tossing two unbiased coins simultane ously. ; number

of tails observed in each tossing. Then range set of ; ^ 0, 1 ,2 `. Although we

cannot in advance predict what value ; will take, we can certa inly state the

probabilities with which ; will take the three values 0,1,2. Th e following table

helps to determine such probabilities. Outcome Probability of Outcome Value of ; HH 1/4 0 TH 1/4 1 HT 1/4 1 TT 1/4 2 Following events are associated with the distinct values of ;

(; 0) ֜ ^ HH`

(; 1) ֜ ^ TH,HT`

(; 2) ֜ ^ TT`

Therefore, probabilities of various values of ; are nothing bu t the probabilities of

the events with which the respective values are associated.

P (; 0) ֜ P ^ HH` 1/4

P (; 1) ֜ P ^ TH, HT` 1/4 1/4 1/2

P (; 2) ֜ P ^TT` 1/4

Example2:

A random variable is number of tails when a coin is flipped thr ice. Find probability

distribution of the random variable.

Solution:

Sample space Ω = { HHH , THH, HTH, HHT, TTH, THT, HTT, TTT `

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168NUMERICAL AND STATISTICAL METHODSThe required probability distribution is Value of Random Variable ; xi 0 1 2 Probability P(; x i) 1/8 3/8 3/8 1/8

Example :

A random variable is sum of the numbers that appear when a pair of dice is rolled.

Find probability distribution of the random variable.

Solution:

;(1,1) 2, ;(1, 2) ;(2, 1) 3, ;(1, 3) ;(3, 1) ; ( 2, 2) 4 etc..

The probability distribution is, ; xi 2 4 5 6 7 10 11 12 P(; x i) 1/36 2/36 3/36 4/36 5/36 6/35 5/36 4/36 3/36 2/36 1/36

11..1 Probability Mass Function (p.m.f.)

Let X be a discrete random variable defined on a sample space Ω / S . Suppose

^x1, x2, x3, ……. xn` is the range set of ;. With each of xi, w e a s s i g n a n u m b e r

P(xi) P(; xi) called the probability mass function (p.m.f.) such that,

P (xi) ≥ 0 for i 1, 2, 3 ….,n and (AND)

σሺݔሻൌͳ୬

ୀଵ

The table containing, the value of ; along with the probabiliti es given by

probability mass function (p.m.f.) is called probability distri bution of the random

variable ;.

For example, ; xi ;1 ;2 ….. ….. ;i … …. ;n Total P(; x i) P1 P2 …. …. Pi … … Pn 1 Remark: Properties of a random variable can be studied only in terms o f its p.m.f.

We need not have refer to the underlying sample space Ω, once w e have the

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169Chapter 11: Random VariablesExample 4:

A fair die is rolled and number on uppermost face is noted. Fin d its probability

distribution (p.m.f.)

Solution:

; Number on uppermost face.

Therefore, Range set of ; ^1, 2, 3, 4, 5, 6 `

Probability of each of the element is 1 / 6

The Probability distribution of ; is ; xi 1 2 3 4 5 6 Total P(; x i) 1/ 6 1/ 6 1/ 6 1/ 6 1/ 6 1/ 6 1 Example 5:

A pair of fair dice is thrown and sum of numbers on the uppermo st faces is noted.

Find its probability distribution (p.m.f.).

Solution:

; Sum of numbers on the uppermost faces.

Ω contains 36 elements (ordered pairs)

Range set of ; ^ 2, 3, 4, 5, 6, 7, 8,9, 10, 11, 12 `

Since, ; (1, 1) 2 and ; (6, 6) 12 Value of ; Subset of Ω Pi P (; i) 2 ^ (1,1) ` 1/36 3 ^ (1,2), (2,1) ` 2/36 4 ^ (1,3), (2,2), (3,1) ` 3/36 5 ^ (1,4), (2,3), (3,2), (4,1) ` 4/36 6 ^ (1,5), (2,4), (3,3), (4,2), (5,1) ` 5/36 7 ^ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) ` 6/36 8 ^ (2,6), (3,5), (4,4), (5,3), (6,2) ` 5/36 9 ^ (3,6), (4,5), (5,4), (6,3) ` 4/36 10 ^ (4,6), (5,5), (6,4) ` 3/36 11 ^ (5,6) (6,5) ` 2/36 12 ^ (6,6) ` 1/36 munotes.in

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170NUMERICAL AND STATISTICAL METHODSExample 6:

Let ; represents the difference between the number of heads and the number of

tails obtained when a fair coin is tossed three times. What are the possible values

of ; and its p.m.f."

Solution:

Coin is fair. Therefore, probability of heads in each toss is P H 1/2. Similarly,

probability of tails in each toss is P T 1/2.

; can take values n – 2r where n 3 and r 0, 1, 2, 3.

e.g. ; 3 (HHH), ; 1 (HHT, HTH, THH), ; -1 (HTT, THT, HTT) a n d

; -3 (TTT)

Thus the probability distribution of ; (possible values of ; a nd p.m.f) is ; xi -3 -1 1 3 Total p.m.f. P(; x i) 1/ 8 3/ 8 3/ 8 1/ 8 1 Example 7:

Let ; represents the difference between the number of heads and the number of

tails obtained when a coin is toss ed n times. What are the poss ible values of ; "

Solution:

When a coin is tossed n times, number of heads that can be obta ined are n, n-1,

n-2, ……,2, 1, 0. Corresponding number of tails are 0, 1, 2, ……, n-2, n-1, n.

Thus the sum of number of heads and number of tails must be equ al to number of

trials n.

Hence, values of ; are from n to –n as n, n-2, n- 4, ……, n -2r

where r 0,1, 2, 3, ….., n

Note if n is even ; has one of its value as zero also. However, if n is odd ; has

values -1, 1 but not zero.

11.4 Probability Distributions of Continuous Random Variable

In case of discrete random variable using p.m.f. we get probabi lity distribution of

random variable, however in case of continuous random variable probability mass

is not attached to any particular value. It is attached to an i nterval. The probability

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171Chapter 11: Random VariablesFor example, P (a ; b) varies for different values of a and b. In other words, it

will not be uniform. In order to obtain the probability associa ted with any interval,

we need to take into account the concept of probability density .

11.4.1 Probability Density Function (p.d.f.)

Let ; be a continuous random variable. Function f(x) defined for all real

x א-( ∞, ∞) is called probability density function p.d.f. if for any set B of real

numbers, we get probability,

P ^ ; א B ` ݂ሺݔሻݔ݀

All probability statements about ; can be answered in terms of f (x). Thus,

P { a ≤ X ≤ B} = ݂ሺݔሻݔ݀

Note that probability of a continuous random variable at any pa rticular value is

zero, since

P { X = a} = P { a ≤ X ≤ a} = ݂ሺݔሻݔ݀

ൌͲ

11.5 Properties of Random variable and their probability

distributions

Properties of a random variable can be studied only in terms of its p.m.f . or p.d.f.

We need not have refer to the underlying sample space Ω, once w e have the

probability distribution of random variable. Since the random v ariable is a function

relating all outcomes of a random experiment, the probability d istribution of

random variable must satisfy Axioms of probability. These in c ase of discrete and

continuous random variable are stated as,

Axiom I : Any probability must be between zero and one.

For discrete random variable : 0 ≤ p( xi) ≤ 1

For continuous random variable : For any real number a and b

0 ≤ P { a ≤ x ≤ b } ≤ 1 OR 0 ≤ ݂ሺݔሻݔ݀

≤ 1

Axiom II : Total probability of sample space must be one

For discrete random variable : σሺݔஶ

୧ୀଵሻൌͳ

For continuous random variable : ݂ሺݔሻݔ݀ஶ

ିஶൌͳ

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172NUMERICAL AND STATISTICAL METHODSAxiom III : For any sequence of mutually exclusive events E 1, E2, E3,……… i.e

Ei ∩ EM = Φ for i ≠ j, probability of a union set of events is sum of their indiv idual

probabilities. This axiom can also be written as P(E1 ڂE2) P(E 1) P (E 2) where

E1 and E 2 are mutually exclusive events.

For discrete random variable : ሺڂܑஶ

ܑୀ) σሺ୧ஶ

୧ୀଵሻ

For continuous random variable : ݂ሺݔሻݔ݀

ൌ݂ሺݔሻݔ݀

݂ሺݔሻݔ݀

also,

P (a ≤ x ≤ b ڂ c ≤ x ≤ d) ݂ሺݔሻݔ݀

݂ሺݔሻݔ݀ௗ

Axiom IV : P (Φ) 0

11.6 Cumulative Distribut ion Function (c.d.f.)

It is also termed as Must a distribution function. It is the ac cumulated value of

probability up to a given value of the random variable. Let ; b e a random variable,

then the cumulative distribution function (c.d.f.) is defined a s a function F(a) such

that,

F(a) = P { X ≤ a }

11.6.1 Cumulative Distribution Function (c.d.f.) for Discrete R andom

Variable

Let ; be a discrete random variable defined on a sample space S taking values

^x1, x2,…, xn `

With probabilities p( xi), p(x2), …. p(xn) respectively. Then cumulative distribution

function (c.d.f.) denoted as F(a) and expressed in term of p.m. f. as

F ( a ) σሺݔiሻ

௫iஸ

Note:

1. The c.d.f. is defined for all values of xi € R. However , since the random

variable takes only isolated values, c.d.f. is constant between two successive

values of ; and has steps at the points xiI, i = 1, 2, … , n. Thus, the c.d.f for a

discrete random variable is a step function.

2. F (∞) 1 and F (- ∞) 0 munotes.in

## Page 173

173Chapter 11: Random VariablesProperties of a random variable can be studied only in term of its c.d.f. We need

not refer to the underlying sample space or p.m.f., once we hav e c.d.f. of a random

variable.

11.5.2 Cumulative Distribution Function (c.d.f.) for Continuous Random

Variable

Let ; be a continuous random variable defined on a sample spac e S which p.d.f.

f(x). Then cumulative distribution (c.d.f.) denoted as F (a) and ex pressed in term of

p.d.f. as,

F (a) ݂ሺݔሻݔ݀ஶ

ିஶ

Also, differentiating both sides we get,

ௗ

ௗܨሺܽሻൌ ݂ሺܽሻ, thus the density is the derivative of the cumulative

distribution function.

11.7 Properties of Cumulative Di stribution Function (c.d.f.)

1. F (x) is defined for all x € R , real number.

2. 0 ≤ F (x) ≤ 1

3. F (x) is a non-decreasin g function of x. >if a b, then F (a) ≤ F (b).

4. F (∞) 1 and F (- ∞) 0

Where F(- ∞) =

୶՜ିஶ ሺሻ, F(∞) =

୶՜ஶ ሺሻ

5. Let a and b be two real numbers where a b then using distri bution function,

we can compute probabilities of different events as follows.

i) P (a < X ≤ b) = P[X ≤ b] - P[X ≤ a]

F(b) –F(a)

ii) P (a ≤ X ≤ b) = P[X ≤ b] - P[X ≤ a] + P (X=a)

F(b) –F(a) P(a)

iii) P (a ≤ X < b) = P[X ≤ b] - P[X ≤ a] - P (; b) P (; a)

F(b) –F(a) - P(b) P(a)

iv) P (a < X < b) = P[X ≤ b] - P[X ≤ a] - P (; b)

F(b) –F(a) - P(b) munotes.in

## Page 174

174NUMERICAL AND STATISTICAL METHODSv) P (; ! a) 1 - P[X ≤ a] = 1 - F(a)

vi) P (X ≥ a) = 1 - P[X ≤ a] + P[X=a] = 1 – F(a) P(a)

vii) P (; a) F(a) -

՜ஶܨቀܽെଵ

ቁ

viii) P(; a)

՜ஶቀܽെଵ

ቁ

Example :

The following is the cumulative distribution function of a disc rete random variable.

i) Find the p.m.f of ; ii) P(0 ; 2) iii) P(1 ≤ X ≤ 3) iv) P(-3 < X ≤ 2 ) v) P(1 ≤ X < 1) vi) P(; even) vii) P(; ! 2) viii) P(X ≥ 3) Solution:

i) Since F(x i) σܲj

ୀଵ

F(x i-1) σܲjିଵ

ୀଵ

Pi σܲj

ୀଵ - σܲjିଵ

ୀଵ F(x i) – F(x i-1)

The p.m.f of ; is given by

ii) P(0 ; 2) F(2) –F(0) - P(2) 0.75 – 0.45-0.1 0.2

iii) P(1 ≤ X ≤ 3) = F(3) – F(1) P(1) 0.9-0.650.2 0.45

iv) P(-3 < X ≤ 2 ) F(2) –F(-3) 0.75 – 0.1 0.65 ; xi -3 -1 0 1 2 3 5 8 F(x) 0.1 0.3 0.45 0.65 0.75 0.90 0.95 1.00

; xi -3 -1 0 1 2 3 5 8 F(x) 0.1 0.2 0.15 0.2 0.1 0.15 0.05 0.05 munotes.in

## Page 175

175Chapter 11: Random Variablesv) P(1 ≤ X < 1) = F(1) – F(-1) – P(1) P(-1) 0.65 – 0.3- 0.2 0.2 035

vi) P(; even) P(x 0) P(x 2) P(x 8) 0.15 0.1 0.05 0 .3

vii) P(; ! 2) 1 – F(2) 1 – 0.75 0.25 OR P(x 3) P(x 5) P(x 8) 0.15 0.05 0.05 0.25

viii) P(X ≥ 3) 1 – F(3) P(3) 1- 0.9 0.15 0.25

Example :

A random variable has the following probability distribution Values of ; 0 1 2 3 4 5 6 7 8 P(x) a 3 a 5 a 7 a 9 a 11 a 13 a 15 a 17 a (1) Determine the value of a

(2) Find (i) P(x 3) (ii) P(x d 3) (iii) P(x !7) (iv)P(2 d x d 5) (v) P(2 x 5)

(3) Find the cumulative distribution function of x.

Solution:

1. Since p i is the probability mass function of discrete random variable ;,

We have 6pi 1

? a + 3 a + 5a + 7a + 9a +11a + 13a + 15a + 17a = 1

81 a = 1

a = 1/81

2. (i) P(x 3) P(x 0) P(x 1) P(x 2)

a 3 a 5a

9 a = 9 * (1 / 81) = 1 / 9

(ii) P( x d 3) P (x 0) P(x 1) P(x 2) P(x 3)

a 3 a 5 a 7 a

= 16 a = 16 * (1 / 81) = 16 / 81

(iii) P(x !7) P(x 8) 17 a =17 * (1 / 81)= 17 / 81

(iv) P (2 ≤ x ≤ 5) P(x 2) P(x 3) P(x 4) P(x 5)

5a 7a 9a 11a = 32 a = 32 * (1 / 81) = 32 / 81

(v) P(2 x 5) P(x 3) P(x 4) 7 a 9a = 16a = 16 *

(1 / 81) = 16 / 81 munotes.in

## Page 176

176NUMERICAL AND STATISTICAL METHODS3. The distribution function is as follows: ; x 0 1 2 3 4 5 6 7 8 F(x)

P (;d x) a 4a 9a 16a 25a 36a 49a 64a 81a (or)

F(x) 1

81 4

81 9

81 16

81 25

81 36

81 49

81 64

81 81 1

81

Example 10:

Find the probability between ; 1 and 2 i.e. P( 1 ≤ X ≤ 2 ) for a continuous random

variable whose p.d.f. is given as

݂ሺݔሻൌሺଵ

ݔ ݇ሻ for 0 ≤ x ≤ 3

Solution:

Now, p.d.f must satisfy the probability Axiom II. Thus

݂ሺݔሻݔ݀ஶ

ିஶൌͳ

ቀଵ

ݔ݇ቁݔ݀ ൌ ଷ

ቂ௫మ

ଵଶݔ݇ቃ͵

Ͳ ቂଷమ

ଵଶ͵݇ቃͲൌͳ

12k 1

k ଵ

ଵଶ

Now, P (1 ≤ X ≤ 2) = ݂ሺݔሻݔ݀ ൌଶ

ଵ ቀଵ

ݔଵ

ଵଶቁݔ݀ൌଵ

ଷଶ

ଵ

Example 11:

A continuous random variable whose p.d.f. is given as

݂ሺݔሻൌቄݔ݇ሺʹെݔሻ

Ͳ൏ݔ൏ʹ

݁ݏ݅ݓݎ݄݁ݐ

i) Find k

ii) Find P ( x ଵ

ଶ)

Solution:

i) Now, p.d.f must satisfy the proba bility Axiom II. Thus

݂ሺݔሻݔ݀ஶ

ିஶൌͳ

ݔ݇ሺʹെݔሻݔ݀ ൌଶ

ቈݔ݇2െ௫3ଷʹ

Ͳ 1

k ଷ

ସ munotes.in

## Page 177

177Chapter 11: Random Variablesii)

Now, P ( x ଵ

ଶ) ݂ሺݔሻݔ݀ଵȀଶ

ିஶ

ൌ ଷ

ସݔሺʹെݔሻݔ݀భ

మିஶ ൌቈଷ

ସݔ2െ௫3ଷͳȀʹ

Ͳ ହ

ଵଶ

Example 12:

A random variable is a number of tails when a coin is tossed th ree times. Find

p.m.f. and c.d.f of the random variable.

Solution:

S / Ω = { TTT, HTT, THT, TTH, HHT, HTH,THH, HHH }

n(Ω) = 8 ; xi 0 1 2 3 Total p.m.f. P(; x i) 1/ 8 3/ 8 3/ 8 1/ 8 1 c.d.f F(a) P^ ; x i ≤ a` 1/ 8 4 / 8 7 / 8 1 c.d.f. will be describe as follows:

Example 1:

A c.d.f. a random variable is as follows

Find i) P( X < 3 ) ii) P ( X = 1) F(a) 0 -∞ < a < 0 ଵ

଼ 0 ≤ a < 1 ସ

଼ 1 ≤ a < 2

଼ 2 ≤ a < 3 1 3 ≤ a < -∞ F(a) 0 -∞ < a < 0 ଵ

ଶ 0 ≤ a < 1 ଶ

ଷ 1 ≤ a < 2 ଵଵ

ଵଶ 2 ≤ a < 3 1 3 ≤ a < -∞ munotes.in

## Page 178

178NUMERICAL AND STATISTICAL METHODSSolution:

i) P(X < 3 )

՜ஶܨቀ͵െଵ

ቁൌଵଵ

ଵଶ

ii) P (X =1) P ( X ≤ 1) - P ( X < 1)

F(1) -

՜ஶܨቀ͵െଵ

ቁൌଶ

ଷെଵ

ଶൌଵ

11. Expectation or Expected Value of Random Variable :

Expectation is a very basic concep t and is employed widely in decision theory,

management science, system analysis, theory of games and many o ther fields. It

is one of the most important concepts in probability theory of a random variable.

Expectation of ; is denoted by E( X). The expected value or mathematical

expectation of a random variable ; is the weighted average of the values that ; can

assume with probabilities of its various values as weights. Ex pected value of random

variable provides a central point for the distribution of value s of random variable. So

expected value is a mean or aver age value of the probability di stribution of the random

variable and denoted as ‘μ’ (read as ‘ mew’). Mathematical expecta tion of a random

variable is also known as its arithmetic mean.

E( X)

11..1 Expected Val ue of a Discrete Ra ndom Variable :

If X is a discrete random variable with p.m.f. P( xi), the expectation of X, denoted by

E(X), is defined as,

E ( X) σݔܲכሺݔሻ

ୀଵ Where xi for i = 1, 2, …..n (values of X)

11..2 Expected Value of a Continuous Random Variable :

If X is a discrete random variable with p.d.f. f( x), the expectation of X, denoted by

E(X), is defined as,

E(X) ݂ሺݔሻݔ݀ஶ

ିஶ

11.. Properties of Expectation

1. For two random variable ; and < if E(;) and E(<) exist, E(; <) E(;)

E(<) . This is known as addition theorem on expectation.

munotes.in

## Page 179

179Chapter 11: Random Variables2. For two independent random variable ; and <, E(;<) E(;) .E(<)

provided all expectation exist. This is known as multiplication t h e o r e m o n

expectation.

3. The expectation of a constant is the constant it self. ie E(C) C

4. E(c;) cE(;)

5. E(a;b) aE(;) b

11. Expectation of a Function

Let < g (;) is a function of random variable X, then Y is also a random variable

with the same probability distribution of X.

For discrete random variable X and Y, probability distribution of Y is also P( xi).

Thus the expectation of Y is,

E ( Y) E>(g ( xi)@ σ݃ሺݔሻܲכሺݔ ሻ

ୀଵ

For continuous random variable X and Y, probability distribution of Y is also

f(x). Thus the expectation of Y is,

E( Y) E>(g ( xi)@ ݃ሺݔሻ݂כሺݔሻݔ݀ஶ

ିஶ

Example 14:

A random variable is number of tails when a coin is tossed thre e times. Find

expectation (mean) of the random variable.

Solution:

S / Ω = { TTT, HTT, THT, TTH, HHT, HTH,THH, HHH }

n(Ω) = 8 ; xi 0 1 2 3 p.m.f. P(; xi) 1/ 8 3/ 8 3/ 8 1/ 8 כሺሻ 0 3 / 8 6 / 8 3 / 8

E(X) σݔܲכሺݔሻସ

ୀଵ Where xi for i = 1, 2, …..n (values of X)

E(X) 0 ଷ

଼

଼ଷ

଼ൌଵଶ

଼ൌଷ

ଶ

munotes.in

## Page 180

180NUMERICAL AND STATISTICAL METHODSExample 15:

; is random variable with probability distribution ; xi 0 1 2 p.m.f. P(; xi) 0.3 0.3 0.4 < g(;) 2; 3

Find expected value or mean of < . (i.e. E( Y))

Solution:

< g(;) 2; 3

When ; 0, < """

< 2; 3 2(0) 3 3,

Similarly,

when ; 1, < 5, when ; 2, < 7 ; xi 0 1 2 Y yi 3 5 7 p.m.f. P(Y yi) 0.3 0.3 0.4 E(Y) E>(g ( xi)@ σ݃ሺݔሻܲכሺݔ ሻ

ୀଵ σݕܲכሺݔሻ

ୀଵ

E(Y) 3 uuu

11.10 Variance of a Random Variable

The expected value of ; (i.e. E( X)) provides a measure of central tendency of the

probability distribution. However it does not provide any idea regarding the spread

of the distribution. For this purpos e, variance of a random var iable is defined .

Let ; be a discrete random variable on a sample space S. The va riance of ; denoted

by Var(X) or σ2 (read as ‘sigma square ) is defined as,

Var(X) E >(; - )2@ E >(; – E(;))2@ σሺݔെρሻ2ܲሺݔሻ

ୀଵ

Var(X) σሺݔሻ2ܲሺݔሻെʹρ

ୀଵ σݔܲሺݔሻ

ୀଵ ρ2σܲሺݔሻ

ୀଵ

E(;2) – ʹρ E(;) 2 …………….. munotes.in

## Page 181

181Chapter 11: Random Variables>ܧሺܺሻൌσݔܲሺݔሻ

ୀଵ ݀݊ܽσܲሺݔሻ

ୀଵ ൌͳ@

E(;2) – ʹρ.ρ 2

E(;2) – 2 E(;2) – >E(;)@2

For continuous random variable,

ݎܸܽሺܺሻൌ නሺݔ െρሻ2݂ሺݔሻݔ݀ஶ

ିஶൌන ሺ ݔ2െʹρݔρ2ሻ݂ሺݔሻݔ݀ஶ

ିஶ

ൌන ݔ2݂ሺݔሻݔ݀ஶ

ିஶെනʹ ρ ݂ݔ ሺݔሻݔ݀ න ρ2݂ሺݔሻݔ݀ஶ

ିஶஶ

ିஶ

E ( ;2) – ʹρ E(;) – >E(;)@2

E ( ;2) – ʹρ.ρ 2

E(;2) – 2 E(;2) – >E(;)@2

Since dimensions of variance are square of dimensions of ;, foe comparison, it is

better to take square root of variance. It is known as standard deviation and

denoted be S.D.(X) or σ (sigma)

S.D.= σ = ඥሺሻ

11.10.1 Properties of Variance:

1. Variance of constant is zero. ie Var(c) 0

2. Var(;c) Var ;

N o t e : This theorem gives that varian ce is independent of change of origin.

3. Var ( a;) a2 var(;)

N o t e : This theorem gives that change of scale affects the variance.

4. Var ( a;b) a2Var(;)

5. Var (b- ax) a2 Var(x)

Example 16:

Calculate the variance of ;, if ; denote the number obtained on the face of fair

die. munotes.in

## Page 182

182NUMERICAL AND STATISTICAL METHODSSolution:

; is random variable with probability distribution ; xi 1 2 3 4 5 6 p.m.f. P(; xi) 1/6 1/6 1/6 1/6 1/6 1/6 כሺሻ 1/6 2/6 3/6 4/6 5/6 6/6 2כሺሻ 1/6 4/6 9/6 16/6 25/6 36/6 E(X) σݔܲכሺݔሻ

ୀଵ Where xi for i 1, 2, 3, 4, 5,6 (values of X)

E(X) ଵ

ଶ

ଷ

ସ

ହ

ൌଶଵ

3.5

E(X2) σݔ2ܲכሺݔሻ

ୀଵ Where xi for i 1, 2, 3, 4, 5,6 (values of X)

E(X2) ଵ

ସ

ଽ

ଵ

ଶହ

ଷ

ൌଽଵ

σ2 Var (;) E(;2) – >E(;)@2 ଽଵ

െቀଶଵ

ቁ2ൌଵହ

ଷൌʹ Ǥ ͻ ͳ

Example 17:

Obtain variance of r.v. ; for the following p.m.f. ; xi 0 1 2 3 4 5 p.m.f. P(; xi) 0.05 0.15 0.2 0.5 0.09 0.01

Solution:

; is random variable with probability distribution ; xi 0 1 2 3 4 5 p.m.f. P(; xi) 0.05 0.15 0.2 0.5 0.09 0.01 כሺሻ 0 0.15 0.40 1.50 0.36 0.05 2כሺሻ 0 0.15 0.80 4.50 1.44 0.25

E(X) σݔܲכሺݔሻହ

ୀ Where xi for i 0,1, 2, 3, 4, 5 (values of X)

E(X) 2.46

E(X2) σݔ2ܲכሺݔሻହ

ୀ Where xi for i 0,1, 2, 3, 4, 5 (values of X)

E(X2) 7.14

σ2 Var (;) E(;2) – >E(;)@2 ൌ Ǥ ͳ Ͷെሺ ʹ Ǥ Ͷ ሻ2 1.0884 munotes.in

## Page 183

183Chapter 11: Random VariablesExample 1:

Obtain variance of r.v. ; for the following probability distri bution

P(x) ௫మ

ଷ, x = 0,1, 2,3, 4

Solution:

; is random variable with probability distribution P( x) ௫మ

ଷ, x = 0,1, 2,3, 4

E(X) σݔܲכሺݔሻସ

ୀ Where xi for i 0,1, 2, 3, 4 (values of X)

σݔכସ

ୀ௫మ

ଷ ଵ

ଷσݔଷସ

ୀൌଵ

ଷሺͲͳͺʹͶ ሻൌଵ

ଷ ଵ

ଷ 3.33

E(X) ଵ

ଷ

E(X2) ଵ

ଷσݔସସ

ୀ Where xi for i 0,1, 2, 3, 4, 5 (values of X)

ሺܺʹሻൌଵ

ଷσݔସସ

ୀ ଵ

ଷሺͳͳͺͳʹͷ ሻൌଷହସ

ଷ 11.8

E(X2) σ2 Var (;) E(;2) – >E(;)@2 ൌͳ ͳ Ǥ ͺെሺ ͵ Ǥ ͵ ͵ ሻ2 0.6889

11.11 Summary

In this chapter, random variables, its types with its Probabili ty Distributions,

expected value and variance is discussed.

Random variable : Let S / Ω be the sample space corresponding to the outcomes

of a random experiment . A function X: S → R (where R is a set of real numbers)

is called as a random variable.

A random variable ; is said to be discrete if it takes finite o r countably infinite

number of possible values. A sample space which is finite or co untably infinite is

called as denumerable of countable. If the sample space is not countable then it is

called continuous.

The probability distribution (or simply distribution ) of a ran dom variable ; on a

sample space Ω is set of pairs (; xi, P (; xi)) for all xi : א x(Ω), where P (;

xi) is the probability that ; takes the value xi .

Let X be a discrete random variable defined on a sample space Ω / S . Suppose { x1,

x2, x3, ……. xn ` is the range set of ;. With each of xi, we assign a number P( xi) P(;

xi) called the probability mass function (p.m.f.) such that, munotes.in

## Page 184

184NUMERICAL AND STATISTICAL METHODS P ( xi) ≥ 0 for i 1, 2, 3 ….,n and (AND)

σሺݔሻൌͳ୬

ୀଵ

Probability Density Function (p.d.f.) Let ; be a continuous r andom variable.

Function f(x) defined for all real x א-( ∞, ∞) is called probability density function

p.d.f. if for any set B of real numbers, we get probability,

P ^ ; א B ` ݂ሺݔሻݔ݀

Axiom I : Any probability must be between zero and one.

Axiom II : Total probability of sample space must be one

Axiom III : For any sequence of mutually exclusive events E 1, E2, E3,……… i.e

Ei ∩ EM = Φ for i ≠ j

Cumulative Distribution Function (c.d.f.)

It is also termed as Must a distribution function. It is the ac cumulated value of

probability up to a given value of the random variable. Let ; b e a random variable,

then the cumulative distribution function (c.d.f.) is defined a s a function F(a) such

that,

F(a) = P { X ≤ a }

Expected Value of Random Variable expected value is a mean or average value

of the probability dis tribution of t he random variable E( X)

The variance of ; denoted by Var(X) S.D.= σ = ඥሺሻ

11.12 Reference

Fundamentals of Mathematical Statistics S. C. Gupta, V. K. Kapo or munotes.in

## Page 185

185Chapter 11: Random Variables

11.1 Unit End Exercise

1 An urn contains 6 red and 4 white balls. Three balls are drawn at random.

Obtain the probability distribution of the number of white ball s drawn.

Hints and Answers :

; xi01 2 3

p.m.f. P(; xi ) 5/30 15/30 9/30 1/30

2 Obtain the probability distribution of the number of sixes in two tosses of a

die

Hints and Answers :

; xi01 2

p.m.f. P(; xi ) 25/36 10/36 1/36

3 If the variable ; denotes the maximum of the two numbers , whe n a pair of

unbiased die is rolled , ﬁ nd the probability distribution of ;.

Hints and Answers:

; xi123456

p.m.f. P(; xi ) 1/36 3/36 5/36 4/36 9/36 11/36

4 A box of 20 mangoes contain 4 bad mangoes. Two mangoes are dra wn

at random without replacement from this box. Obtain the probabi lity

distribution of the number of bad mangoes in sample

Hints and Answers:

; xi01 2

p.m.f. P(; xi ) 95/138 40/138 3/138

5 Three cards are drawn at random successively, with replacement , from a well

shu൷ ed pack of 52 playing cards. Getting ‘a card of diamonds ‘ is termed as

a success. Obtain the probability distribution of the number of the successes.

Hints and Answers:

; xi01 2 3

p.m.f. P(; xi ) 27/64 27/64 9/64 1/64munotes.in

## Page 186

186NUMERICAL AND STATISTICAL METHODS

6 Determine ‘k’ such that the following functions are p.m.f.s

i) f(x) kx , x 1 , 2 , 3 ,««.,10. > Ans: 1 /55 @

ii) f(x) k. x 0 , 1 , 2 , 3 > Ans: 3/19 @

7 A random variable ; has the following probability distribution .

; x i0123456

P(; xi )k 3k 5k 7k 9k 11k 13k

Find k, > Ans: 1/49 @

i) Find P( ; 2 ) > Ans: 45/49 @

ii) Find P( 0 ; 5) >Ans: 24/49 @

8 Given the following distribution function of a random variable ;.

; xi-3 -2 -1 0 1 2 3

F(;) 0.05 0.15 0.38 0.57 0.72 0.88 1

Obtain:

i) p.m.f. of ; >Ans :

; xi-3 -2 -1 0 1 2 3

P(;) 0.05 0.1 0.23 0.19 0.15 0.16 0.12

@

ii) P ( -2 ; 1) > Ans: 0.67 @

iii) P ( ; ! 0) > Ans: 0.43 @

iv) P(-1 ; 2) > Ans: 0.34 @

v) P ( -3 ; -1) > Ans: 0.15 @

vi) P( -2 ; 0) > Ans: 0.42 @

; xi0123456

P(; xi )k 3k 5k 7k 9k 11k 13k

9 Continuous random variable ; assumes values between 2 and 5 wi th p.d.f.

i) Find k > Ans : 2 /27 @

ii) Find P ( x 4) > Ans: 16 /27 @munotes.in

## Page 187

187Chapter 11: Random Variables

10 Following is the c.d.f. of a discrete random variable ;.

; xi12345678

F(; a) 0.08 0.12 0.23 0.37 0.48 0.62 0.85 1

Find:

i) p.m.f. of ; > Ans :

; xi12345678

P(;) 0.08 0.04 0.11 0.08 0.17 0.14 0.23 0.15

@

ii) P ( ; 4) > Ans : 0.31 @

iii) P( 2 ; 6) > A ns : 0.54 @

munotes.in

## Page 188

188NUMERICAL AND STATISTICAL METHODSUnit 5

12

DISTRIBUTIONS: DISCRETE DISTRIBUTIONS

Unit Structure

12.0 ObMectives

12.1 Introduction

12.2 Uniform Distribution

12.2.1 Definition

1 2 . 2 . 2 M e a n a n d V a r i a n c e o f U n i f o r m D i s t r i b u t i o n

1 2 . 2 . 3 A p p l i c a t i o n s o f U n i f o r m D i s t r i b u t i o n

12.3 Bernoulli Distribution

12.3.1 Definition

1 2 . 3 . 2 M e a n a n d V a r i a n c e o f B e r n o u l l i D i s t r i b u t i o n

1 2 . 3 . 3 A p p l i c a t i o n s o f B e r n o u l l i D i s t r i b u t i o n

1 2 . 3 . 4 D i s t r i b u t i o n o f S u m o f i n d e p e n d e n t a n d i d e n t i c a l l y d istributed

Bernoulli Random variables

12.4 Binomial Distribution

12.4.1 Definition

1 2 . 4 . 2 M e a n a n d V a r i a n c e o f B i n o m i a l D i s t r i b u t i o n

1 2 . 4 . 3 A p p l i c a t i o n s o f B i n o m i a l D i s t r i b u t i o n

12.5 Poisson Distribution:

12.5.1 Definition

1 2 . 5 . 2 M e a n a n d V a r i a n c e o f P o i s s o n D i s t r i b u t i o n

1 2 . 5 . 3 A p p l i c a t i o n s o f P o i s s o n D i s t r i b u t i o n

1 2 . 5 . 4 C h a r a c t e r i s t i c s o f P o i s s o n D i s t r i b u t i o n

12.6 Summary

12.7 Reference

12.8 Unit End Exercise munotes.in

## Page 189

189Chapter 12: Distributions: Discrete Distributions12.0 Objectives

After going through this unit, you will be able to:

x Understand the need of standard probability distribution as mod els

x Learn the Probability distributions and compute probabilities

x Understand the specific situati ons for the use of these models

x Learn interrelations among the di fferent probability distributi ons.

12.1 Introduction

In previous unit we have seen the general theory of univariate probability

distributions. For a discrete random variable, p.m.f. can be ca lculated using

underlying probability structure o n the sample space of the ran dom experiment.

The p.m.f. can be expressed in a mathematical form. This probab ility distribution

can be applied to variety of real-life situations which possess s o m e c o m m o n

features. Hence these are also called as ‘Probability Models’.

Discrete distributions:

Uniform, Bernoulli, Binomial, Poisson Continuous distributions:

Uniform distributions, Exponential

In this unit we will study some probability distributions. Viz. Uniform, Binomial,

Poisson and Bernoulli distributions.

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## Page 190

190NUMERICAL AND STATISTICAL METHODS12.2 Uniform Distribution

Uniform distribution is the simplest statistical distribution. When a coin is tossed

the likelihood of getting a tail or head is the same. A good ex ample of a discrete

uniform distribution would be the possible outcomes of rolling a 6-sided fair die.

Ω = { 1, 2, 3, 4, 5, 6 }

In this case, each of the six numbers has an equal chance of a ppearing. Therefore,

each time the fair die is thrown, each side has a chance of 1/6 . The number of

values is finite. It is impo ssible to get a value of 1.3, 4.2, or 5.7 when rolling a fair

die. However, if another die is added and they are both thrown, the distribution

that results is no longer uniform because the probability of th e sums is not equal.

A deck of cards also has a uniform distribution. This is becaus e an individual has

an equal chance of drawing a spade, a heart, a club, or a diamo nd i.e. 1/52.

Consider a small scale company with 30 employees with employee id 101 to 130.

A leader for the company to be selected at random. Therefore a employee id is

selected randomly from 101 to 130. If ; denotes the employee i d selected then

since all the id’s are equally likely , the p.m.f. of ; is given by,

P(x) ଵ

ଷ x 101, 102, …….150

0 otherwise

Such distribution is called as a discrete uniform distribution. The discrete uniform

distribution is also known as the equally likely outcomes dis tribution.

The number of values is finite. It is impossible to get a value of 1.3, 4.2, or 5.7

when rolling a fair die. Howeve r, if another die is added and t hey are both

thrown, the distribution that results is no longer uniform beca use the probability

of the sums is not equal. Another simple example is the probabi lity distribution of

a coin being flipped. The possible outcomes in such a scenario can only be two.

Therefore, the finite value is 2.

12.2.1 Definition: Let X be a discrete random taking values 1, 2, ……, n. Then ; is said to follow uniform discrete uniform distribution if it s p.m.f is given by P (; x) ͳ

݊ x = 1, 2,,….n

0 otherwise munotes.in

## Page 191

191Chapter 12: Distributions: Discrete Distributions‘n’ is called as the parameter of the distribution. Whenever , the parameter value is

known, the distribution is known completely. The name is ‘unifo rm’ as it treats

all the values of the variable ‘uniformly’. It is applicable wh ere all the values of

the random variable are equally likely.

Some examples or the situation where it applied

1. Let ; denote the number on the face of unbiased die, after it is rolled. P (; x) ͳ

x 1, 2, 3,4,5,6

0 otherwise 2. A machine generates a digit randomly from 0 to 9 P (; x) ͳ

ͳͲ x = 0, 1, 2, ….9

0 otherwise

12.2.2Mean and Variance of Uniform Distribution

Let ; is said to follow Uniform discrete uniform distribution a nd its p.m.f is

given by P (; x) ͳ

݊ x = 1, 2,,….n

0 otherwise

Mean E(;) σݔܲכሺݔሻ

ୀଵ ଵ

σݔ

ୀଵ ሺାଵሻ

ଶൌାଵ

ଶ

Variance (X) = E(;2) – >E(;)@2

E(;2) σݔ2ܲכሺݔሻ

ୀଵ ൌଵ

σݔ2

ୀଵൌሺାଵሻሺଶାଵሻ

Variance (X) = E(;2) – >E(;)@2 ሺାଵሻሺଶାଵሻ

െሺାଵሻమ

ସൌమିଵ

ଵଶ

Standard Deviation (S.D.) (;) = σ ඥݎܸܽሺܺሻ ටమିଵ

ଵଶ

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192NUMERICAL AND STATISTICAL METHODS12.2. Applications of Uniform Distribution

Example1: Find the variance and standard deviation of ;, where ; is the square

of the score shown on a fair die.

Solutions: Let ; is a random variable which shows the square of the score on a

fair die.

; ^ 1, 4, 9,16, 25, 36 ` each on have the probability ଵ

Mean E(;) σݔܲכሺݔሻ

ୀଵ ଵ

ሺͳͶͻͳʹͷ͵ሻ ଽଵ

Variance (X) = E(;2) – >E(;)@2

E(;2) σݔ2ܲכሺݔሻ

ୀଵ ଵ

ሺͳͳͺͳʹͷʹͷͳʹͻሻ ଶଶହ

Variance (X) = E(;2) – >E(;)@2 ଶଶହ

െቂଽଵ

כଽଵ

ቃൌହଷଽ

ଷ

Standard Deviation ( S.D.) (X)= σ = ඥݎܸܽሺܺሻൌටହଷଽ

ଷ

12. Bernoulli Distribution

A trial is performed of an experiment whose outcome can be clas sified as either

success or failure. The probability of success is p ( 0 ≤ p ≤ 1) and probability of

failure is (1-p). A random variable ; which takes two values 0 and 1 with

probabilities ‘q’ and ‘p’ i.e. P(x 1) p and P(x 0) q, p q 1 (i.e q 1 –p),

is called a Bernoulli variate and is said to be a Bernoulli Dis tribution, where p and

q takes the probabilities for success and failure respectively . It is discovered by

Swiss Mathematician James or Jacq ues Bernoulli (1654-1705). It is applied

whenever the experiment results in only two outcomes. One is su ccess and other

is failure. Such experiment is called as Bernoulli Trial.

12..1 Definition: Let ; be a discrete random tak ing values either success (1/ True / p) or failure (0 / False / q). Then ; is said to fol low Bernoulli discrete

distribution if its p.m.f is given by P (; x) ௫ݍଵି௫ x 0, 1 0 otherwise Note: 0 ≤ p ≤ 1, p q 1 This distribution is known as Bernoulli distribution with param eter ‘p’ munotes.in

## Page 193

193Chapter 12: Distributions: Discrete Distributions12..2 Mean and Variance of Bernoulli Distribution

Let X follows Bernoulli Distribution with parameter ‘p’. Theref ore its p.m.f. is

given by P (; x) ௫ݍଵି௫ x 0, 1 0 otherwise Note: 0 ≤ p ≤ 1 , p q 1 Mean E(;) σݔܲכሺݔሻଵ

ୀ ൌσݔ௫ݍଵି௫ ଵ

ୀ

S u b s t i t u t e t h e v a l u e o f x= 0, and x 1, we get

E(;) (0 î p0 î q1-0) (1 î p1 î q1-1) p

Similarly, E(;2) σݔଶ

ܲכሺݔሻଵ

ୀ ൌσݔଶ௫ݍଵି௫ ଵ

ୀ

Substitute the value of x= 0, and x 1, we get

E(;2) p

Variance (X) = E(;2) – >E(;)@2 p – p2 p (1- p) = pq ………… (p q 1)

Standard Deviation ( S.D.) (X)= σ = ඥݎܸܽሺܺሻ ඥݍ

Note: if p q ଵ

ଶ the Bernoulli distribution is reduced to a Discrete Uniform

Distribution as P (; x) ͳ

ʹ x 0,1

0 otherwise 12.. Applications of Bernoulli Distribution

Examples of Bernoulli’ s Trails are:

1) Toss of a coin (head or tail)

2) Throw of a die (even or odd number)

3) Performance of a student in an examination (pass or fail)

4) Sex of a new born child is recorded in hospital, Male 1, Female 0

5) Items are classified as ‘defective=0’ and ‘non -defective = 1’. munotes.in

## Page 194

194NUMERICAL AND STATISTICAL METHODS12..4 Distribution of Sum of independent and identically dis tributed

Bernoulli Random variables

Let < i, i= 1,2,…n be ‘n’ independent Bernoulli random variables with pa rameter

‘p’ (‘p’ for success and ‘q’ for failure p + q = 1)

i.e. P >< i 1@ p and P >< i 0@ = q, for i = 1,2,…n.

Now lets define, X which count the number of ‘1’s (Successes) in ‘n’

independent Bernoulli trials,

; σ୧

ୀଵ

In order to derive probability of ‘ x’ successes in ‘n’ trials i.e. P [ X = x@

Consider a particular sequence of ‘ x’ successes and remaining (n – x) failures as

1 0 1 1 1 0 1 0 ……… 1 0

Here ‘1’ (Success p) occurs ‘x’ times and ‘0’ (Failure q) occurs (n – x) times.

Due to independence, probability of such a sequence is given as follows:

ǥǤǤᇣᇧᇧᇧᇤᇧᇧᇧᇥ௫௧௦ ݍݍݍݍǥǥǥǤݍᇣᇧᇧᇧᇧᇤᇧᇧᇧᇧᇥൌ௫

ሺି௫ሻ௧௦ݍሺି௫ሻ

However, the successes ( 1’s) can occupy any ‘x’ places out of ‘ n’ places in a

sequence in ൫୬

୶൯ ways, therefore P (; x) ቀ݊

ݔቁ௫ݍି௫ x 0, 1, 2…..n 0 otherwise Note: 0 ≤ p ≤ 1 , p q 1 This gives us a famous distribution called as ‘ Binomial Distribution’

12.4 Binomial Distribution

This distribution is very useful in day to day life. A binomia l random variable

counts number of successes when ‘n ’ Bernoulli trials are perfor med. A single

success/failure experiment is al so called a Bernoulli trial or Bernoulli experiment,

and a sequence of outcomes is called a Bernoulli process for a single trial,

i.e., n 1, the binomial distribution is a Bernoulli distribut ion.

Binomial distribution is denoted by X → B (n, p)

Bernoulli distribution is Must a binomial distribution with n = 1

i.e. parameters (1, p). munotes.in

## Page 195

195Chapter 12: Distributions: Discrete Distributions12.4.1 Definition:

A discrete random variable ; taking values 0, 1, 2, …….n is said to follow a

binomial distribution with parameters ‘n’ and ‘p’ if its p.m.f. is given by P (; x) ቀ݊

ݔቁ௫ݍି௫ x 0, 1, 2…..n 0 otherwise Note: 0 ≤ p ≤ 1 , p q 1 Remark:

1) The probabilities are term in the binomial expansion of ( p q )n, hence the

name ‘ Binomial Distribution ‘ is given

2) σሺሻൌ୬

୶ୀ σൌ୬

୶ୀ൫୬

୶൯୶୬ି୶ൌሺ ሻ୬ൌͳ

3) The binomial distribution is frequently used to model the numbe r of

successes in a sample of size n drawn with replacement from a p opulation

of size N. If the sampling is carried out without replacement, the draws are

not independent and so the resulting distribution is a hypergeo metric

distribution, not a binomial one

12.4.2 Mean and Variance of Binomial Distribution

Let ; follows Binomial Distribution with parameters ‘n’ and ‘p’ if its p.m.f. is

given by P (; x) ቀ݊

ݔቁ௫ݍି௫ x 0, 1, 2…..n 0 otherwise Note: 0 ≤ p ≤ 1 , p q 1 Mean E(;) σݔܲכሺݔሻଵ

ୀ ൌσݔ൫

௫൯௫ݍି௫ ଵ

ୀ Mean E(;) σ୧כሺ୧ሻ୬

୧ୀ

σ୧൫୬

୶൯୶୬ି୶ ୬

୧ୀ

σ୧Ǥ୬Ǩ

୶Ǩሺ୬ି୶ሻǨ୶୬ି୶ ୬

୧ୀ

Substitute the value of x= 0, we get first term as ‘0’

σǤሺ୬ିଵሻǨ

ሺ୶ିଵሻǨሺ୬ି୶ሻǨ୶୬ି୶ ୬

୧ୀଵ munotes.in

## Page 196

196NUMERICAL AND STATISTICAL METHODS ݊σǤሺ୬ିଵሻǨ

ሺ୶ିଵሻǨሺ୬ିଵିሺ୶ିଵሻሻ ሻǨ୶ିଵ୬ି୶ ୬

୧ୀଵ

݊σǤ൫୬ିଵ

୶ିଵ൯୶ିଵ୬ିଵିሺ୶ିଵሻ ୬

୧ୀଵ

݊ሺݍሻିଵ ----- Using Binomial Expansion

݊ǥǥǥǥǥǥǤǤሺݍ ൌ ͳሻ

Mean E (;) ݊(------------------------------ 1) E(;2) E>; (;-1)@ E>;@ -------------------------(2)

E>; (;-1)@ σ୧ሺ୧െͳሻ൫୬

୶൯୶୬ି୶ ୬

୧ୀ

σ୶ሺ୶ିଵሻ୬Ǩ

୶Ǩሺ୬ି୶ሻǨ୶୬ି୶ ୬

୧ୀ

݊ሺ݊െͳሻଶσǤ൫୬ିଶ

୶ିଶ൯୶ିଶ୬ିଶିሺ୶ିଶሻ୬

୧ୀଶ

݊ሺ݊െͳሻଶሺݍሻሺିଶሻ

݊ሺ݊െͳሻଶ---------------------------------------(3)

Using (1), (2) and (3)

E(;2) ݊ሺ݊െͳሻଶ݊(------------------------ 4)

Variance (X) = E(;2) – >E(;)@2

Variance (X) = ሾ݊ሺ݊െͳሻଶ݊ሿെሺ݊ሻଶൌݍ݊

Standard Deviation ( S.D.) (X)= σ = ඥݎܸܽሺܺሻ ඥݍ݊ NOTE: Binomial Theorem (Binomial Expansion) it states that, where n is a positive integer: (a b)n an (nC1)an-1b (nC2)an-2b2 + … + (nCn-1)abn-1 bn

൫

൯ൌǨ

ሺିሻǨǨൌ nCr 12.4. Applications of Binomial Distribution

We get the Binomial distribution under the following experiment al conditions.

1) The number of trials ‘ n’ is finite. (not very large)

2) The trials are independent of each other. munotes.in

## Page 197

197Chapter 12: Distributions: Discrete Distributions3) The probability of success in any trial‘ p’ is constant for eac h trial.

4) Each trial (random experiment) must result in a success or a fa ilure

(Bernoulli trial).

Following are some of the real li fe examples of Binomial Distri bution

1. Number of defective items in a lot of n items produced by a mac hine

2. Number of mail births out of ‘n’ births in a hospital

3. Number of correct answers in a multiple choice test.

4. Number of seeds germinated in a row of ‘n’ planted seeds

5. Number of rainy days in a month

6. Number of re- captured fish in a sample of ‘n’ fishes.

7. Number of missiles hitting the targets out of ‘n’ fired.

In all above situations, ‘p’ is the probability of success is assumed to be constant.

Example 1:

Comment on the following: “ The mean of a binomial distribution is 5 and its

variance is 9”

Solution:

The parameters of the binomial distribution are n and p

We have mean np 5

Variance npq 9

ݍൌ୬୮୯

୬୮ ଽ

ହͳ

Which is not admissible since q c annot exceed unity. (p q 1 ) Hence the given

statement is wrong.

Example 2:

Eight coins are tossed simultaneously. Find the probability o f getting atleast six

heads.

Solution:

Here number of trials, n 8, p denotes the probability of gett ing a head.

P 1 / 2 and q 1 – p 1 / 2 munotes.in

## Page 198

198NUMERICAL AND STATISTICAL METHODSIf the random variable ; denotes the number of heads, then the probability of a

success in n trials is given by

P r o b a b i l i t y o f g e t t i n g a t l e a s t 6 h e a d s i s g i v e n b y

P ( X ≥ 6) P(; 6) P(; 7) P(; 8)

ଵ

ଶఴ൫଼

൯ ଵ

ଶఴ൫଼

൯ଵ

ଶఴ൫଼

଼൯

ଵ

ଶఴሾ൫଼

൯ ൫଼

൯ ൫଼

଼൯ሿ

ଵ

ଶହሾʹͺͺͳ ሿൌଷ

ଶହ

Example :

Ten coins are tossed simultaneously. Find the probability of ge tting (i) at least

seven heads (ii) exactly seven heads (iii) at most seven heads

Solution:

Here number of trials, n 10, p denotes the probability of get ting a head.

P 1 / 2 and q 1 – p 1 / 2

If the random variable ; denotes the number of heads, then the probability of a

success in n trials is given by P (; x) ቀ݊

ݔቁ௫ݍି௫ x 0, 1, 2…..n

൬ͺ

ݔ൰൬ͳ

ʹ൰௫

൬ͳ

ʹ൰଼ି௫

ൌ൬ͺ

ݔ൰൬ͳ

ʹ൰଼

ൌͳ

ʹ଼൬ͺ

ݔ൰ Note: 0 ≤ p ≤ 1 , p q 1

P (; x) ቀ݊

ݔቁ௫ݍି௫ x 0, 1, 2…..n

൬ͳͲ

ݔ൰൬ͳ

ʹ൰௫

൬ͳ

ʹ൰ଵି௫

ൌ൬ͳͲ

ݔ൰൬ͳ

ʹ൰ଵ

ൌͳ

ʹଵ൬ͳͲ

ݔ൰ Note: 0 ≤ p ≤ 1 , p q 1

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## Page 199

199Chapter 12: Distributions: Discrete Distributionsi) Probability of getting at least 7 heads is given by

P( X ≥ 7) P(; 7) P(; 8) P(; 9) P(; 10)

ଵ

ଶభబ൫ଵ

൯ ଵ

ଶభబ൫ଵ

଼൯ଵ

ଶభబ൫ଵ

ଽ൯ଵ

ଶభబ൫ଵ

ଵ൯

ଵ

ଶభబሾ൫ଵ

൯ ൫ଵ

଼൯൫ଵ

ଽ൯ ൫ଵ

ଵ൯ሿ

ଵ

ଵଶସሾͳʹͲͶͷͳͲͳ ሿൌଵ

ଵଶସ

ii) Probability of getting exactly 7 heads is given by .

P(; 7) ଵ

ଶభబ൫ଵ

൯ൌଵଶ

ଵଶସ

iii) Probability of getting at most 7 heads is given by .

P( X ≤ 7) 1 - P(; ! 7)

1 – ^ P(; 8) P(; 9) P(; 10) `

1 - ଵ

ଶభబሾ ൫ଵ

଼൯൫ଵ

ଽ൯ ൫ଵ

ଵ൯ሿ

1 - ଵ

ଵଶସሾͶͷͳͲͳ ሿൌͳെହ

ଵଶସ ଽ଼

ଵଶସ

Example 4:

20 wrist watches in a box of 100 are defective. If 10 watches are selected at

random, find the probability that (i) 10 are defective (ii) 10 are good (iii) at least

one watch is defective (iv)at most 3 are defective.

Solution:

20 out of 100 wrist watches are defective, so Probability of d efective wrist watch

p 20 /100

p ଶ

ଵൌଵ

ହ q 1 – p ͳെଶ

ଵൌ଼

ଵൌସ

ହ

Since 10 watches are selected at random, n 10 P ( ; x ) ቀ݊

ݔቁ௫ݍି௫ x = 0 , 1, 2…..n

൬ͳͲ

ݔ൰൬ͳ

ͷ൰௫

൬Ͷ

ͷ൰ଵି௫

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## Page 200

200NUMERICAL AND STATISTICAL METHODSi) Probability of selecting 10 defective watches

P(; 10) ൫ଵ

ଵ൯ቀଵ

ହቁଵ

ቀସ

ହቁଵିଵ

ൌͳ Ǥଵ

ହభబǤͳൌଵ

ହభబ

ii) Probability of selecting 10 good watches (i.e. no defective)

P(; 0) ൫ଵ

൯ቀଵ

ହቁ

ቀସ

ହቁଵି

ൌͳ Ǥ ͳ Ǥସభబ

ହభబൌሺସ

ହሻ10

iii) Probability of selecting at leas t one defective watch

P(X ≥ 1) = 1 – P(; 1) 1 - P(; 0)

1 - ൫ଵ

൯ቀଵ

ହቁ

ቀସ

ହቁଵି

ൌͳെሺସ

ହሻ10

iv) Probability of selecting at most 3 defective watches

P(X ≤ 3) = P(; 0) P(; 1) P(; 2) P(; 3)

൫ଵ

൯ቀଵ

ହቁ

ቀସ

ହቁଵ

൫ଵ

ଵ൯ቀଵ

ହቁଵ

ቀସ

ହቁଽ

൫ଵ

ଶ൯ቀଵ

ହቁଶ

ቀସ

ହቁ଼

൫ଵ

ଷ൯ቀଵ

ହቁଷ

ቀସ

ହቁ

ͳǤͳǤቀସ

ହቁଵ

ͳͲǤଵ

ହସవ

ହవͶͷǤଵ

ହమସఴ

ହఴͳʹͲǤଵ

ହయସళ

ହళ

0.859 (Approx.)

Example 5:

With the usual notation find p for binomial random variable ; i f n 6 and

>9
P(; 4) P(; 2)@

Solution:

The probability mass function of binomial random variable ; is given by

Here n 6,

P (; x) ൫

௫൯ሺሻ௫ሺݍሻି௫

P (; 4) ൫

ସ൯ሺሻସሺݍሻିସ ൫

ସ൯ሺሻସሺݍሻଶ

P (; 2) ൫

ଶ൯ሺሻଶሺݍሻିଶ ൫

ଶ൯ሺሻଶሺݍሻସ

P (; x) ቀ݊

ݔቁ௫ݍି௫ x 0, 1, 2…..n Note: 0 ≤ p ≤ 1 , p q 1 munotes.in

## Page 201

201Chapter 12: Distributions: Discrete DistributionsGiven that, 9
P(; 4) P(; 2)

9
൫

ସ൯ሺሻସሺݍሻଶ ൫

ଶ൯ሺሻଶሺݍሻସ

9
15
ሺሻସሺݍሻଶ 15
ሺሻଶሺݍሻସ

9 ሺሻସሺݍሻଶ ሺሻଶሺݍሻସ

9 p2 = q2

Taking positive square root on both sides we get,

3 p = q

3p = 1- p

4p = 1 ൌଵ

ସ= 0.25

12.5 Poisson Distribution

Poisson distribution was discovered by a French Mathematician-c um-Physicist

Simeon Denis Poisson in 1837. Poisson distribution is also a di screte distribution.

He derived it as a limiting case of Binomial distribution. For n-trials the binomial

distribution is (p q)n the probability of x successes is given by P(; x) ൌ

σ୧൫୬

୶൯୶୬ି୶ ୬

୧ୀ If the number of trials n is very large and the probability of

success ‘ p’ is very small so that the product np = m is non – negative and finite.

12.5.1 Definition:

A discrete random variable ;, taking on one of the countable i nfinite values 0, 1,

2,……. is said to follow a Binomial distribution with parameters ‘ λ’ or ‘m’, if for

some m ! 0, its p.m.f. is given by P ( ; x ) ݁ି݉௫

ݔǨ x 0 , 1, 2…..

m ! 0 0 otherwise Note: ݁ିͲ ǡ݉ܦܰܣ ௫Ͳ ǡݔǨͲ ǡ

Hence, షೣ

௫ǨͲ Note: ݁ൌబ

Ǩభ

ଵǨమ

ଶǨെെെെെൌ σೣ

௫Ǩ∞

௫ୀ

݁ൌଵబ

Ǩଵభ

ଵǨଶమ

ଶǨെെെെെൌʹǤͳͺʹͺ , 0 1, 1 1

Since, ݁ିͲ ܲ ሺሻͲ

σሺሻൌσஶ

௫ୀషೣ

௫Ǩ ݁ିǡσஶ

௫ୀೣ

௫Ǩൌ݁ିǤ݁ൌͳ munotes.in

## Page 202

202NUMERICAL AND STATISTICAL METHODSIt is denoted by X → P(m) or X → P(λ)

Since number of trials is very large and the probability of suc cess p is very small,

it is clear that the event is a rare event. Therefore Poisson d istribution relates to

rare events.

12.5.2 Mean and Variance of Poisson Distribution

Let X is a Poisson random variable with parameter ‘m’ (or ‘λ’), if its p.m.f. is

given as P (; x) ݁ି݉௫

ݔǨ x 0, 1, 2…..

m ! 0 0 otherwise Note: ݁ିͲ ǡ݉ܦܰܣ ௫Ͳ ǡݔǨͲ ǡ

Hence, షೣ

௫ǨͲ Mean E(;) σ୧ሺ୧ሻஶ

୧ୀ

σݔ୧ஶ

୧ୀషೣ

௫Ǩ

The term corresponding to x 0 is zero.

ൌσஶ

୧ୀଵషೣషభ

ሺ௫ିଵሻǨൌ݁݉ ିσஶ

୧ୀଵషೣషభ

ሺ௫ିଵሻǨ ݁݉ିǤ݁ൌ݉

Mean E(;) m

But, E(;2) E>; (;-1)@ E>;@

E>; (;-1)@ σݔሺݔെͳሻ

ஶ

ୀ ܲሺݔሻൌ σݔሺݔെͳሻ

ஶ

ୀషೣ

௫Ǩ

ൌ݉ଶ݁ିσஶ

୧ୀଶೣషమ

ሺ௫ିଶሻǨൌ ݉ଶ݁ିǤ݁ൌ݉ଶ

E>; (;-1)@ ݉ଶ

E(;2) E>; (;-1)@ E>;@ ݉ଶ ݉

Variance (X) = E(;2) – >E(;)@2

Variance (X) = ሺ݉ଶ ݉ሻ - (݉)2 ݉

Standard Deviation ( S.D.) (X)= σ = ඥݎܸܽሺܺሻ ξ݉ munotes.in

## Page 203

203Chapter 12: Distributions: Discrete DistributionsThus the mean and variance of Poisson distribution are equal an d each is equal to

the parameter of distribution ( Ԣ݉ԢݎԢߣԢሻ

12.5. Applications of Poisson Distribution

Some examples of Poisson variates are :

1) The number of blinds born in a town in a particular year.

2) Number of mistakes committed in a typed page.

3) The number of students scoring very high marks in all subMects.

4) The number of plane accidents in a particular week.

5) The number of defective screws in a box of 100, manufactured by a reputed

company.

6) Number of accidents on the express way in one day.

7) Number of earthquakes occurring in one year in a particular sei smic zone.

8) Number of suicides reported in a particular day.

9) Number of deaths of policy holders in one year.

Conditions:

Poisson distribution is the limiting case of binomial distribu tion under the

following conditions:

1. The number of trials n is indefinitely large i.e., n → ∞

2. The probability of success ‘ p’ for each trial is very small; i .e., p → 0

3. np m (say) is finite, m ! 0

Characteristics of Poisson Distribution:

Following are the characteristics of Poisson distribution

1. Discrete distribution: Poisson distribution is a discrete distr ibution like

Binomial distribution, where the random variable assume as a co untably

infinite number of values 0,1,2 ….

2. The values of p and q: It is applied in situation where the pro bability of

success p of an event is very small and that of failure q is ve ry high almost

equal to 1 and n is very large.

3. The parameter: The parameter of the Poisson distribution is m. If the value

of m is known, all the probabilities of the Poisson distributio n can be

ascertained. munotes.in

## Page 204

204NUMERICAL AND STATISTICAL METHODS4. Values of Constant: Mean m variance so that standard devia tion ξ݉

Poisson distribution may have either one or two modes.

5. Additive Property: If ; 1 and ; 2 a r e t w o i n d e p e n d e n t P o i s s o n d i s t r i b u t i o n

variables with parameter m 1 and m 2 respectively. Then (; 1 ; 2) also

follows the Poisson distribution with parameter (m 1 m 2) i.e. (; 1 ; 2) → P

(m1 m 2)

6. As an approximation to binomial distribution: Poisson distribut ion can be

taken as a limiting form of Binomial distribution when n is lar ge and p is

very small in such a way that product np m remains constant .

7. Assumptions: The Poisson distribution is based on the following

assumptions.

i) The occurrence or non- occurrence of an event does not influenc e the

occurrence or non-occurrence of any other event.

ii) The probability of success for a short time interval or a sma ll region

of space is proportional to the length of the time interval or space as

the case may be.

iii) The probability of the happening of more than one event is a ve ry

small interval is negligible.

Example 1:

Suppose on an average 1 house in 1000 in a certain district has a fire during a

year. If there are 2000 houses in that district, what is the p robability that exactly

5 houses will have a fire during the year" >given that e-2 0.13534@

Solution:

Mean np , n 2000, p ଵ

ଵ

m np ʹͲͲͲכଵ

ଵ 2

m 2, now According to Poisson distribution P (; x) ݁ି݉௫

ݔǨ x 0, 1, 2…..

m ! 0 ۾ ሺ܆ ൌ ሻൌష

Ǩൌ܍ష

Ǩ

۾ሺ܆ ൌ ሻൌሺǤଵଷହଷସሻכଷଶ

ଵଶ 0.36 munotes.in

## Page 205

205Chapter 12: Distributions: Discrete DistributionsExample 2:

In a Poisson distribution 3P(; 2) P(; 4) Find the parameter ‘ m’ .

Solution: P (; x) ݁ି݉௫

ݔǨ x 0, 1, 2…..

m ! 0 Given, 3P(; 2) P(; 4)

͵ି୫ଶ

ʹǨൌି୫ସ

ͶǨ

ଶൌଷǤସǨ

ଶǨ 36

݉ ൌ േ

Since mean is always positive m 6

Example :

If 2 of electric bulbs manufactured by a certain company are d efective. Find the

probability that in a sample of 200 bulbs i) less than 2 bulbs ii) more than 3 bulbs

are defective.>e-4 0.0183@

Solution :

The probability of a defective bulb p 2 / 100 0.02

Given that n 200 since p is small and n is large, we use Poisson Distribution,

mean m np

m np 200
0.02 4

Now, Poisson Probability Function P (; x) ݁ି݉௫

ݔǨ x 0, 1, 2…..

m ! 0 i) Probability of less than 2 bulbs are defective

P (; 2) P (; 0) P (; 1)

ୣషరସబ

Ǩୣషరସభ

ଵǨൌିସሺͳͶሻൌିସכͷൌͲǤͲͳ͵ͺכͷൌͲǤͲͻͳͷ

munotes.in

## Page 206

206NUMERICAL AND STATISTICAL METHODSii) Probability of getting more than 3 defective bulbs

P (; ! 3) 1 - P (X ≤ 3)

1 –^ P (; 0) P (; 1) P (; 2) P (; 3) `

1 – ିସ (1 4 ସమ

ଶǨସయ

ଷǨ)

1 – 0.0183
(14810.67)

0 . 5 6 7

Example 4:

In a company previous record show that on an average 3 workers are absent

without leave per shift. Find the probability that in a shift

i) Exactly 2 workers are absent

ii) More than 4 workers will be absent

iii) At most 3 workers will be absent

Solution :

This is a case of Poisson dist ribution with parameter ‘m=3’

i) P (; 2) ୣషయଷమ

ଷǨ 0.2241

ii) P (; ! 4) 1 - P (X ≤ 4)

1 –^ P (; 0) P (; 1) P (; 2) P (; 3) P (; 4) `

1 – ^ ୣషయଷబ

Ǩୣషయଷభ

ଵǨୣషయଷమ

ଶǨ ୣషయଷయ

ଷǨୣషయଷర

ସǨ ` 0.1845

iii) P (X ≥ 3) = 1 - P (X ≤ 3)

1 –^ P (; 0) P (; 1) P (; 2) P (; 3) `

1 – ^ ୣషయଷబ

Ǩୣషయଷభ

ଵǨୣషయଷమ

ଶǨ ୣషయଷయ

ଷǨ ` 0.5767

Example 5:

Number of accidents on Pune Mumbai express way each day is a Po isson random

variable with average of three accidents per day. What is the p robability that no

accident will occur today"

Solution :

This is a case of Poisson distribution with m 3,

P (; 0) ୣషౣଷబ

Ǩ ିଷ 0.0498 munotes.in

## Page 207

207Chapter 12: Distributions: Discrete DistributionsExample 6:

Number of errors on a single page has Poisson distribution with average number

of errors one per page . Calculate probability that there is at least one error on a

page.

Solution :

This is a case of Poisson distribution with m 1,

P (X ≥ 1) 1 - P (; 0) ͳെୣషభଵబ

Ǩ ͳെିଵ 0.0632

12.6 Summary

In this chapter, Discrete distribution, its types Uniform, Bern oulli, Binomial and

Poisson with its mean , variance and its application is discuss ed. Distributi

on Definition Mean

E(;) Varianc

e (;) Uniform P ( ; x ) ͳ

݊ x = 1, 2,,….n

0 otherwise

݊ͳ

ʹ ඨ݊ଶെͳ

ͳʹ Bernoulli P ( ; x ) ௫ݍଵି௫ x 0 , 1 0 otherwise Note: 0 ≤ p ≤ 1 , p + q = 1

p

pq Binomial P ( ; x ) ቀ݊

ݔቁ௫ݍି௫ x = 0 , 1,2,…n 0 otherwise Note: 0 ≤ p ≤ 1 , p + q = 1

݊ ݍ݊ Poisson P ( ; x ) షೣ

௫Ǩ x = 0 , 1, 2….. m > 0

0 Otherwise m ݉ 12.7 Reference

Fundamentals of Mathematical Statistics S. C. Gupta, V. K. Kapo or munotes.in

## Page 208

208NUMERICAL AND STATISTICAL METHODS

12. Unit End Exercise

1 A random variable ; has the following discrete uniform distri bution

If P(;) , ; 0, 1 , «..n, Find E(;) and Var(;)

>Hints and Answers : n/2 , n(n2/ 12 @

2 Let ; follow a discrete uniform distribution over 11, 12, ««., 20.

Find i) P(; ! 15) ii) P(12 ; 18 ) iii) P(; 14) iv) Mean and S.D. of ;

>Hints and Answers : i)0.5 ii) 0.6 iii)0.4 iv) Mean 15.5 and S.D. 2.8723 @

3 Let ; be the roll number of a student selected at random from 20 students

bearing roll numbers 1 to 20. Write the p.m.f. of ;.

>Hints and Answers:

P(;) 1 /20 x 1,2,3,«.,20

0 otherwise @

4 Let ; and < be two discrete uniform r.v.s assuming values 1,2, «..,10, ;

and < are independent. Find p.m.f. of = ;< . Also obtain i) P(= 13)

ii) P(= 12 ).

Hints and Answers:

0.08 ii) 0.64

5 A radar system has a probability of 0.1 of detecting a certain target during a

single scan. Find the probability that the target will be detec ted i) at least twice

in four scans ii) at most once in four scans.

>Hints and Answers: i) 0.0523 ii) 0.9477 @

6 If the probability that any person 65 years old will be dead w ithin a year is

0.05. Find the probability that out of a group of 7 such person s (i) Exactly one,

(ii) none, (iii) at least one, (iv) that more than one,

>Hints and Answers: i) 0.2573 ii) 0.6983 iii) 0.3017 iv) 0.9556 @

7 For a B (5, p ) distribution , P(; 1) 0.0768 , P(; 2) 0.2304. Find the value

of p.

>Hints and Answers: 0.6 @munotes.in

## Page 209

209Chapter 12: Distributions: Discrete Distributions

8 Suppose ; ĺ B( n, p ),

i) If E(;) 6 and Var(;) 4.2 , ﬁ nd n and p.

ii) If p 0.6 E(;) 6 ﬁ nd n and Var(;).

iii) If n 25 , E(;) 10, ﬁ nd p and Var(;).

>Hints and Answers: i) 20, 0.3 ii) 10 , 2.4 iii) 0.4 , 6 @

9 In a summer season a truck driver experiences on an average on e puncture is

1000 km . Applying Poisson distribution, ﬁ nd the probability that there will

be

i) No Puncture ii) two punctures in a Mourney of 3000 km

>Hints and Answers: i) 0.049787 ii) 0.224042 @

10 A book contains 400 misprints distributed randomly throughout its 400 pages.

What is probability that a page observed at random, contains at least two

misprints"

>Hints and Answers: 0.264 @

munotes.in

## Page 210

210NUMERICAL AND STATISTICAL METHODSUnit 5

1

DISTRIBUTIONS: CONTINUOUS

DISTRIBUTIONS

Unit Structure

13.0 ObMectives

13.1 Introduction

13.2 Uniform Distribution

13.2.1 Definition

1 3 . 2 . 2 M e a n a n d V a r i a n c e o f U n i f o r m D i s t r i b u t i o n

1 3 . 2 . 3 A p p l i c a t i o n s o f U n i f o r m D i s t r i b u t i o n

13.3 Exponential Distribution

13.3.1 Definition

1 3 . 3 . 2 M e a n a n d V a r i a n c e o f E x p o n e n t i a l D i s t r i b u t i o n

1 3 . 3 . 3 D i s t r i b u t i o n F u n c t i o n

1 3 . 3 . 4 A p p l i c a t i o n s o f E x p o n e n t i a l D i s t r i b u t i o n

13.4 Normal Distribution

13.4.1 Definition

1 3 . 4 . 2 P r o p e r t i e s o f N o r m a l D i s t r i b u t i o n

13.4.3 Properties and Applications of Normal Distribution

1 3 . 4 . 4 = t a b l e a n d i t s U s e r M a n u a l

13.5 Summary

13.6 Reference

13.7 Unit End Exercise munotes.in

## Page 211

211Chapter 13: Distributions: Continuous Distributions1.0 Objectives

After going through this unit, you will be able to:

x Make familiar with the concept of continuous random variable

x Understand how to study a continuous type distribution

x Make Students to feel it very interesting and can find applica tions.

x Make the students familiar with applications and properties of exponential

distribution.

x To understand why life time distributions of electronic equipme nts can be

assumed to be exponential distributed.

x To understand the importance of normal distributions in the the ory of

distribution

x To study different properties of normal distribution useful in practice to

compute probabilities and expected values.

1.1 Introduction

In continuous set up, many times it is observed that the variab le follows a specific

pattern. This behavior of the random variable can be described in a mathematical

form called ‘ Probability Models’.

In this unit we will study important continuous probability dis tributions such as

Uniform, Exponential, Normal

1.2 Uniform Distribution

In this distribution, the p.d.f. of the random variable remains constant over the

range space of the variable.

1.2.1 Definition:

Let ; be a continuous random v ariable on an interval ( c, d) . ; is said to be a

uniform distribution with if the p.d.f. is constant over the ra nge ( c, d) . Further

the probability distribution must satisfy the Axioms of probabi lity.

Notation: X → U >c, d@

This distribution is also known as ‘Rectangular Distribution’ , as the graph of for

this describes a rectangle over the ; axis and between the ordi nates at ; c and

; d munotes.in

## Page 212

212NUMERICAL AND STATISTICAL METHODSProbability Density Function (p.d.f.)

p.d.f. of Uniform random variable is given by, f (x) ͳ

݀െܿ

if c < x < d

0 otherwise Note that, 0 ≤ f(x) ≤ 1 Axiom I is satisfied

݂ሺݔሻݔ݀ ൌଵ

ௗିௗ

ஶ

ିஶdx =ଵ

ௗିሾݔሿ

ௗൌௗି

ௗିൌͳ Axiom II is satisfied

(c, d) are parameters of the distribution.

Note: Whenever the value of parameter is known, the distributio n is known

completely. Knowing the parameter or parameters we can compute probabilities

of all events, as well as quantities such as mean, variance etc . Different values of

parameters specify different probability distributions of same kind.

A special case of uniform distribution very commonly used in di gital electronics

is a step function. This is a uniform distribution with paramet ers (0, 1) and its

p.d.f. is

݂ሺݔሻൌቄͳ݂݅Ͳ ൏ ݔ ൏ ͳ

Ͳ݁ݏ݅ݓݎ݄݁ݐܱ

1.2.2 Mean and Variance of Uniform Distribution

Mean of Uniform random variable with parameter ( c, d) is,

Mean E(;) ݂ݔሺݔሻݔ݀ ൌଵ

ௗିݔௗ

ஶ

ିஶdx =ଵ

ௗିቂ௫మ

ଶቃ

ௗൌௗమିమ

ଶሺௗିሻൌௗା

ଶ

Variance of Uniform random variable with parameter ( c, d) is,

Variance (X) = E(;2) – >E(;)@2

E(;2) ݔଶ݂ሺݔሻݔ݀ ൌଵ

ௗିݔଶௗ

ஶ

ିஶdx =ଵ

ௗିቂ௫య

ଷቃ

ௗൌௗయିయ

ଷሺௗିሻൌௗమାௗାమ

ଷ

Variance (X) = E(;2) – >E(;)@2 ௗమାௗାమ

ଷെሺௗାሻమ

ସൌሺௗିሻమ

ଵଶ

Standard Deviation (S.D.) (;) = σ ඥݎܸܽሺܺሻ ටሺௗିሻమ

ଵଶൌ ௗି

ଶξଷ munotes.in

## Page 213

213Chapter 13: Distributions: Continuous Distributions1.2. Applications of Uniform Distribution

1. It is used to represent the di stribution of rounding-off errors .

2. It is used in life testing and traffic flow experiments

3. It is being the simplest conti nuous distribution is widely appl ied in research.

4. If a random variable < follows any continuous distribution then i s

distribution function ; F (<) can be shown to follow U>0, 1@. This result

is useful in sampling from any continuous probability distribut ions.

Example1: If mean and variance of a U >c, d@ random variable are 5 and 3

respectively, determine the values of c and d.

Solutions: Let ; is a random variable which follows Uniform distribution with

parameters (c,d)

X → U >c,d@

Mean E(;) ௗା

ଶ 5 c d 10

Variance (X) = ሺௗିሻమ

ଵଶ= 3 ( d – c)2 36 d –c 6

c d 10 and d –c 6 after solving these equations we get c 2 and d 8

Example2: ; is continuous random variable with f(x) constant ( k) over a ≤ X ≤ b

and 0 elsewhere. Find i) p.d.f ii) Mean

Solutions:

i) P.d.f. is f (x) k For a ≤ X ≤ b 0 Elsewhere f ( x ) must satisfy Axioms of probability, Hence

݂ݔሺݔሻݔ݀ ൌݔ݀ݔ݇ ൌ ݇ ሺܾെܽሻൌͳ

ஶ

ିஶ

k ൌଵ

ሺିሻ

Thus the p.d.f. is f (x) ͳ

ܾെܽ For a ≤ X ≤ b 0 Elsewhere munotes.in

## Page 214

214NUMERICAL AND STATISTICAL METHODSii) Mean

න݂ݔሺݔሻݔ݀ ൌනͳ

ܾെܽݔ݀ݔ ൌͳ

ܾെܽ

ஶ

ିஶනݔ݀ݔ ൌͳ

ܾെܽ

ቈݔଶ

ʹ

ൌܾଶെܽ

ʹሺܾ െ ܽሻൌܾܽ

ʹ

Example: Let X → U >-a, a@. Find the value of a such that P>_;_@ ! 1

Solutions: P>_;_@ ! 1 1 - P[|X|] ≤ 1

1 - P >- 1 ≤ X ≤ 1]

P >-1 ≤ X ≤ 1] = ݂ݔሺݔሻݔ݀ ൌଵ

ଶݔ݀ଵ

ିଵଵ

ିଵൌଵ

a 7

Example4: Amar gives a birthday party to his friends. A machine fills the ice

cream cups. The quantity of ice cream per cup is uniformly dist ributed over 200

gms to 250 gms. i) what is the probability that a friend of Ama r gets a cup with

more than 230gms of ice cream" ii) If in all twenty five peop le attended the

party and each had two cups of ice cream, what is the expected quantity of ice

cream consumed in the party"

Solutions: Let X = Quantity of ice cream per cup, X → U >c, d@ → >200, 250@

i) P >; ! 230@ 1 – P >X ≤ 230@

1 – ሺଶଷିଶሻ

ሺଶହିଶሻൌͳെͲ Ǥ ൌͲ Ǥ Ͷ

ii) On an average, quantity per cup is given by

Mean (200 250) / 2 225 gms Total quantity consumed 225 î 2 î 25

11250 gms 11.25 kg

Example5: ; is Uniform continuous random variable with p.d.f. given as, f (x) ͳ

ͺ For 0 ≤ x ≤ 8 0 Elsewhere Determine i) P( 2 ≤ X ≤ 5 ) ii) P( 3 ≤ X ≤ 7 ) iii) P( X ≤ 6)

iv) P(; ! 6) v) P( 2 ≤ X ≤ 12 ) munotes.in

## Page 215

215Chapter 13: Distributions: Continuous DistributionsSolutions: i) P( 2 ≤ X ≤ 5 ) ଵ

଼ݔ݀ ൌଵ

଼ሾݔሿ

ଶହൌହ

ଶଷ

଼

ii) P( 3 ≤ X ≤ 7 ) ଵ

଼ݔ݀ ൌଵ

଼ሾݔሿ

ଷൌ

ଷସ

଼ൌଵ

ଶ

iii) P( X ≤ 6) ଵ

଼ݔ݀ ൌଵ

଼ݔ݀ ൌଵ

଼ሾݔሿ

ൌ

ିஶ

଼ൌଷ

ସ

iv) P(; ! 6) 1 – P (X ≤ 6) 1 - ଷ

ସൌଵ

ସ

v) P( 2 ≤ X ≤ 12 ) ൌଵ

଼ݔ݀ ൌଵ

଼ݔ݀ሺͲሻݔ݀ ൌଵଶ

଼଼

ଶଵଶ

ଶ

଼െͲൌଷ

ସ 1. Exponential Distribution

Exponential distribution is one of the important distributions and has wide

applications in operation research, life testing experiments, i n reliability theory

and survival analysis. Life time of an electronic component, ti me until decay of a

radioactive element is modeled by the exponential distribution. E x p o n e n t i a l

distribution is closely related t o Poisson distribution. If the number of occurrences

of the event follows Poisson distribution then the distribution o f t h e t i m e t h a t

elapses between these successive occurrences follows an exponen tial distribution.

For example, if the number of patients arriving at hospital fol lows Poisson

distribution, then the time gap between the current arrival and the next arrival is

exponential. This has specific application in 4ueuing Theory.

1..1 Definition: Let ; be a continuous random variable taking non-negative

values is said to follow expone ntial distribution with mean θ if its probability

density function (p.d.f.) is given by f (x) ͳ

ߠ݁ି௫Ȁఏ x ≥ 0, θ > 0 0 otherwise Notation: X → Exp(θ)

Note : (1) f(x) ≥ 0 x ≥ 0, θ > 0 and ݁ିೣ

ഇͲ

(2) ݂ݔሺݔሻݔ݀ ൌଵ

ఏ݁ି௫Ȁఏݔ݀ ൌଵ

ఏஶ

ஶ

ିஶቂെషೣȀഇ

ଵȀఏቃ

ஶൌൣ െ ݁ି௫Ȁఏ൧

ஶൌͳ

From (1) and (2) it is clear that f(x) is p.d.f. munotes.in

## Page 216

216NUMERICAL AND STATISTICAL METHODS () if θ = 1, then the distribution is called standard exponential distribu tion.

Hence probability density function is given by f (x) ݁ି௫ x ≥ 0, θ > 0 0 otherwise 1..2 Mean and Variance of Exponential Distribution

If X → Exp(θ) , then its p.d.f is given by f (x) ͳ

ߠ݁ି௫Ȁఏ x ≥ 0, θ > 0 0 otherwise Mean of Exponential random variable with parameter (0 ,∞) is,

Mean E(;) ݂ݔሺݔሻݔ݀ ൌଵ

ఏ݁ି௫Ȁఏݔ݀ݔஶ

ஶ

=௫

ఏ݁ି௫Ȁఏݔ݀ஶ

ൌ ߠ

Variance of Exponential random variable with parameter (0 ,∞) is,

Variance (X) = E(;2) – >E(;)@2

E(;2) ݔଶ݂ሺݔሻݔ݀ ൌଵ

ఏ݁ି௫Ȁఏݔݔଶݔ݀ஶ

ஶ

=௫మ

ఏ݁ି௫Ȁఏݔ݀ஶ

ൌʹ ߠଶ

Variance (X) = E(;2) – >E(;)@2 ʹߠଶെߠଶൌߠଶ

Standard Deviation ( S.D.) (X)= σ = ඥݎܸܽሺܺሻ ξߠଶൌ ߠ

Note: For Exponential distribution, mean and standard deviation are same.

13.3.3 Distribution Function of Exponential Distribution i.e. E xp (θ)

Let X → Exp(θ). Then the distribution function is given by

Fx (;) P > X≤ x@ ݂ሺݐሻݐ݀ ൌଵ

ఏ݁ି௧Ȁఏݐ݀ ൌ௫

௫

ଵ

ఏ݁ି௧Ȁఏݐ݀ ൌ௫

ଵ

ఏቂെషȀഇ

ଵȀఏቃ

௫

ൣെ݁ି௧Ȁఏ൧

௫

Fx (;) 1- ݁ି௫Ȁఏ x ! 0, θ > 0

Note: When ; is life time of a component, P >; ! x@ is taken as reliability

function or survival function.

Here P >; ! x@ 1 – P >X ≤ x@ 1 - F x (;) 1- ݁ି௫Ȁఏ x ! 0, θ > 0 munotes.in

## Page 217

217Chapter 13: Distributions: Continuous Distributions1..4 Applications of Exponential Distribution

Exponential distribution is use d as a model in the following si tuations:

1) Life time of an electronic component

2) The time between successive r adioactive emissions or decay

3) Service time distribution at a service facility.

4) Amount of time until interrupt occurs on server

5) Time taken to serve customer at ATM counter.

Example1: Suppose that the life time of a ;<= company T.V. tube is

exponentially distributed with a mean life 1600 hours. What is the probability that

i) The tube will work upto 2400 hours"

ii) The tube will survive after 1000 hours"

Solutions: Let ; : Number of hours that T.V. tube work

X → Exp(θ) where θ 1600

We know that if X → Exp(θ) then

P >X ≤ x@ 1- ݁ି௫Ȁఏ x ! 0, θ > 0

i) P >X ≤ 2400 @ 1- ݁ିଶସȀଵ 1 - ݁ିଵǤହ 1 – 0.223130 0.77687

ii) P >; ! 1000@ - ݁ିଵȀଵൌ݁ିǤଶହൌͲ Ǥ ͷ ͵ ͷ ͵ ͵ ͳ ͻ

( If X → Exp(θ) , P >; ! x@ 1- ݁ି௫Ȁఏൌ݁ି௫Ȁఏ)

Example2: T h e l i f e t i m e i n h o u r s o f a c e r t a i n e l e c t r i c c o m p o n e n t f o l l o w s

exponential distribution with distribution function.

F(x) 1- ݁ିǤସ௫; x ≥ 0

i) What is the probability that the component will survive 200 hou rs"

ii) What is the probability that it will fail during 250 to 350 hou rs"

iii) What is the expected life time of the component"

Solutions: Let ; : Life time (in hours) of the electric component, X → Exp(θ)

and

F(x) 1- ݁ିǤସ௫ ; x ≥ 0 ݁ିǤସ௫ൌ݁ି௫Ȁఏ munotes.in

## Page 218

218NUMERICAL AND STATISTICAL METHODS െͲǤͲͲͶݔൌെݔȀߠ ͲǤͲͲͶൌ ͳȀߠ ߠൌଵǤସൌʹ ͷ Ͳ ݏݎݑ݄ i) P >; ! 200@ ݁ି௫Ȁఏൌ݁ିమబబ

మఱబൌ݁ିǤ଼ൌͲ Ǥ Ͷ Ͷ ͻ ͵ ʹ ͻ ii) P >250 ; 350@ P>; ! 250@ – P >; ! 350@ ݁ିమఱబ

మఱబെ݁ିయఱబ

మఱబ ݁ିଵ - ݁ିଵǤସ

0.367879 – 0.246597

iii) Expected life time of the component = E(X) = θ = 250 hours

Example: The life time of a microprocessor is exponentially distributed with

mean 3000 hours. What is the probability that

i) The microprocessor will fail within 300 hours"

ii) The microprocessor will function for more than 6000 hours"

Solutions: Let ; : Life time (in hours) of the microprocessor,

X → Exp(θ) and θ = 3000

i) P (Microprocessor will fail within 300 hours)

P ( X ≤ 300)

ଵ

ଷ݁ିೣ

యబబబݔ݀ ൌ ͲǤͲͻͷͳଷ

ii) P (Microprocessor will function for more than 6000 hours)

P ( X ≥ 6000 )

ଵ

ଷ݁ିೣ

యబబబݔ݀ ൌ ݁ିଶൌͲ Ǥ ͳ ͵ ͷ ͵ஶ

1.4 Normal Distribution

In this section we deal with the most important continuous dist ribution, known as

normal probability distribution or simply normal distribution. It is important for

the reason that it plays a vital role in the theoretical and applied statistics. It is

one of the commonly used distribution. The variables such as in telligence

quotient, height of a person, weight of a person, errors in mea surement of

physical quantities follow normal distribution. It is useful in s t a t i s t i c a l q u a l i t y munotes.in

## Page 219

219Chapter 13: Distributions: Continuous Distributionscontrol, statistical inferen ce, reliability theory, operation r esearch, educational and

psychological statistics. Normal distribution works as a limiti ng distribution to

several distributions such as Binomial, Poisson.

The normal distribution was first discovered by DeMoivre (Engli sh

Mathematician) in 1733 as limiting case of binomial distributio n. Later it was

applied in natural and social science by Laplace (French Mathem atician) in 1777.

The normal distribution is also known as Gaussian distribution in honour of Karl

Friedrich Gauss(1809).

1.4.1 Definition:

A continuous random variable ; is said to follow normal distrib ution with

parameters mean μ and standard deviation σ2, if its probability density function

(p.d.f.) is

݂ሺݔሻൌͳ

ߪξʹߨ݁ିଵ

ଶఙమሺ௫ିρሻమ

Ǣെλ ൏ ݔ ൏ λǢെλ ൏ ρ ൏ λǢߪ Ͳ

݂ሺݔሻൌͳ

ߪξʹߨ݁ିሺ௫ିρሻమ

ଶఙమǢെλ ൏ ݔ ൏ λǢെλ ൏ ρ ൏ λǢߪ Ͳ

Note:

1. The mean μ and standard deviation σ2 are called the parameters of Normal

distribution.

N o t a t i o n : T h e n o r m a l d i s t r i b u t i o n i s e x p r e s s e d b y X → N(μ, σ2)

2. If 0 and σ2 1,then the normal variable is called as standard normal

variable.

X → N(μ, σ2) => X → N(0, 1) . Generally it is denoted by =. The p.d.f.

of = is

݂ሺݖሻൌͳ

ξʹߨ݁ି௭మ

ଶǢെλ ൏ ݖ ൏ λǢߨ݄ݐ݅ݓ ൌ ͵ǤͳͶͳͳͷͻ݁݀݊ܽ ൌ ʹǤͳͺʹͺ

The advantage of the above function is that it doesn’ t contain any

parameter. This enable us to compute the area under the normal probability

curve.

3. The probability density curve of N(μ, σ2) is bell shaped, symmetric about

and mesokurtic . Naturally, the curve of standard normal distri bution is

symmetric around zero. munotes.in

## Page 220

220NUMERICAL AND STATISTICAL METHODS

4. The maximum height of probability density curve is 1 / ( σ ξʹߨሻ

5. As the curve is symmetric about , the mean, median and mode c oincide

and all are equal to .

6. The parameter σ2 is also the variance of ;, hence the standard deviation

(s.d.) (X) = σ

Relation between N(μ, σ2) and N(0, 1)

If X → N(μ, σ2) then = ሺିρሻ

՜ۼሺǡሻǤ This result is useful while computing

probabilities of N(μ, σ2) variable. The statistical table give probabilities of a

standard normal i.e. N(0,1) variable.

1.4.2 Properties of Normal Distribution: ¾ The normal curve is bell shaped a nd is symmetric at x P

¾ Mean, median, and mode of the distribution are coincide

i.e.,Mean Median Mode

¾ It has only one mode at x (i.e., unimodal)

¾ The points of inflection are at x PrV

¾ The x axis is an asymptote to the curve (i.e. the curve continu es to approach

but never touches the x axis)

¾ The first and third quartiles are equidistant from median.

¾ The mean deviati on about m ean is 0.8 V

¾ 4uartile deviation 0.6745 V

¾ If ; and < are independent normal variates with mean P1 and P2, and

variance V12 and V22 respectively then their sum (; <) is also a normal

variate with mean PP and variance VV)

munotes.in

## Page 221

221Chapter 13: Distributions: Continuous Distributions¾ Area Property P PVuPV 0.6826

P PVuPV 0.9544

P PVuPV 0.9973 1.4. Properties and Applications of Normal Distribution Let ; be random variable which follows normal distribution with m e a n P a n d

variance V2 .The standard normal variate is defined as ܼൌሺିρሻ

which follows

standard normal distribution with mean 0 and standard deviation 1 i.e. = → N

(0,1). The p.d.f. of = is

݂ሺݖሻൌͳ

ξʹߨ݁ି௭మ

ଶǢെλ ൏ ݖ ൏ λǢߨ݄ݐ݅ݓ ൌ ͵ǤͳͶͳͳͷͻ݁݀݊ܽ ൌ ʹǤͳͺʹͺ

The advantage of the above function is that it doesn’t contain any parameter. This

enable us to compute the area under the normal probability curv e.

Area Properties of Normal Curve:

The total area under the normal probability curve is 1. The cur ve is also called

standard probability curve. The area under the curve between th e ordinates at x

a and x b where a b, represents the probabilities that x li es between x a and

x b

i.e., P(a d x d b)

To find any probabil ity value of ;, we first standardize it by using ܼൌሺିρሻ

, and

use the area probability normal table.

For e.g. the probability that the normal random variable ; to l ie in the interval

(- σ, μ + σ) is given by

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## Page 222

222NUMERICAL AND STATISTICAL METHODS

PPV[PV Pd]d

2P(0 z 1) (Due to symme try)

2 (0.3413) (from the area ta ble)

0.6826 PPV[PV P(-2 z 2)

2P(0 z 2) (Due to s ymmetry)

2 (0.4772)

0.9544

PPV[PV P]

2P(0 z 3) (Due to s ymmetry)

2 (0.49865) 0.9973

munotes.in

## Page 223

223Chapter 13: Distributions: Continuous DistributionsThus we expect that the values in a normal probability curve will lie between the

range

Pr 3V, though theoretically it range from f to f.

Example 1: Find the probability that the standard normal variate lies betw een 0

and 1.56

Solution: P(0 z 1.56) Area between z 0 and z 1.56

0.4406 (f rom table)

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## Page 224

224NUMERICAL AND STATISTICAL METHODSExample2: Find the area of the standard normal variate from –1.96 to 0.

Solution: Area between z 0 z 1.96 is same as the area z 1.96 to z 0

P(-1.96 z 0) P(0 z 1.96) (by symmetry)

0.4750 (from the table)

Example : Find the area to the right of z 0.25

Solution: P(z ! 0.25) P(0 z f) – P(0 z 0.25)

0.5000 - 0.0987 (from the table) 0.4013

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## Page 225

225Chapter 13: Distributions: Continuous DistributionsExample 4: Find the area to the left of z 1.5

Solution: P(z 1.5) P(- ∞ < z < 0 ) P(0 z 1.5)

0.5 0.4332 (from the tab le)

0.9332

Example 5: Find the area of the standard normal variate between –1.96 and 1.5

Solution: P(-1.96 z 1.5) P (-1.96 z 0) P(0 z 1.5)

P (0 z 1.96) P(0 z 1.5)

0.4750 0.4332 (from the table)

0.9082

Example 6: Given a normal distribution with P 50 and V 8, find the

probability that x assumes a value between 42 and 64

Solution: Given that P 50 and V 8 The standard normal variate ܼൌሺିρሻ

munotes.in

## Page 226

226NUMERICAL AND STATISTICAL METHODSIf ; 42, ܼଵൌሺିρሻ

ൌሺସଶିହሻ

଼ൌି଼

଼ -1

If ; 64, ܼଶൌሺିρሻ

ൌሺସିହሻ

଼ൌଵସ

଼ 1.75

? P(42 x 64) P( 1 z 1.75)

P( 1 z 0) P(0 z 1.95)

P(0z1) P (0 z 1.75) (by s ymmetry)

0.3413 0 .4599 (from the table)

0 .8012

Example 7: Students of a class were given an aptitude test. Their marks we re

found to be normally distributed with mean 60 and standard devi ation 5. What

percentage of students scored.

i) More than 60 marks (ii) Le ss than 56 marks (iii) Between 45 and 65 marks

Solution: Given that P 60 and V 5 The standard normal variate ܼൌሺିρሻ

i) If ; 60, ܼൌሺିρሻ

ൌሺିሻ

ହൌ

ହ 0

? P(x ! 60) P(z ! 0)

P( < z < ∞) = 0.5000

Hence percentage () of students scored more than 60 marks is

0.5000
100 50

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## Page 227

227Chapter 13: Distributions: Continuous Distributions

ii) If ; 56, ܼൌሺିρሻ

ൌሺହିሻ

ହൌିସ

ହ - 0.8

? P(x 56) P(z -0.8)

P(- f z 0) – P(0.8 z 0) (by symmetry)

P(0 z f) – P(0 z 0.8)

0.5 0.2881 (from the table)

0.2119

Hence the percentage of students score less than 56 marks is

0.2119
(100) 21.19

iii) If ; 45, ܼൌሺିρሻ

ൌሺସହିሻ

ହൌିଵହ

ହ -3

; 65, ܼൌሺିρሻ

ൌሺହିሻ

ହൌହ

ହ 1

P(45 x 65) P( 3 z 1)

P( 3 z 0) P (0 z 1)

P(0 z 3) P(0 z 1) (by symmetry)

0.4986 0.3413 (from the table)

0.8399

Hence the percentage of students scored between 45 and 65 marks i s

0.8399
(100) 83.99

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## Page 228

228NUMERICAL AND STATISTICAL METHODS

Example : ; is normal distribution with mean 2 and standard deviation 3. Find

the value of the variable x s uch that the probability of the in terval from mean to

that value is 0.4115

Solution: Given 2, σ 3

Suppose z 1 is required standard value,

Thus P (0 z z 1) 0.4115

From the table the value co rresponding to the area 0.4115 is 1. 35 (i.e. z 1 1.35)

ൌሺെρሻ

ોൌሺ܆െሻ

ൌ Ǥ

(; – 2) 3
1.35

; 4.05 2 6.05 Example : In a normal distribution 31 of the items are under 45 and 8 are

over 64. Find the mean a nd variance of the distribution.

Solution: Let x denotes the items are given and it follows the normal d istribution

with mean and standard deviation σ

The points x 45 and x 64 are located as shown in the figure .

Since 31 of items are under x 45,

position of x into the left of the ordinate x

Since 8 of items are above x 64,

position of this x is to the right of ordinate x

munotes.in

## Page 229

229Chapter 13: Distributions: Continuous DistributionsWhen ; 45, ܼൌሺିρሻ

ൌሺସହିρሻ

- z 1 (Say)

Since x is left of x P, =1 is taken as negative

When ; 64, ܼൌሺିρሻ

ൌሺସିρሻ

z 2 (Say)

From the diagram P(x 45) 0.31

P(z - z 1) 0.31

P(- z 1 z 0) P(- f z 0) – P(- f z z 1)

P(- z 1 z 0) 0.5 - 0.31 0.19

P(0 z z 1) 0.19 (by symmetry)

z 1 0.50 (from the table)

Also from the diagram P(; ! 64) 0.08

P(0 z z 2) P(0 z f) – P(z 2 z f)

0.5 - 0.08 0.42

z 2 1.40 (from the table)

Substituting the values of z 1 and z 2 we get ܼൌሺିρሻ

ൌሺସହିρሻ

ൌെͲǤͷͲ

ܼ ൌሺିρሻ

ൌሺସିρሻ

ൌͳ Ǥ Ͷ Ͳ

Solving P - 0.50 V 45 (1)

P 1.40 V 64 ----- (2)

(2) – (1) 1.90 V 19 V 10

Substituting V 10 in (1)

P 45 0.50 (10) 45 5.0 50.0

Hence mean 50 and variance V2 100 1.4.4 = table and its User Manual

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## Page 230

230NUMERICAL AND STATISTICAL METHODS

Patterns for Finding Areas Under the Standard Normal (a) Area between

a given z value

d0Use Table A in Appendix I 0 z (b) Area between

z values on

either side of 0 z1 0z1 0

Af t0Area from z1 to Area from 0 to (c) Area between

z values on

same side of 0 —0 z10 z2 Area between z1 and z2 A r e a f r o m 0

t0 z1

Af 0 t(d) Area to the right

of a positive z value

or to the left of a

negative z value —0 z1 0 0 z1

Area to the right of z1 Area to the right of 0* Area from 0 to

z1 (e) Area to the right

of a negative z value

or to the left of a

positive z1 0

At t h i h t fz1 0

Af t0

Area to the right of 0

hi h i 0 5000FIGURE 2 munotes.in

## Page 231

231Chapter 13: Distributions: Continuous DistributionsTABLE A Areas of a Standard Normal Distribution

(Alternate Version of Table)

The table entries represent the area under the standard normal curve from

0 to the specified value of z. z .00 .01 .02 .0 .04 .05 .06 .07 .0 .0 0.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359 0.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753 0.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141 0.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517 0.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879 0.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224 0.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2517 .2549 0.7 .2580 .2611 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852 0.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133 0.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389 1.0 .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3577 .3599 .3621 1.1 .3643 .3665 .3686 .3708 .3729 .3749 .3770 .3790 .3810 .3830 1.2 .3849 .3869 .3888 .3907 .3925 .3944 .3962 .3980 .3997 .4015 1.3 .4032 .4049 .4066 .4082 .4099 .4115 .4131 .4147 .4162 .4177 1.4 .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .4319 1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441 1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545 1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633 1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706 1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767 2.0 .4772 .4778 .4783 .4788 .4793 .4798 .4803 .4808 .4812 .4817 2.1 .4821 .4826 .4830 .4834 .4838 .4842 .4846 .4850 .4854 .4857 2.2 .4861 .4864 .4868 .4871 .4875 .4878 .4881 .4884 .4887 .4890 2.3 .4893 .4896 .4898 .4901 .4904 .4906 .4909 .4911 .4913 .4916 2.4 .4918 .4920 .4922 .4925 .4927 .4929 .4931 .4932 .4934 .4936 2.5 .4938 .4940 .4941 .4943 .4945 .4946 .4948 .4949 .4951 .4952 2.6 .4953 .4955 .4956 .4957 .4959 .4960 .4961 .4962 .4963 .4964 2.7 .4965 .4966 .4967 .4968 .4969 .4970 .4971 .4972 .4973 .4974 munotes.in

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232NUMERICAL AND STATISTICAL METHODSz .00 .01 .02 .0 .04 .05 .06 .07 .0 .0 2.8 .4974 .4975 .4976 .4977 .4977 .4978 .4979 .4979 .4980 .4981 2.9 .4981 .4982 .4982 .4983 .4984 .4984 .4985 .4985 .4986 .4986 3.0 .4987 .4987 .4987 .4988 .4988 .4989 .4989 .4989 .4990 .4990 3.1 .4990 .4991 .4991 .4991 .4992 .4992 .4992 .4992 .4993 .4993 3.2 .4993 .4993 .4994 .4994 .4994 .4994 .4994 .4995 .4995 .4995 3.3 .4995 .4995 .4995 .4996 .4996 .4996 .4996 .4996 .4996 .4997 3.4 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4998 3.5 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 3.6 .4998 .4998 .4998 .4999 .4999 .4999 .4999 .4999 .4999 .4999 For values of z greater than or equal to 3.70, use 0.4999 to approximate the

shaded area under the standard normal curve.

1.5 Summary

In this chapter, Continuous Distributions , its types Uniform, Exponential with

its mean , variance and its application is discussed.

A special and very useful distribution called as Normal distrib ution and its

application in day to day life also discussed. Distribution Definition Mean

E(;) Variance

(;) Uniform X → U [ c, d ] f ( x ) ͳ

݀െܿ

if c < x < d

0 otherwise

ሺ݀ ܿሻ

ʹ

ሺ݀ െܿሻଶ

ʹ Exponential f ( x ) ͳ

ߠ݁ି௫Ȁఏ

x ≥ 0 , θ > 0 0 otherwise

ߠ

ߠଶ

Normal distribution with paramete rs mean and standard deviati on

σ2, X → N(μ, σ2)

݂ሺݔሻൌͳ

ߪξʹߨ݁ିଵ

ଶఙమሺ௫ିμሻమ

Ǣെ∞൏ݔ൏∞Ǣെ∞൏μ൏∞ǢߪͲ munotes.in

## Page 233

233Chapter 13: Distributions: Continuous Distributions

1.6 Unit End Exercise

1 If ; ĺ U (2, 8 ) State Mean and Variance of ;.

>Hints and Answers :Mean 5, Variance 3 @

2 If ; is a continuous random variable having uniform distributi on with parameters

‘a’ and ‘b’ such that mean 1 and variance 3.Find P(; 0)

>Hints and Answers : 1/3 @

3 On a route buses travels at an half hourly intervals, starting at 8.00 am. On

a given day, a passenger arrives at the stop at a time ;, which is uniformly

distributed over the interval >8.15am, 8.45am @. What is the probability that the

passenger will have to wait for more than 15 minutes for the bu s"

>Hints and Answers: 1/2 @

4 The mileage which car owners get with certain kind of radial t yres ( measured

in ‘000’ km) is a random variable with mean 40. Find the probab ilities that one

of these tyres will last.

i) At least 20000 km

ii) At most 30000 km

>Hints and Answers: i) 0.6065 ii) 0.5276 @

5 The lifetime of a microprocessor is exponentially distributed with mean 3000

hours. Find the probability that

i) The microprocessor will fail within 300 hours

ii) The microprocessor will function for more than 6000 hours

>Hints and Answers: i) 0.0951 ii) 0.1353 @

6 The time until next earthquake occurs in a particular region i s assumed to be

exponentially distributed with mean 1/2 per year. Find the prob ability that the

next earthquake happens

i) Within 2 years

ii) After one and half year

>Hints and Answers: i) 0.9817 ii) 0.0497 @munotes.in

## Page 234

234NUMERICAL AND STATISTICAL METHODS

7 Let ; be normally distributed random variable with parameters ( 100,25)

Calculate:

i) P( ; 108 ) ii) P ( 90 ; 110)

>Hints and Answers: i) 0.27425 ii) 0.9545 @

8 If ; is a standard normal variable > ; ĺN(0,1)@, determine th e following

probabilities using normal probability tables

(i) P(; 1.3) (ii) P(0;1.3) (iii) P(; 1.3) (iv) P(-2;1. 3) (v)P(0.5;1.3)

>Hints and Answers: (i)0.09680 (ii) 0.4032 (iii)0.9032 (iv)0.88045

(v)0.21174 @

9 The annual rainfall (in inches ) in a certain region is normal ly distributed with

mean 40 and standard deviation 4. What is the probability t hat in the current

year it will have rainfall of 50 inches or more

>Hints and Answers: i) P(;50) 0.0062097 @

10 The intelligent quotient (I4) of adults is known to be normal ly distributed with

mean 100 and variance 16. Calculate the probability that a rand omly selected

adult has I4 lying between 90 and 110.

>Hints and Answers: 0.98758 @

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