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CONTENTS

Unit No. Title Page No.

SEMESTER - II

1. Functions, Derivatives and theirApplications 01

2. Simple Interest and Compound Interest 34

3. Annuities and EMI 47

4. Correlation and Regression 62

5. Time Series 84

6. Index Numbers 110

7. Probability Distributions 129

munotes.in

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1

UNIT I

Unit -1

FUNCTIONS, DERIVATIVES AND

THEIR APPLICATIONS

Unit Structure :

1.0 Objectives

1.1 Introduction

1.2 Derivatives

1.3 Second Order Derivatives

1.4 Applications of Derivatives

1.5 Maxima and Minima

1.0OBJECTIVE S

After reading this chapter you will be able to recognize.

1)Defin ition of function.

2)Standard Mathematical function .

3)Definition of derivative.

4)Derivatives of standard function s.

5)Second order derivatives.

6)Application of derivatives.

7)Maxim a and Minima .

1.1 FUNCTION S

If y= f(x) is a function then the set of all values of x for which this

function is defined is called the domain of the function ƒ. Here xis called

an independent variable and y is called the dependent variable. The set of

all co rresponding values of y for xin the domain is called the range of the

functionƒ.

The function ƒis defined from the domain to the range.

We shall discuss only those functions where the domain and the

range are subsets of real numbers. Such fun ctions are called 'real valued

functions'.munotes.in

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1.1.1 Standard Mathematical Functions:

(1) Constant function:

The constant function is defined by

y=ƒ(x)= C where C is a constant.

The c onstants are denoted by real numbers or alphabets. The graph

of a constant function is a straight line parallel to x -axis.

Examples:

y=ƒ(x)= 5

y=ƒ(x)=-10

y=ƒ(x)=K

y=ƒ(x)=a

(2) Linear function :

The linear function is defined by y =ƒ(x)= ax+ b where a and b are

constants.

Examples: y= ƒ(x)=2x+5

y=ƒ(x)=-3x+10

y=ƒ(x)= 5x-7

(3) Functions with power of x:

A function ƒ(x)=xnis called power function or function with

power of x. Here xis called base and n is called power .

Examples : ƒ(x) =x2

ƒ(x)=x-5

ƒ(x)=x-4/3ƒ(x)=x3/2

(4) Exponential functions :

These functions are of the type .

ƒ(x) = exand

ƒ(x) = ax, a >0

(5) Lo garithmic function : The logarithmic function is defined by

y=ƒ(x) = log ex,x>0

1.1.2. Standard functions from Economics :

(1) Demand : It refers to the quantity of a product is desired by the

buyers . The demand depends on the pric e. T herefore , there is a

relationship between the price and the quantity demanded. Such

relationship is called a demand function.

Hence the demand function is defined as

D=g(p) where D= demand and p= price .munotes.in

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Here demand is a dependent variable and the price is an

independent variable.

For example, D= 50 + 4 p -3p2

(2) Supply : It refers to the quantity of a product , the market can offer .

The supply depends on the price. Therefore, There is a relationship

between the price and the quantity supplied. Such relationship is called a

supply function.

Hence the supply function is defined as S = ƒ(p) where S = supply and

p = price.

Here supply is a dependent variable and price is an independent

variable.

For example , S = 2p2-6p + 25

(3) Break -even Point : Equilibrium point.

(i) By the law of demand, the demand decreases when the price increases,

the demand curve is a decreasing curve as shown in the figure :

D

D=ƒ(p)

The demand curve

P

(ii)By the law of supply, the supply increases when the price increases,

the supply curve is an increasing curve as shown in the figure.

S = g(p)

The supply curve

(iii) The demand and supply curves D= ƒ(p) and S= g(p) are intersecting

at a point . The point of intersection of the demand and supply curves

represents that specific price at which the demand and supply are equal.

This point is called the Break -even point or equ ilibrium point. The

corresponding price at which this point occu rs is called an equilibrium

price and is denoted by p e

At equilibrium price, the amount of goods supplied is equal to the

amount of goods demanded.munotes.in

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q D = f(p) S = (p)

Break –even point

q=ƒ(p) =D

q= g(p) = S

o pe p

equilibrium price -Break -even point

(4) The total cost function :

The total cost function or cost function is denoted by C and it is

expressed in terms of x. If C is the cost of producing xunits of a p roduct ,

then C is generally a function of xand is called the total cost function .

i.e.C=ƒ(x)

For example , C= 2 x2-5x+ 10

(5) Average cost function :

The ratio between the cost function a nd the number of units produced is

called average cost function . i.e. AC =C

x

For example , AC = x2+2x+5

x

(6) Total Revenue function :

The total revenue function is defin ed as in terms of the demand

and the price per item. If D units are demanded with the selling price of

p per unit , then the total revenue function R is given by

R = p x D where p = price a nd D= demand

For example ,

If D = p2+2p +3 th en R = p x (p2+2p +3) = p3+2p2+3p

(7) Average revenue :

Average revenue is defined as the ratio between the revenue and the

demand and is denoted by AR.

i.e. , AR =R/D AR = p x D/D (as R= p x D)

AR = p

Average revenue is nothing but the selling price per unit.

(8) The Profit function :

The profit function or the total profit function is denoted by P and

is defined by the difference between the total revenue and the total cost.

Total Profit = Total Revenue -Total cost

i.e. P = R -C

Example 1:

Find the total profit function if the cost function C= 40 + 15 x-x2,

x= number of items produced and the demand function is p= 200 -x2munotes.in

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Solution :

Given C= 40 + 15 x-x2

p = 200 -x2

R = p x D (D= x)

= (200 -x2)x

R = 200 x-x3

Profit = Revenue -Cost

P = R -C

= (200 x-x3)-( 40 + 15 x-x2)

= 200 x-x3-40-15x+x2

P= 185 x+x2-x3-40

Example 2:The total cost function is C= 20 -3x2and the demand

function is p= 5 + 6 x. Find the profit when x=100.

Solution :

Given : C= 20 -3x2

R= p x D (D= x)

= p x x

= (5 + 6 x)x

= 5x+ 6x2

Profit = Revenue -Cost

=R-C

= (5x+ 6x2)-( 20-3x2)

= 5x+ 6x2-20 + 3 x2

= 5x+ 9x2-20

When x= 100 , P =5(100) + 9 (100)2-20

=500 + 90000 -20

= 90480.

1.2DERIVATIVES

1.2.1 Derivativ e as rate measure :

Definition : Let y= ƒ(x) be the given function .

If lim [ƒ(x+h)-ƒ(x)]exists ,

h→0 h

then we say that the function ƒ(x) has derivative at xand is denoted by ƒ'

(x).

i.e. ,ƒ' (x) = lim [ƒ(x+h)-ƒ(x) ]

h→0 h

The rate of change is called the " derivative " of y= ƒ(x) with

respect to xand is denoted by dyorƒ' (x).

dx

dy= the rate of change of y with respect to xor the derivative of y

dx with respect to x.

Note : (1) Derivative means " rate of change "

(2)The process of finding the derivative of a fu nction is called "

differentiation".munotes.in

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dC= the rate of change cost with respect to x.

dx

For example dD=therateof change of demand with respect to p .

dp

1.2.2 Derivatives of Standard functions :

(1)Ify=xn, where n is a real number , then

dy= nxn-1

dx

i.e. , dy=d (xn)= nxn-1

dx dx

(2)If y = C , where C is a constant ,

then dy=0

dx

i.e. , dy=d (C) = 0

dxdx

(3) If y = ex, then dy=ex

dx

i.e. , dy=d (ex)=ex

dxdx

(4)If y = ax, where a isapositive real number , then

dy= axlog a

dx

i.e. , dy=d (ax)= axlog a

dx dx

(5) If y= log x, then dy=1

dxx

where x> 0

i.e. , dy=d (log x)=1

dx dx x

Examples :

(1) y= x=x'dy= 1.x1-1= 1.x0=1

dx

(2) y= x4dy= 4.x4-1= 4.x3

dx

(3) y= x10dy= 10. x10-1= 10x9

dx

(4) y= 1=x-1dy=-1.x-1-1=-1x-2=-1

x dx x2munotes.in

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(5) y= 1=x-3dy=-3x-3-1=-3x-4=-3

x3dx x4

(6) y=√x=x1/2dy=1x1/2-1=1x-1/2=1 =1

dx2 2 2 x1/22√x

(7) y= x5/2dy=5x5/2-1=5x3/2

dx2 2

(8) y= x-3/2

dy=-3x-3/2-1=-3x-5/2

dx 2 2

(9) y= x-7/2

dy=-7x-7/2-1

dx2

=-7x-9/2

2

(10) y=5 , 5 is a constant

dy= 0

dx

(11) y=K , dy=0

dx

(12) y= log2 , dy=0

dx

(13) y= -10dy= 0

dx

(14) y= exdy= ex

dx

(15) y= 2xdy= 2xlog 2

dx

(16) y= 10xdy= 10xlog 10

dx

(17) y= log xdy=1

dxx

Exercise: 1.1

Find dyfor the following :

dx

(1) y = x6

(2) y = 1

x2

(3) y = x7/2munotes.in

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(4) y = x-5/2

(5) y = 3

(6) y = log 10

(7) y = -8

(8) y = 4x

(9) y = 9x

(10)y = 15x

Answers :

(1) 6 x5(2)-2(3)7x5/2(4)-5x-7/2(5) 0 (6) 0 (7) 0 (8) 4xlog 4

x32 2

(9) 9xlog 9 (10) 15xlog 15

1.2.3 Rules of derivatives :

Rule : 1 Addit ion Rule (or) Sum rule :

If y= u + v where u and v are differentiable functions of xthen

dy=du+dv

dx dxdx

i.e,dy=d(u+v) =du+du

dx dx dxdx

Examples:

(1) If y = x2+ ex, find dy

dx

Solution : Given : y = x2+ ex

dy=d (x2+ ex)

dx dx

=d(x2)+d (ex)

dx dx

dy= 2x+ ex

dx

(2) If y= x10+ log x, find dy

dx

Solution: Given : y= x10+ log x

dy=d[x10+logx]

dxdx

=d(x10) +d(logx)

dx dx

dy= 10 x9+1

dx x

Rule:2 Subtraction Rule (or) Difference rule:

If y= u -v where u and v are differentiable functions of xthenmunotes.in

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dy=du-dv

dxdxdx

i.e,dy=d(u-v) = du-dv

dxdx dxdx

Examples :

(1) If y= x5-2x, find dy

dx

Solution : Given : y = x5-2x

dy=d (x5-2x)

dx dx

=d(x5)-d(2x)

dx dx

dy= 5x4-2xlog2

dx

(2) If y = 100 -logx, find dy

dx

Solution : Given : y = 100 -logx

dy=d(100 -logx)

dx dx

=d(100) -d(logx)

dx dx

= 0 -1

x

dy=-1

dx x

Rule : 3 Product Rule :

If y = uv where u and v are differentiable functions of x,

then dy= udv+ vdu

dx dx dx

i.e. , dy=d(uv) = u dv+ vdu

dx dx dx dx

Examples :

(1) If y = x4logx, find dy

dx

Solution : Given y = x4logx

u =x4, v = log x

dy=d(x4logx)

dxdxmunotes.in

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=x4d(logx)+logxd(x4)

dx dx

=x4(1/x) + log x(4x3)

=x3+ 4x3logx

dy=x3[ 1+ 4 log x]

dx

(2) If y = x2ex, find dy

dx

Solution : Given y = x2ex

dy=x2d(ex) + exd(x2)

dx dx dx

=x2ex+ex(2x)

dy= ex[x2+ 2x]

dx

Rule :4 Quotient Rule :

If y = u, v≠0where u and v are differentiable functions ofx, then

2du dvv udy dx dx

dx v

Examples :

(1) If y = x+ 4, find dy

logx dx

Solution: Given y= x+ 4

logx

Here u= x+ 4

v= log x

dy= log xd(x+ 4 ) -(x+ 4 ) d( logx)

dx dx dx

(logx)2

=(logx)(1+0) -(x+ 4) ( ¼)

(logx)2

dy=( logx)-(x+ 4) ( ¼)

dx (logx)2

(2) If y = ex+5, find dy

x6-10 d x

Solution :

Given : y = ex+5

x6-10munotes.in

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Here u = ex+ 5 ; v= x6-10

dy= (x6-10)d(ex+5)-(ex+5)d(x6-10)

dx dx dx

(x6-10 )2

dy=(x6-10 ) (ex)-( ex+5) (6 x5)

dx (x6-10)2

(3) If y = x3-1, find dy

x3+1 dx

Solution :

Given : y =x3-1

x3+1

Here u = x3-1

v =x3+1

dy= (x3+1)d(x3-1)-(x3-1)d(x3+1)

dx dx dx

(x3+1)2

=(x3+ 1) (3 x2)-(x3-1) (3 x2)

(x3+ 1)2

=3x2[ (x3+ 1) -(x3-1)]

(x3+1)2

=3x2[x3+ 1-x3+ 1]

(x3+1)2

=3x2[2]

(x3+ 1)2

dy

dx=2

26

( 1)x

x

Rule 5 : Scalar multiplication rule or constant mul tiplied by a function

rule.

If y = cu , c is a constant , where u is a differentiable function of x,then

dy= cdu

dx dx

i.e.,dy=d(cu) = c du

dxdx dx

Examples :

(1) If y= 5 x3, find dy

dx

Solutio n : Given : y= 5 x3

dy=d(5x3)munotes.in

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dxdx

= 5d(x3) = 5(3 x2)= 15 x2

dx

(2)If y= 10 log x, find dy

dx

Solution : Given : y = 10 log x

dy=d(10 log x)

dxdx

= 10 d(logx)

dx

= 10 (1/ x)

=10

x

1.2.4 List of formulae :

y =ƒ(x) dy=ƒ' (x)

dx

1. xnnxn-1

2. C , C = constant 0

3. exex

4. axaxlog a

5. log x 1/x

6. x 1

7. √x 1/ 2√x

8. 1/ x -1/x2

1.2.5 List of Rules :

(1) d(u+v) = du+dv

dx dx dx

(2) d(u-v) = du-dv

dx dxdx

(3) d(uv) =u dv+vdu

dx dx dx

(4) d(u/v) =v du-udv

dx dx dx

v2

(5) d(cu) = c du, c = constant.

dx dx

1.2.6 Examples :

Find dyfor each of the following :

dx

Ex: (1) y= x6+ 4 ex+ log x+10

Solution:

dy=d(x6+ 4 ex+ log x+10)munotes.in

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dxdx

=d(x6) + d(4ex) +d(logx) +d(10)

dx dx dx dx

=d(x6) + 4 d(ex) +d(logx) +d(10)

dx dx dx dx

=6x5+4ex+ 1/x+ 0

dy=6x5+4ex+ 1/x

dx

Ex : ( 2) y= 5 x4-3ex+4√x+ 2x

Solution :

dy=d(5x4-3ex+4√x+ 2x)

dxdx

=d(5x4)-d(3ex) +d(4√x) +d(2x)

dx dx dx dx

=5d(x4)-3d(ex) + 4 d(√x) +d(2x)

dx dx dx dx

= 5(4 x3)-3ex+ 4(1/2√x) + 2Xlog 2

= 20 x3-3ex+ 2/√x+2Xlog 2

Ex: (3) y= x3/2+ 4 log x-10x2+ 15

Solution:

dy=d(x3/2+ 4 log x-10x2+ 15)

dxdx

=d(x3/2) +d(4 log x)-d(10x2)+d(15)

dx dx dx dx

=3x3/2-1+ 4 ( 1)-10(2x) + 0

2 x

dy=3x1/2+4-20x.

dx2 x

Ex: (4) y= ( x+ ex) (5+log x)

Solution: Here u= x+ ex

v= 5+ log x

dy= (x+ ex)d(5+log x) + (5+log x)d(x+ ex)

dx dx dx

= (x+ ex) (0+1/ x) + (5+ log x) (1+ex)

dy= (x+ ex) (1/x) + (5+ log x) (1+ex)

dxmunotes.in

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Ex: (5) y= ( x10) (10x)

Solution : Here u= x10, v = 10x

dy = x10d(10x) + 10xd(x10)=x10(10xlog 10) + 10X(10x9)

dx dx dx

Ex:(6) y= (√x+ex)(2x3+7)

Solution: Here u=√x+ex, v= 2x3+7

dy=(√x+ex)d(2x3+7) + (2 x3+7)d(√x+ex)

dx dx dx

dy = (√x+ex) (6x2) + (2 x3+7) ( 1/2√x+ ex)

dx

Ex: (7) y = x2+ 5x+6

x+ 7

Solution : Here u= x2+ 5x+ 6

v=x+ 7

dy=(x+ 7) d(x2+ 5x+ 6) -(x2+ 5x+ 6) d(x+ 7)

dx dx dx

(x+ 7)2

=(x+ 7) (2 x+5)-(x2+ 5x+ 6) (1)

(x+ 7)2

=2x2+ 19x+ 35 -x2-5x-6

(x+ 7)2

dy=x2+ 14x+ 29

dx (x+ 7)2

Ex: 8y=10ex+ 5logx

x3+ 12

Solution : Here u= 10ex+ 5 log x

v=x3+ 12

dy=(x3+ 12) d(10 ex+ 5 log x)-(10 ex+ 5 log x)d(x3+ 12)

dx dx dx

(x3+ 12)2

dy=(x3+ 12) (10ex+ 5/x)-(10 ex+ 5 log x) ( 3x2)

dx (x3+ 12)2

Ex:9 y=4x+ 6

2x2+5

Soluti on : Here u = 4x+ 6 , v=2 x2+5munotes.in

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dy= (2x2+5)d(4x+ 6) -(4x+ 6)d(2x2+5)

dx dx dx

(2x2+5)2

=(2x2+5)(4xlog 4) -(4x+6)(4 x)

(2x2+5)2

Ex: 10 y= 2 √x+ 16ex+(6x)+ 20x

Solution:

dy=d(2√x+16ex+ 6x+ 20x)

dxdx

= 2 ( 1/2√x) + 16ex+ 6xlog 6 + 20

dy= 1/√x+ 16ex+ 6xlog 6 + 20

dx

Exercise : 1.2

Differentiate the f ollowing with respect to x.

(1) y = x8-6ex+ 4x3/2-3x2+ 5

(2) y = 6 log x-3x+ 2ex+ 10√x+2

(3) y = 5 x4-12x3+ 18 ex+ 10x-25

(4) y = 8x(5x3+3x+1)

(5) y = (10 x2+ 2x+5) (√x+ex)

(6) y = (2 x3+3x2) (5log x+ 14)

(7) y = (x+ log x) (x5-4x2+ 10)

(8) y = (8 x5-6x5/2+ 1) (40√x+ 2ex)

(9) y = (ex+ 2 log x+ 2) (6x+ 2x2+5)

(10)y = x2+ 1

x4-1

(11)y = 2x+4√x

2ex+ 5

(12)y = x3-x2+ 2

x2-4

(13)y = x+√x

√x-1

(14)y= 3log x+5

x5+ 2x

(15)y = ex-√x

2√x+1munotes.in

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Answers :

(1) 8 x7-6ex+ 6x1/2-6x

(2) 6/ x-3xlog3 + 2ex+5/√x

(3) 20 x3-36x2 + 18ex+ 10xlog 10

(4) 8x(15x2+3) + (5 x3+ 3x+1) (8xlog8)

(5) (10 x2+2x+5)(1/2√x+ ex) +(√x+ex)(20x+2)

(6) (2 x3+3x2) (5/x) + (5log x+14) (6 x2+ 6x)

(7) (x+logx) (5x4-8x) + (x5-4x2+10) (1+1/ x)

(8) (8 x5-6x5/2+1) (20/√x+2ex) + (40√x+2ex) (40 x4-15x3/2)

(9)(ex+ 2logx+2) (6xlog6+4x) + (6x+ 2x2+5) ( ex+2/x)

(10) -2x5-4x3-2x

(x4-1)2

(11) (2ex+5) (2xlog2+2/√x)-(2x+ 4√x) (2ex)

(2ex+5)2

(12) x4-12x2+ 4x

(x2-4)2

(13)√x-1-(1+x)/2√x

(√x-1)2

(14) (x5+2x) (3/x)-(3logx+5) (5 x4+2)

(x5+2x)2

(15) (2√x+1) (ex-1/2√x)-(ex-√x) (1/√x)

(2√x+1)2

1.3 SECOND ORDER DERIVATIVES

If y=ƒ(x) is differentiable function with respect to x, then

dy=ƒ'(x) is called first order derivative of y with respect to xand

dx

d2y=ddy is called the second order derivative of y with

dX2 dx dx

respect to X

Notation :

(i) First order derivative :

dy=ƒ'(x) = y 1= y'

dx

(ii) Second order derivative :

d2y=ƒ"(x) = y 2=y"

dx2munotes.in

## Page 18

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1.3.1 Examples :

(1) If y = x3-6x2+ 19 x+ 100 , find d2y

dx2

Solution :

Given : y = x3-6x2+ 19 x+ 100

dy=3x2-12x+19

dx

d2y=d(3x2-12x+19)

dx2dx

d2y= 6x-12

dx2

(2) If y = ex+2x3+5x2+4 , find d2y

dx2

Solution :

Given : y = ex+2x3+5x2+4

dy= ex+ 6x2+10x

dx

d2y=d(dy/d x)

dx2dx

=d(ex+ 6x2+10x)

dx

d2y= ex+ 12x+10

dx2

(3) If y= ƒ(x)=x5-6x3+ 2x2+10x+5, findƒ"(x)

Solution:

Givenƒ(x) =x5-6x3+2x2+ 10x+5

ƒ'(x) = 5 x4-18x2+4x+10

ƒ"(x)= 20 x3-36x+ 4

(4) If y=x+1/x,x≠ 0, find d2y/dx2

Solution: Given: y= x+1/x

dy/dx= 1+( -1/x2) = 1-1/x2

d2y=0-(-2/x3) =2

dx2x3

Exercise:9.3

Find d2yfor each of the following:

dx2

(1) y= 8 x5-16x4+4x3+x+2

(2) y= 2 x3-5x2+12x+15

(3) y= x+25/x

(4) y= 2 x2+ex+5x+12

(5) y= x2+x+logxmunotes.in

## Page 19

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Answers:

1) 160 x3-192x2+24x

2) 12 x-10

3)50

x3

4) 4+ex

5) 2-1

x2

1.4 APPLICATIONS OF DERIVATIVES

Applications to Economics:

1.4.1 The Total Cost function:

(i) Total cost function C= ƒ(x)

(ii) Average cost function AC= C

x

(iii) Marginal cost function:

The rate of cha nge of cost with respect to the number of units

produced is called the Marginal cost and is denoted by MC.

i.e, MC= dC

dx

1.4.2 The Total Revenue Function:

(i) Total Revenue Function R = p x D

(ii) Average Revenue Function AR= p

(iii) Marginal Revenue function:

The rate of change of total revenue with respect to the

demand D is called

the Marginal revenue function and is denoted by MR.

i.e MR = d R

d D

1.4.3 Elasticity:

Let D be the demand and p be the price. The quantity -PdDis

D dP

calledelasticity of demand with respect to the price and is denoted by η

(η = eta : Greek alphabet).

i.e, η = -pdD

D d pmunotes.in

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If η = 0, demand D is Constant and the demand is said to be perfectly

elastic.

If 0< η < 1, the demand is said to be inelastic

If η = 1, the demand is directly proportional to the price.

If η > 1, the demand is said to be elastic.

1.4.4 Relation between the Marginal Revenu e and elasticity of

demand.

Let R = Total revenue

p = price

D = demand

R = pD

MR = dR

dD

=d(pD)

dD

= p d(D) + D. d(p) (by product rule)

dD dD

= p (1) + D dp

dD

MR = p + D dp

dD

η = -pdD

D dp

Ddp= -p

dD η

MR = p + ( -p)

η

MR = p [ 1 -1]

η

MR = AR [ 1 -1]

η (AR = p )

1.4.5 Examples:

Ex: (1) The cost of producing ҳ items is given bymunotes.in

## Page 21

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2 x2+ 5x + 20. Find the total cost, average cost and marginal cost

when x= 10.

Solution:

Let C = ƒ(ҳ) = 2x2+ 5x+20

AC = C=2x2+ 5x+20

x x

MC = dC

dx

=d(2x2+ 5x+20)

dx

MC = 4x + 5

when x = 10

C = 2 ( 10)2+ 5 (10 ) + 20

C = 270

AC = C=270 = 27

x 10

MC = 4 ( 10 ) + 5 = 45

Ex: ( 2)

The demand function is given by ρ = 50 + 6 D + 4D2.

Find t he total revenue, average revenue and the marginal revenue when

the demand is 5 units.

Solution:

Given : ρ = 50 + 6 D + 4 D2

R = ρ x D

= ( 50 + 6 D + 4 D2) ( D)

R = 50 D + 6 D2+ 4 D3

AR= ρ = 50 + 6 D + 4 D2

MR =dR

dDmunotes.in

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=d( 50 D + 6 D2+ 4D3)

dD

MR = 50 + 12 D + 12 D2

When D = 5

R = 50 ( 5) + 6 ( 5 )2+ 4 ( 5 )3

= 250 + 150 + 500

R = 900

AR = 50 + 6 (5) + 4 (5)2

= 50 + 30 + 100

= 180

MR = 50 + 12 ( 5 ) + 12 ( 5 )2

=50 + 60 + 300

= 410

Ex : (3)

The total revenue function is given by R = 20 D -D2, D =

Demand. Find the demand function. Also find AR when MR = 0 .

Solution :

Given R = 20 D -D2

ρD = 20 D -D2( R = ρD )

ρ = 20D -D2

D

ρ = 20 -D

The demand function is ρ = 20 -D .

Now, MR = dR

dD

=d( 20D -D2)

dD

MR = 20 -2Dmunotes.in

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Given that MR = 0.

0 = 20 -2D

2D = 20

D = 10

AR = ρ = 20 -D

AR = 20 -10

AR = 10

Ex : ( 4 )

The demand function is given by

D = 25 - 2 ρ - ρ2

Find the elasticity of demand when the price is 4.

Solution :

Given : D = 25 - 2 ρ - ρ2

dD= 0-2- 2 ρ

dp

dD=-2- 2 ρ

dp

η = - ρ dD

D dρ

=(- ρ) (-2- 2 ρ )

25- 2ρ - ρ2

η = p( 2 + 2 p)

25-2p-p2

When p = 4,

η = 4 ( 2 + 8 )

25-8-16

=4 ( 10 )

1

η = 40munotes.in

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Ex: 5

The demand function is given by D = p + 3

2p-1

where D = Demand and p = price. Find the elasticity of demand whe n the

price is 8.

Solution:

Given D = p + 3

2p-1

( 2p-1 )d( p + 3 ) -( p + 3 ) d( 2p-1 )

dD dp dp

dp =

(2p-1)2

= ( 2p-1) (1) -(p +3) (2)

(2p-1)2

= 2p-1-2p-6

(2p-1)2

dD= -7

dp (2p-1)2

η = -p dD

D dp

(-p) [ -7 ]

= (2p-1)2

[(p +3)/(2p -1)]

=7 p ( 2 p -1 )

( p + 3 ) ( 2 p -1 )2

η = 7 p

( p + 3 ) ( 2 p -1 )

When p = 8

η = 7 ( 8)

( 8 + 3 ) [ 2 ( 8 ) -1 ]

= 56

( 11) (15)munotes.in

## Page 25

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=56

165

η = 0.34

Ex : (6) If MR = 45, AR = 75, Find η

Solution:

Given MR = 45

AR = 75

η = ?

MR =AR [ 1- 1/η ]

45 = 75 [ 1 - 1/ η ]

45= 1 - 1/ η

75

3= 1-1 / η

5

0.6 = 1 - 1 / η

1 / η = 1 -0.6

1 / η = 0.4

η = 1 / 0.4

= 2.5

Ex : ( 7)

If AR = 95 and η = 7/ 2 , Find MR.

Solution : Given AR = 95

η = 7/2 = 3.5

MR = ?

MR = AR [ 1 - 1 / η ]

= 95 [ 1-1 / 3.5 ]

= 95 [ 1 -0.29 ]

= 95 [ 0.71]

MR = 67.45munotes.in

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Exercise : 1.4

(1) The cost of producing x items is given by x3+ 4x +15. Find the total

cost, average cost and marginal cost when x = 6.

(2) The total cost function is given by C= x3+2 x2+ 5 x + 30. Find the

total cost, average cost and marginal cost when x = 10.

(3) The demand function is given by p= 20 -8 D + 3 D2. Find the total

revenue, average revenue and marginal revenue when the demand is

4 units.

(4) The total revenue f unction is given by R= 30 D -2 D2+ D3.Find

the demand function. Also find total revenue, average revenue and

marginal revenue when the demand is 5 units.

(5) The demand function is given by D = -28-5p + 2 p2Find the

elasticity of demand when t he price is 3.

(6) The demand function is given by D = 2p + 5 Where D = Demand

and p-3

p = price. Find the elasticity of demand when price is 6.

(7) IfAR = 65 and η = 3, find MR.

(8)If MR = 85 and η = 4.5 , find AR.

(9)If MR = 55 and AR = 98 , find η .

(10) If the price is 65 and the elasticity of demand is 5.2, find the

marginal revenue.

Answers:

(1) C = 255 ; AC = 42.5 ; MC = 112

(2) C = 1280 ; AC = 128 ; MC = 345

(3) R = 144 ; AR = 36 ; MR = 100

(4) p = 30 -2D + D2

R = 225 ; AR = 45 ; MR = 85

(5)η = 0.64

(6)η = 1.29

(7) MR = 43.3munotes.in

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(8) AR = 108.97

(9)η = 2.28

(10) MR = 52.5

1.5 MAXIMA AND MINIMA

Let y =ƒ(x) be th e given function. A curve ƒ(x) is said to have a

maximum or minimum point (extreme point), if ƒ(x) attains either a

maximum or minimum of that point.

Y

y = f(x)

f(a)

0 x=a X

Maximum at x = a

Y

y = f (x)

(f(c) b

0 x=b X

Maximum at x = b

In the first figure, x = a is the point where the curve ƒ(x) attains a

maximum. In the second figure, x = b is the point where the curve ƒ(x)

attains a minimum

1.5.1 Conditions for Maximum & Minimum:

1. Condition for Maximum:

(i)ƒ' (x) = 0

(ii)ƒ" (x) < 0 at x = a

2. Condition for Minimum:

(i) ƒ' (x) = 0

(ii)ƒ" (x) > 0 at x = bmunotes.in

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1.5.2. To find the Maximum and Minimum v alues of ƒ(x) :

Steps:

(i) Findƒ'(x) andƒ"(x) .

(ii) putƒ'(x) = 0 , solve and get the values of x.

(iii) Substitute the values of x in ƒ"(x) .

Ifƒ"(x) < 0 , then ƒ(x) has maximum value at x = a. If ƒ"(x) > 0, then

ƒ(x) has minimum valu e at x= b.

(iv) To find the maximum and minimum values, put the points

x = a and x = b in ƒ(x).

Note: Extreme values of ƒ(x) = Maximum and minimum values of ƒ(x).

1.5.3. Examples :

Ex: (1) Find the extreme values of ƒ(x) = x3-3x2-45x + 25.

Solution:

Given : ƒ(x) = x3-3x2-45x + 25

ƒ'(x) = 3x2-6x-45

ƒ"(x) = 6x -6.

Sinceƒ(x) has maximum or minimum value,

ƒ'(x) = 0

3x2-6x-45 = 0

3 (x2-2x-15) = 0

x2-2x-15 = 0

x2-5x + 3x -15 = 0

x ( x -5 ) + 3 (x -5 ) = 0

( x-5 ) ( x + 3 ) = 0

x-5 = 0 or x + 3 = 0

x = 5 or x = -3.munotes.in

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When x = 5, ƒ"(5)=6(5) -6=24>0

ƒ(x) has minimum at x = 5.

When x = -3,ƒ"(-3 )= 6 ( -3 )-6

=-24 < 0

ƒ(x) has maximum at x = -3.

To find the maximum and minimum values of ƒ(x) :

putx = 5 and x = -3 inƒ(x) .

ƒ(5) = 53-3 ( 5 )2-45(5) + 25

= 125 -75-225 + 25

=-150

ƒ(-3 ) = ( -3 )3-3 (-3 )2-45(-3 ) + 25

=-27-27 + 135 + 25

= 106

Maximum value = 106 at x = -3

Minimum value = -150 at x = 5.

Ex: ( 2 ) Find the maximum and minimum values of ƒ(x) = x +16

x,x≠ 0.

Solution : Given :ƒ( x ) = x + (16 / x )

ƒ'( x ) = 1 + 16 ( -1 / x2)

ƒ'( x ) = 1 -16

x2

ƒ"( x ) = 0 -16 ( -2)

x3

ƒ"( x ) = 32

x3

Sinceƒ( x ) has maximum or minimum value, ƒ'( x ) = 0.

1-16= 0

x2

x2-16= 0

x2munotes.in

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x2-16 = 0

x2-42= 0

( x-4 ) ( x + 4 ) = 0

x = 4 or x = -4

when x = 4 , ƒ"( 4 ) = 32=32=1> 0

( 4 )364 2

ƒ( x ) has minimum at x = 4.

when x = -4 ,ƒ"(-4 ) = 32 =32=-1< 0

(-4 )3-64 2

ƒ( x ) has maximum at x = -4.

Now to find the extreme values of ƒ( x ) : put x = 4 and x = -4 inƒ( x )

ƒ( 4 ) = 4 + 16 = 4 + 4 = 8

4

ƒ(-4 ) = -4 + 16=-4-4 = -8

(-4)

Maximum value = -8 at x = -4

Minimum value = 8 at x = 4.

Ex: ( 3) Divide 80 into two parts such that the sum of their squares is a

minimum.

Solutio n:

Let x and 80 -x be the two required numbers.

By the given condition,

ƒ(x) = x2+ ( 80 -x)2

ƒ(x)= x2+ 802-2(80) (x) + x2

ƒ(x) = 2x2-160x + 6400

ƒ'(x)= 4x -160

ƒ"(x) = 4

Sinceƒ(x) has minimum,

ƒ'(x)= 0

4x-160= 0

4x=160munotes.in

## Page 31

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x=160=40.

4

ƒ" (x) = 4 > ƒ" (40)= 4>0.

ƒ(x) has minimum at x= 40.

The required numbers are

40 and 80 -40=40

The required parts of 80 are

40 and 40.

Ex: (4)

A manufacturer can sell x items at a price of Rs. (330 -x) each. The

cost of p roducing x items is Rs. (x2+ 10x+12). Find x for which the profit

is maximum.

Solution:

Given that the total cost function is

C= x2+ 10x+ 12.

Selling price p= 330 -x

Revenue is R= p x D

= p x x (D= x)

= (330 -x) x

= 330x -x2

Profit= Revenue -Cost

P= R -C

= (330x -x2)-(x2+10x+ 12)

= 330x -x2-x2-10x-12

P= 320x -2x2-12

dp= 320 -4x

dx

d2p=-4 < 0

dx2

The profit is maximum.

Since the profit is maximum,

dp= 0

dxmunotes.in

## Page 32

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320-4x= 0

4x=320

x=80.

Hence the profit is maximum when 80 items are sold.

Ex: (5) The total cost function is

C= x3-9x2+ 24x+ 70.

Find x for whic h the total cost is minimum.

Solution:

Let C =ƒ(x)= x3-9x2+ 24x+ 70

C'=ƒ' (x) = 3x2-18x+24

C"=ƒ" (x) = 6x -18

Sinceƒ(x) has minimum, ƒ' (x) = 0

3x2= 18x+ 24=0

3(x2-6x+8) = 0

x2-6x +8=0

x2-2x-4x+8=0

x(x-2)-4(x-2)=0

(x-2)(x-4)= 0

x-2=0 or x -4=0

x=2 or x=4

When x=4

ƒ"(x) = 6x -18

ƒ" (4) = 6(4) -18

= 6>0

ƒ(x) has minimum at x=4.

The t otal cost is minimum at x=4.

Ex: (6) The total revenue function is given by

R = 4x3-72x2+ 420x+ 800.

Find x for which the total revenue is maximum.

Solution:

Let R =ƒ(x)= 4x3-72x2+ 420x+800

R'=ƒ'(x) = 12x2-144x+420

R" =ƒ" (x) = 24x -144

Sinceƒ(x) has maximum,munotes.in

## Page 33

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ƒ'(x)= 0

12x2-144x+420=0

12(x2-12x+35)=0

x2-12x+35=0

x2-5x-7x+35=0

x(x-5)-7(x-5)=0

(x-5)(x-7)=0

x-5=0 or x -7=0

x=5 or x=7

When x=5

R"=ƒ"(x) = 24x -144

= 24(5) -144

=-24<0

ƒ(x) has maximum at x=5

The total revenue is maximum at x=5

Exercise : 1.5

(1) Find the extreme values of ƒ(x)=2x3-6x2-48x+90.

(2) Find the maximum and minimum values of ƒ(x) = x+ (9/x), x0

(3) Find the extreme values of ƒ(x)= 4x3-12x2-36x+25

(4) Find the extreme values of ƒ(x) =x + (36/x), x0.

(5) Divide 120 into two parts such that their product is maximum.

(6) Divide 70 into two parts such that the sum of their squares is a

minimum.

(7) A manufacturer sells x items at a price of Rs. (400 -x) each. The cost of

producing x items is Rs (x2+ 40x+52). Find x for which the profit is

maximum.

(8) The cost function is given by C= x3-24x2+ 189x+120. Find x for

which the cost is minimum,

(9) The total revenue function is given by

R= 2x3-63 x2+ 648x+250.

Find x for which the total revenue is maximum.

(10) The total cost of producing x units is Rs.(x2+ 2x+5) and the price is

Rs.(30-x) per unit. Find x for which the profit is maximum.munotes.in

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Answers:

(1) Maximum value= 146 at x= -2

Minimum value = -70 at x=4

(2) Maximum value= -6 at x= -3

Minimum value = 6 at x= 3

(3) Maximum value= 45 at x= -1

Minimum value= -83 at x =3

(4) Maximum value = -12 at x= -6

Minimum value = 12 at x=6

(5) 60,60

(6) 35,35

(7) 90

(8) 9

(9) 9

(10) 7

munotes.in

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Unit II

Unit -2

SIMPLE INTEREST AND

COMPOUND INTEREST

Unit Structure :

2.0 Objectives

2.1 Introduction

2.2 Definitions of Terms Used In This Chapter

2.3 Simple Interest

2.4 Compound Interest

2.0OBJECTIVES

After reading this chapter you will be abl e to:

Define interest, principal, rate of interest, period.

Find simple interest (SI), rate of S.I., period of investment.

Find Compound Interest (CI), rate of C.I., Amount accumulated at the

end of a period.

Compound interest compounded yearly, half -yearl y, quarterly or

monthly.

2.1INTRODUCTION

In every day life individuals and business firms borrow money

from various sources for different reasons. This amount of money

borrowed has to be returned from the lender in a stipulated time by paying

some interest on the amount borrowed. In this chapter we are going to

study the two types of interests viz. simple and compound interest. We

start with some definitions and then proceed with the formula related to

both the types of interests.

2.2DEFINITIONS OF T ERMS USED IN THIS CHAPTER

Principal : The sum borrowed by a person is called its principal . It is

denoted by P.

Period : The time span for which money is lent is called period . It is

denoted by n.munotes.in

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Interest : The amount paid by a borrower to the lender for the use of

money borrowed for a certain period of time is called Interest . It is

denoted by I.

Rate of Interest : This is the interest to be paid on the amount of Rs. 100

per annum (i.e. per year). This is denoted by r.

Total Amount : The sum of the princ ipal and interest is called as the total

amount (maturity value )and is denoted by A. Thus, A=P + I .

i.e. Interest paid I=A–P.

2.3SIMPLE INTEREST

The interest which is payable on the principal only is called as

simple interest (S.I.). For ex ample the interest on Rs. 100 at 11% after one

year is Rs.11 and the amount is 100 + 11 = Rs. 111.

It is calculated by the formula: I =100Pnr=Pxnx100r

Amount at the end of nthyear = A =P + I =P +100Pnr=P1100nr

Remark : The period nis always taken in ‘years’. If the period is given in

months/days, it has to be converted into years and used in the above

formula. For example, if period is 4 months then we take n=4/12 = 1/3 or

if period is 60 days then n= 60/365.

Example 1:If Mr. Sagar borrows Rs. 500 for 2years at 10% rate of

interest, find (i) simple interest and (ii) total amount.

Ans: Given P= Rs. 500,n=2andr=10 %

(i)I=100Pnr=500 x 2 x 10

100= Rs. 100

(ii)A = P + I = 500 + 100 = Rs. 600

3.3.1Problems involving unknown factors in the formula I =100Pnr

The formula I =100Pnrremaining the same, the unknown factor in

the formula is taken to the LHS and its value is computed. For example, if

rate of interest is unknown then the formula is rewritten asx 100

P xIrn .

Example 2:If Mr. Prashant borrows Rs. 10 00 for 5 years and pays an

interest of Rs. 300, find rate of interest.

Ans: Given P=1000,n= 5 and I= Rs. 300Simple Interest = Prinicpal x period x rate of interest

munotes.in

## Page 37

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Now, I=100Pnr x 100

P xIrn =300 x 100

1000 x 5=6

Thus, the rate of interest is 6%.

Example 3:Find the period for Rs. 2500 to yield Rs. 900 in simple

interest at 12 %.

Ans: Given P= Rs. 2500, I= 900, r= 12%

Now, I=100Pnr x 100

xInP r =900 x 100

2500 x 12= 3

Thus, the period is 3 years.

Example 4:Find the period for Rs. 1000 to yield Rs. 50 in simple interest

at 10%.

Ans:Given P= Rs. 1000, I= 50, r= 10%

Now, I=100Pnr x 100

xInP r =50 x 100

1000 x 10= 0.5

Thus, the period is 0.5 years i.e. 6 months.

Example 5:Mr. Ak ashlent Rs. 5000 to Mr. Prashantand Rs. 4000 to Mr.

Sagarfor 5 years and received total simple interest of Rs. 4950. Find (i)

the rate of interest and (ii) simple interest of each.

Ans: Let the rate of interest be r.

S.I. for Prashant=5000 x 5 x

100r=250r … (1)

and S.I. for S agar =4000 x 5 x

100r= 200 r … (2)

from (1) and (2), we have,

total interest from both = 250 r+ 200 r

= 450 r

But total interest received be Mr. Ak ash= Rs. 4950

450r= 4950 r=4950

450= 11

the rate of interest = 11%

Example 6:The S.I. on a sum of money is one -fourth the principal. If the

period is same as that of the rate of interest then find the rate of interest.

Ans: Given I=4Pandn = r

Now, we know that I=100Pnr

4P=x x

100P r r100

4=r2

r2= 25 r= 5.

the rate of interest = 5%munotes.in

## Page 38

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Example 7:If Rs. 8400 amoun t to Rs. 11088 in 4 years, what will Rs.

10500 amount to in 5 years at the same rate of interest?

Ans:

(i) Given n= 4,P= Rs. 8400, A= Rs. 11088

I=A–P= 11088 –8400 = Rs. 2688

Letrbe the rate of interest.

Now, I=100Pnr2688 =8400 x 4 x

100r

r= 8%

(ii) To find Awhen n= 5,P= Rs. 10500, r= 8

A = P 1100nr= 10500 x5 x 81100 = 10500 x140

100= 14700

the required amount = Rs. 14,700

Example 8: Mr. Shirish borrowed Rs. 12,000 at 9% interest from Mr.

Girish on January 25, 2007. The interest and principal is due on August

10, 2007. Find the interest and total amount paid by Mr. Shirish.

Ans: Since the period is to be taken in years, we first count number of

days from 25thJanuary t o 10thAugust, which is 197days.

Now, I=100Pnr= 12000 x197

365x9

100

I= Rs. 582.9

Total amount = P + I = 12000 + 582.9

A= Rs. 12 ,582.9

Check your progress 10.1

1.Find the SI and amount for the following data giving principal, rate

of interest and number of years:

(i) 1800, 6%, 4 years. (ii) 4500, 8%, 5 years

(iii) 7650, 5.5%, 3 years. (iv) 6000, 7.5%, 6 years

(v) 25000, 8%, 5 years (vi) 20000, 9.5%, 10 years.

Ans: (i) 432, 2232 (ii) 1800, 6300, (iii) 1262.25, 8912.25

(iv) 2700, 8700 (v) 10000, 35000 (vi) 19000, 39000

2.Find the S.I. and the total amount for a principal of Rs. 6000 for 3

years at 6% rate of interest.

Ans: 1080, 7080

3.Find the S.I. and the total amount for a principal of Rs. 3300 for 6

years at 3½ % rate of interest.

Ans:693, 3993January 6

February 28

March 31

April 30

May 31

June 30

July 31

August 10

Total 197

munotes.in

## Page 39

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4.Find the S.I. and the total amount for a principal of Rs. 10550 for 2

years at 10¼ % rate of interest.

Ans:2162.75, 1271 2.75

5.Find the rate of interest if a person invests Rs. 1000 for 3 years and

receives a S.I. of Rs. 150.

Ans: 5%

6.Find the rate of interest if a person invests Rs. 1200 for 2 years and

receives a S.I. of Rs. 168.

Ans: 7%

7.A person invests Rs. 4050 in a bank which pays 7% S.I. What is the

balance of amount of his savings after ( i) six months, ( ii) one year?

Ans: 141.75, 283.5

8.A person invests Rs. 3000 in a bank which offers 9% S.I. After how

many years will his balance of amount will be Rs. 3135?

Ans: 6 mont hs

9.Find the principal for which the SI for 4 years at 8% is 585 less than

the SI for 3½ years at 11%.

Ans: 9000

10.Find the principal for which the SI for 5 years at 7% is 250 less than

the SI for 4 years at 10%.

Ans: 5000

11.Find the principal for which t he SI for 8 years at 7.5% is 825 less

than the SI for 6½ years at 10.5%.

Ans: 10000

12.Find the principal for which the SI for 3 years at 6% is 230 more

than the SI for 3½ years at 5%.

Ans: 46000

13.After what period of investment would a principal of Rs. 12,3 50

amount to Rs. 17,043 at 9.5% rate of interest?

Ans: 4 years

14.A person lent Rs. 4000 to Mr. Xand Rs. 6000 to Mr. Yfor a period

of 10 years and received total of Rs. 3500 as S.I. Find ( i) rate of

interest, ( ii) S.I. from Mr. X, Mr. Y.

Ans: 3.5%, 1400, 2 100

15.Miss Pankaj Kansra lent Rs. 2560 to Mr. Abhishek and Rs. 3650 to

Mr. Ashwin at 6% rate of interest. After how many years should he

receive a total interest of Rs. 3726?

Ans: 10 yearsmunotes.in

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16.If the rate of S.I. on a certain principal is same as that of the p eriod

of investment yields same interest as that of the principal, find the

rate of interest.

Ans: 10%

17.If the rate of S.I. on a certain principal is same as that of the period

of investment yields interest equal to one -ninth of the principal, find

the rate of interest.

Ans: 31

3years

18.Find the principal and rate of interest if a certain principal amounts

to Rs. 2250 in 1 year and to Rs. 3750 in 3 years.

Ans: 1500, 50%

19.Find the principal and rate of interest if a certain principal amounts

to Rs. 3340 in 2 years and to Rs. 4175 in 3 years. Ans: 1670, 50%

20.If Rs. 2700 amount Rs. 3078 in 2 years at a certain rate of interest,

what will Rs. 7200 amount to in 4 years at the same rate on interest?

Ans: 7%, 9216

21.At what rate on interest will certain sum of money amount to three

times the principal in 20 years?

Ans: 15%

22.Mr. Chintan earns as interest Rs. 1020 after 3 years by lending Rs.

3000 to Mr. Bhavesh at a certain rate on interest and Rs. 2000 to Mr.

Pratik at a rate on interest 2% m ore than that of Mr. Bhavesh. Find

the rates on interest.

Ans: 6%, 8%

23.Mr. Chaitanya invested a certain principal for 3 years at 8% and

received an interest of Rs. 2640. Mr. Chihar also invested the same

amount for 6 years at 6%. Find the principal of Mr. Chaitanya and

the interest received by Mr. Chihar after 6 years.

Ans: 11000, 3960

24.Mr. Ashfaque Khan invested some amount in a bank giving 8.5%

rate of interest for 5 years and some amount in another bank at 9%

for 4 years. Find the amounts invested in bot h the banks if his total

investment was Rs. 75,000 and his total interest was Rs. 29,925.

Ans: 45000, 30000

25.Mrs. Prabhu lent a total of Rs. 48,000 to Mr. Diwakar at 9.5% for 5

years and to Mr. Ratnakar at 9% for 7 years. If she receives a total

interest of Rs. 25,590 find the amount she lent to both.

Ans: 18000, 30000munotes.in

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2.4COMPOUND INTEREST

The interest which is calculated on the amount in the previous year

is called compound interest .

For example, the compound interest on Rs. 100 at 8% after one

year is Rs. 8 and after two years is 108 + 8% of 108 = Rs. 116.64

IfPis the principal, ris the rate of interest p.a. then the amount at the end

ofnthyear called as compound amount is given by the formula:

Thecompound interest is given by the form ula:

Note :

1.The interest may be compounded annually (yearly), semi -annually

(half yearly), quarterly or monthly. Thus, the general formula to calculate

the amount at the end of nyears is as follows:

Here p: number of times the interest is compou nded in a year.

p= 1 if interest is compounded annually

p= 2 if interest is compounded semi -annually (half-yearly )

p= 4 if interest is compounded quarterly

p= 12 if interest is compounded monthly

2.It is easy to calculate amount first and then the compo und interest as

compared with finding interest first and then the total amount in case of

simple interest.

Example 9:Find the compound amount and compound interest of Rs.

1000 invested for 10 years at 8% if the interest is compounded annually.

Ans: Give nP=1000,r =8,n= 10.

Since the interest is compounded annually, we have

A= 1100nrP=1000 x1081100= 1000 x 2.1589 = Rs.2158.9

Example 10:Find the principal which will amount to Rs. 11,236 in 2

years at 6% co mpound interest compounded annually.

Ans: Given A= Rs. 11236, n= 2,r= 6 and P= ?A= P 1100nr

CI = A–P

A = P 1x 100npr

p

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Now, A= 1100nrP

11236 = P261100=Px 1.1236

P=11236

1.1236= 10,000

the required principal is Rs. 10,000.

Exam ple11:Find the compound amount and compound interest of Rs.

1200 invested for 5 years at 5% if the interest is compounded ( i) annually,

(ii) semi annually, ( iii) quarterly and ( iv) monthly.

Ans: Given P= Rs. 1 200,r=5,n= 5

Let us recollect the formu laA=P1x 100npr

p

(i)If the interest is compounded annually ,p= 1:

A= 1100nrP= 1200 x551100= 1200 x 1. 2763 =Rs. 1 531.56

CI= A–P= 1531.56 –1200 = Rs. 33 1.56

(ii)If the interest is compounde d semi -annually ,p= 2:

A=2

12 x 100nrP = 1200 x1051200= 1200 x 1. 28=Rs. 1 536

CI= A–P= 1536–1200 = Rs. 3 36.

(iii)If the interest is compounded quarterly ,p= 4:

A=4

14 x 100nrP = 1200 x2051400= 1200 x 1. 2820 =

Rs. 1 538.4

CI= A–P= 1538.4 –1200 = Rs. 3 38.4

(iv)If the interest is compounded monthly ,p= 12:

A=12

112 x 100nrP = 1200 x60511200 = 1200 x 1. 2834 =Rs. 1 540

CI= A–P= 1540–1200 = Rs. 340

Example 12:Mr. Santosh wants to invest some amount for 4 years in a

bank. Bank Xoffers 8% interest if compounded half yearly while bank Y

offers 6% interest if compounded monthly. Which bank should Mr.

Santosh select for better benefits?

Ans: Givenn= 4.

Let the principal Mr. Santosh wants to invest be P= Rs. 100

From Bank X:r= 8and interest is compounded half -yearly, so p= 2.

A=2

12 x 100nrP = 100 x481200= 116.9858 … (1)munotes.in

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From Bank Y:r= 6,p= 12

A=12

112 x 100nrP = 100 x48611200 = 127.0489 … (2)

Comparing (1) and ( 2), Dr. Ashwinikumar should invest his amount in

bank Yas it gives more interest at the end of the period.

Example 13:In how many years would Rs. 75,000 amount to Rs.

1,05,794.907 at 7% compound interest compounded semi -annually?

Ans: Given A= Rs. 1057 94.907, P= Rs. 75000, r= 7,p= 2

A=2

12 x 100nrP

105794.907 = 75000 x271200n

105794.907

75000=(1.035)2n

1.41059876 = (1.035)2n

(1.035)10= (1.035)2n2n= 10

Thus, n= 5

Example 14:A certa in principal amounts to Rs. 4410 after 2 years and to

Rs.4630.50 after 3 years at a certain rate of interest compounded annually.

Find the principal and the rate of interest.

Ans: Let the principal be Pand rate of interest be r.

Now, we know that A=P1100nr

From the given data we have,

4410 = P2

1100rand 4630.5 = P3

1100r

4410 =P(1 + 0.01 r)2… (1)

4630.5 = P(1 + 0.01 r)3… (2)

Dividing (2) by (1), we haveDo not write ‘1 + 0.01 r’

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4630.5

4410=3

2(1 0.01 )

(1 0.01 )P r

P r

1.05 = 1 + 0.01 r

0.05 = 0.01 r

Thus, r= 5%

Example 15:Find the rate of interest at which a sum of Rs. 2000 amounts

to Rs. 2690 in 3 years given that the interest is compounded half yearly.

(61.345 = 1.05)

Ans: Given P= Rs. 2000, A= Rs. 2680, n= 3,p= 2

Now, A=2

12 x 100nrP

2690 = 2000 x6

1200r

2690

2000=6

1200r1.345 =6

1200r

61.345 = 1+200r1.05 = 1 +200r

r= 0.05 x 200 = 10%

Thus, the rate of compound interest is 10 % .

Example 16:If the interest compounded half yearly on a certain principal

at the e nd of one year at 8% is Rs. 3264, find the principal.

Ans: Given CI= Rs. 3264, n= 1,p= 2 and r= 8

Now, CI=A–P=P281200–P

i.e. 3264 = P[ (1.04)2–1] = 0.0816 P

P=3264

0.0816= 40000

Thus, the principal is Rs. 40,000.

Check your progress 10.2

1.Compute the compound amount and interest on a principal of Rs.

21,000 at 9% p.a. after 5 years.

Ans:32,311.10, 11,311.10

2.Compute the compound amount and interest on a principal of Rs.

6000 at 11% p.a. after 8 years.

Ans:13827.23, 7827.23

3.Compute the compound amount and compound interest of Rs. 5000

if invested at 11% for 3 years and the interest compounded i)

annually, ( ii) semi annually, ( iii) quarterly and ( iv) monthly.

Ans: (i)6838.16, 1838.16 (ii) 6894.21, 18942 1

(iii) 6923.92, 1923.92 (iv) 6944.39, 1944.39munotes.in

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4.Compute the compound amount and compound interest of Rs. 1200

if invested at 9% for 2 years and the interest compounded i)

annually, ( ii) semi annually, ( iii) quarterly and ( iv) monthly.

Ans: (i) 1425.7 2, 225.72 (ii) 1431.02, 231.02

(iii) 1433.8, 233.8 (iv) 1435.7, 235.7

5.Miss Daizy invested Rs. 25,000 for 5 years at 7.5% with the interest

compounded semi -annually. Find the compound interest at the end

of 5 years. Ans:

11,126.10

6.Mr.Dayanand borrowed a sum of Rs. 6500 from his friend at 9%

interest compounded quarterly. Find the interest he has to pay at the

end of 4 years? Ans: 2779.54

7.Mr. Deepak borrowed a sum of Rs. 8000 from his friend at 8%

interest compounded annually. Fi nd the interest he has to pay at the

end of 3 years? Ans:

2077.70

8.Mr. Deshraj borrowed Rs. 1,25,000 for his business for 3 years at

25% interest compounded half yearly. Find the compound amount

and interest after 3 years. Ans: 2,53,410.82;

12,841 0.82

9.Mrs. Hemlata bought a Sony Digital Camera for Rs. 15,800 from

Vijay Electronics by paying a part payment of Rs. 2,800. The

remaining amount was to be paid in 3 months with an interest of 9%

compounded monthly on the due amount. How much amount did

Mrs. Hemlata paid and also find the interest.

Ans: 13294.70, 294.70

10.Mr. Irshad bought a Kisan Vikas Patra for Rs. 10000, whose

maturing value is Rs. 21,000 in 4½ years. Calculate the rate of

interest if the compound interest is compounded quarterly.

Ans: 16.8%

11.What sum of money will amount to Rs. 11236 in 2 years at 6% p.a.

compound interest? Ans: 10,000

12.Find the principal which will amount to Rs. 13468.55 in 5 years at

6% interest compounded quarterly. [ (1.015)20= 1.346855]

Ans: 10000

13.Find the principal which will amount to Rs. 30626.075 in 3 years at

7% interest compounded yearly. Ans: 25000

14.Find the principal if the compound interest payable annually at 8%

p.a. for 2 years is Rs. 1 664. Ans: 10000munotes.in

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15.If Mr. Sagar wants to earn Rs. 50000 after 4years by investing a

certain amount in a company at 10% rate of interest compounded

annually, how much should he invest? Ans:

34150.67

16.Find after how many years will Rs. 4000 amount to Rs. 4494.40 at

6% rate of interest compounded yearly. Ans: n= 2

17.Find after how many years Rs. 10,000 amount to Rs. 12,155 at 10%

rate of interest compounded half -yearly . Ans: n= 1

18.Find the rate of interest at which a principal of Rs.10000 amounts to

Rs. 11236 after 2 years. Ans:6%

19.Find the rate of interest at which a principal of Rs.50000 amounts to

Rs. 62985.6 after 3 years. (31.259712 = 1.08) Ans: 8%

20.Mrs. Manisha Lokhande deposited Rs. 20,000 in a bank for 5 years.

If she received Rs.3112.50 as in terest at the end of 2 years, find the

rate of interest p.a. compounded annually. Ans: 7.5%

21.A bank Xannounces a super fixed deposit scheme for its customers

offering 10% interest compounded half yearly for 6 years. Another

bank Yoffers 12% simple interest for the same period. Which bank’s

scheme is more beneficial for the customers? Ans: Bank X

22.ABC bank offers 9% interest compounded yearly while XYZ bank

offers 7% interest compounded quarterly. If Mr. Arunachalam wants

to invest Rs. 18000 for 5 years, which bank should he choose?

Ans: Bank ABC

23.Mangesh borrowed a certain amount from Manish at a rate of 9% for

4 years. He paid Rs. 360 as simple interest to Manish. This amount

he invested in a bank for 3 years at 11% rate of interest compounded

quarterly. Find the compo und interest Mangesh received from the

bank after 3 years. Ans:

1384.78

24.On a certain principal for 3 years the compound interest

compounded annually is Rs. 11 25.215 while the simple interest is

Rs. 1050, find the principal an d the rate of interest.

Ans: 5000, 7%

25.On a certain principal for 4 years the compound interest

compounded annually is Rs. 13923 while the simple interest is Rs.

12000, find the principal and the rate of interest.

Ans: 30000, 10%.munotes.in

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26.Which investment is better for Mr. Hariom Sharma (i) 6%

compounded half yearly or (ii) 6.2% compounded quarterly?

Ans:

27.Which investment is better for Mr. Suyog Apte (i) 9% compounded

yearly or (ii) 8.8% compounded quarterly?

Ans:

28.A bank Xoffers 7% interest compounded s emi-annually while

another bank offers 7.2% interest compounded monthly. Which bank

gives more interest at the end of the year?

Ans:

29.Mr. Nitin Tare has Rs. 10000 to be deposited in a bank. One bank

offers 8% interest p.a. compounded half yearly, while the other

offers 9% p.a. compounded annually. Calculate the returns in both

banks after 3 years. Which bank offers maximum return after 3

years?

Ans:

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Unit II

Unit -3

ANNUITIES AND EMI

Unit Structure :

3.0 Objectives

3.1 Introduction

3.2 Annuity

3.3 Types of Annuities

3.4 Sinking Fund

3.5 Equated Monthly Installment (Emi)

3.0OBJECTIVES

After reading this chapter you will be able to:

Define annuit y, future value, present value, EMI, Sinking Fund.

Compute Future Value of annuity due, Present Value of an ordinary

annuity.

Compute EMI of a loan using reducing balance method and flat

interest method.

Compute Sinking Fund (periodic payments).

3.1INTRO DUCTION

In the previous chapter we have seen how to compute compound

interest when a lump sum amount is invested at the beginning of the

investment. But many a time we pay (or are paid) a certain amount not in

lump sum but in periodic installments. This series of equal payments done

at periodic intervals is called as annuity .

Let us start the chapter with the definition of an annuity .

3.2ANNUITY

A series of equal periodic payments is called annuity . The

payments are of equal size andat equal t ime interval .

The common examples of annuity are: monthly recurring deposit

schemes, premiums of insurance policies, loan installments, pension

installments etc. Let us understand the terms related to annuities and then

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Periodic Payment:

The amount of payment made is called as periodic payment .

Period of Payment:

The time interval between two payments of an annuity is called as the

period of payment .

Term of an annuity:

The time of the beginning of the first payment period to t he end of the last

payment period is called as term of annuity . An annuity gets matured at

the end of its term.

3.3TYPES OF ANNUITIES

Though we will be discussing two types of annuities in detail, let

us understand different types of annuities based on the duration of the term

or on the time when the periodic payments are made. On the basis of the

closing of an annuity, there are three types of annuities:

1.Certain Annuity :

Here the duration of the annuity is fixed (or certain), hence called

certain an nuity . We will be learning such annuities in detail.

2.Perpetual Annuity :

Here the annuity has no closing duration, i.e. it has indefinite duration.

Practically there are rarely any perpetuities.

3.Contingent Annuity :

Here the duration of the annuity depen ds on an event taking place. An

example of contingent annuity is life annuity . Here the payments are to

be done till a person is alive, like, pension, life insurance policies for

children (maturing on the child turning 18 years) etc.

Onthe basis of whe n the periodic payments are made we have two

types of annuities: ordinary annuity and annuity due.

3.3.1 Immediate (Ordinary) Annuity:

The annuity which is paid at the end of each period is called as

immediate (ordinary )annuity . The period can be mont hly, quarterly or

yearly etc. For example, stock dividends, salaries etc.

Let us consider an example of an investment of Rs. 5000 each year

is to be made for four years. If the investment is done at the end of each

year then we have the following diagr ammatic explanation for it:munotes.in

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3.3.2 Present Value :

The sum of all periodic payments of an annuity is called its present

value . In simple words, it is that sum which if paid now will give the same

amount which the periodic payments would have given at the end of the

decided period. It is the one time payment of an annuity.

The formula to find the present value ( PV) is as follows:

Leti=x 100r

p, the rate per period, then the above formula can be

rewritten as f ollows:

3.3.3 Future Value (Accumulated value) :

The sum of all periodic payments along with the interest is called

thefuture value (accumulated amount ) of the annuity.

The formula to find the future value ( A) of an immediate annuity is as

follow s:PV=11

1x 100 x 100npP

r r

p p

Where

P: periodic equal payment

r: rate of interest p.a.

p: period of annuity

PV=

11

1npP

i i

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Example 1:Find the future value after 2 years of an immediate annuity of

Rs. 5000, the rate of interest being 6% p.a compounded annually.

Ans: Given n= 2,P= Rs. 5000, r= 6 and p= 1i=6

100= 0.06

A=P

i1 1npi =5000

0.0621 0.06 1 = 50001.1236 1

0.06

A= 5000 x 2.06 = Rs. 10300

Example 2: Find the amount for an ordinary annuity with periodic

payment of Rs. 3000, at 9% p. a. compounded semi -annually for 4 years.

Ans: Given n= 4,P= Rs. 3000, r= 9 and p= 2i=9

2 x 100= 0.045

Now, A =P

i1 1npi =3000

0.0452 x 4[(1 0.045) 1] =3000

0.045x 0.4221

Thus, A= Rs. 28,140

Example 3: Mr. Ravi invested Rs. 5000 in an annuity with quarterly

payments for a period of 2 years at the rate of interest of 10%. Find the

accumulated value of the annuity at the end of 2ndyear.

Ans: Given n= 2,P= Rs.5000, r= 10 and p= 4i=10

4 x 100= 0.025

Now, A =P

i1 1npi =5000

0.0252 x 4(1.025) 1 =5000

0.025x 0.2184

Thus, A= Rs. 43,680

Example 4:Mr. Ashok Rane borrowed Rs. 20,000 at 4% p.a. compounded

annually for 10 years. Find the periodic payment he has to make.

Ans: Given PV= Rs. 20,000; n= 10; p =1 and r= 4i= 0.04

Now to find the periodic payments Pwe use the following formula:

PV=

11

1npP

i i

20000 =

10110.04 1 0.04P

=0.04Px 0.3244

P=20000 x 0.04

0.3244= 2466.09A= 1 1x 100

x 100npP r

r p

p

A=P

i1 1npi Here,

P: periodic equal payment

r: rate of interest p.a.

p: period of annuity i.e. yearly,

half yearly, quarterly or

monthly

and i=x 100r

p

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Thus, Mr. Rane has to make the periodic payments of Rs. 2466.09

Example 5:Find the future value of an immediate annuity after 3 years

with the periodic payment of Rs. 12000 at 5% p.a. if the period of

payments is (i) yearly, (ii) half -yearly, (iii) quarterly and (iv) monthly.

Ans: Given P= Rs. 1200, n= 3,r= 5

(i) period p= 1 then i=5

100= 0.05

A=P

i1 1ni =12000

0.0531 0.05 1 =12000

0.05 1.1576 1

A= 12000 x 3.1525 = Rs. 37,830

(ii) period p= 2 then i=5

2 x 100= 0.025

A=P

i21 1ni =12000

0.0256(1 0.025) 1 =12000

0.025x 0.1597

A= 12000 x 6.388 = Rs. 76,656

(iii) period p=4 then i=5

4 x 100= 0.0125

A=P

i41 1ni =12000

0.0125[(1 + 0.0125)12–1] =12000

0.0125x 0.16075

A= 12000 x 12.86 = Rs. 1,54,320

(iv) period p= 12 then i=5

12 x 100= 0.00417

A=P

i121 1ni =12000

0.00417[(1 + 0.00417)36–1] =12000

0.00417x 0.1615

A= 1200 x 38.729 = Rs. 4,64,748

Example 6:Mr. Nagori invested certain princ ipal for 3 years at 8%

interest compounded half yearly. If he received Rs.72957.5 at the end of

3rdyear, find the periodic payment he made. [(1.04)6= 1.2653]

Ans: Given n= 3,r= 8,p= 2i=8

2 x 100= 0.04

Now, A=P

i1 1npi

72957.5 =0.04P[(1 + 0.04)6–1]=0.04Px 0.2653

72957.5 = P[6.6325]munotes.in

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P=72957.5

6.6325= 11000

Thus, the periodic payment is Rs. 11,000

3.4SINKING FUND

The fund (money) which is kept aside to accumulate a certain sum

in a fixed period through periodic equal payments is called as sinking

fund .

We can consider an example of a machine in a factory which needs

to be replaced after say 10 years. The amount for buying a new machine

10 ye ars from now may be very large, so a proportionate amount is

accumulated every year so that it amounts to the required sum in 10 years.

This annual amount is called as sinking fund . Another common example is

of the maintenance tax collected by any Society from its members.

A sinking fund being same as an annuity, the formula to compute

the terms is same as that we have learnt in section 2.3.3

Example 7:A company sets aside a sum of Rs. 15,000 annually to enable

it to pay off a debenture issue of Rs. 1 ,80,000 at the end of 10 years.

Assuming that the sum accumulates at 6% p.a., find the surplus after

paying off the debenture stock.

Ans: Given P= Rs. 15000, n= 10, r= 6i= 0.06

A=P

i1 1ni =15000

0.06x [(1 + 0.06)10–1] =15000

0.06x 0.7908

A= Rs. 1,97,700

Thus, the surplus amount after paying off the debenture stock is

= 197712 –180000 = Rs. 17712.

Example 8:Shriniketan Co -op Hsg. Society has 8 members and colle cts

Rs. 2500 as maintenance charges from every member per year. The rate of

compound interest is 8% p.a. If after 4 years the society needs to do a

work worth Rs. 100000, are the annual charges enough to bear the cost?

Ans: Since we want to verify whether Rs. 2500 yearly charges are enough

or not we assume it to be Pand find its value using the formula:

A=P

i1 1ni

Here A= Rs. 100000, n= 4,r= 8i= 0.08munotes.in

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P=

x

1 1nA i

i=

4100000 x 0.08

1 0.08 1= 22192

Thus, the annual payment of all the members i.e. 8 members should be Rs.

22192.

the annual payment per member =22192

8= Rs. 2774

This payment is less than Rs. 2500 which the society has decided to take

presently. Th us, the society should increase the annual sinking fund.

3.5EQUATED MONTHLY INSTALLMENT (EMI)

Suppose a person takes a loan from a bank at a certain rate of

interest for a fixed period. The equal payments which the person has to

make to the bank per m onth are called as equated monthly installments in

short EMI.

Thus, EMI is a kind of annuity with period of payment being

monthly and the present value being the sum borrowed .

We will now study the method of finding the EMI using reducing

balance method and flat interest method.

(a)Reducing balance method :

Let us recall the formula of finding the present value of an annuity.

PV=

11

1npP

i i

The equal periodic payment ( P) is our EMI which is denoted it by E.

The present value ( PV) is s ame as the sum ( S) borrowed.

Also the period being monthly p= 12 ,i=1200ras we are interested in

finding monthly installments and nis period in years .

Substituting this in the above formula we have:

S=

1211

1nE

i i

Thus, if Sis the sum borrowed for nyears with rate of interest r% p.a.

then the EMI is calculated by the formula:

E=

12x

11(1 )nS i

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(b) Flat Interest Method:

Here the amount is calculated using Simple Interest for the period and

the EMI i s computed by dividing the amount by total number of monthly

installments.

LetSdenote the sum borrowed, rdenote the rate of interest and n

denote the duration in years, then as we know the amount using simple

interest formula is A=S1100nr.The total number of monthly

installments for duration of nyears is 12n. Hence the EMI is calculated as

E=12A

n

Example 9:Mr. Sudhir Joshi has taken a loan of Rs. 10,00,000 from a

bank for 10 years at 11% p.a. Find his EMI using (a) reducing balance

method and (b) Flat interest method.

Ans: Given S= Rs. 1000000, n= 10, r= 11i=11

1200= 0.0092

(a)Using flat interest method :

A=S1100nr=10000001101100=2100000

Thus, E=12A

n=2100000

120= 17,500 … (1)

(b)Using reducing balance method :

Now, E=

12x

11(1 )nS i

i=

1201000000 x 0.0092

11(1 0.0092)= 13797.65

E= Rs. 13,798 approximately … (2)

Comparing (1)and (2), we can see that the EMI using flat interest method

is higher than by reducing balance method .

Example 10:Mr. Prabhakar Naik has borrowed a sum of Rs. 60,000 from

a person at 6% p.a. and is due to return it back in 4 monthly installments.

Find th eEMI he has to pay and also prepare the amortization table of

repayment.

Ans: Given S= Rs. 60,000; n= 4 months;

r= 6%i=6

1200= 0.005

Now, E=x

11(1 )nS i

i=

460000 x 0.005

11(1 0.005)=300

0.01975

E= Rs. 15,187.97munotes.in

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Now, we will prepare the amortization table i.e. the table of repayment of

the sum borrowed using reducing balance method.

In the beginning of the 1stmonth the outstanding principal is the sum

borrowed i.e. Rs. 60000 and the EMI paid is Rs. 15187.97

The interest on the outstanding principal is 0.005 x 60000 = Rs. 300 … (1)

Thus, the principal repayment is 15187.97 –300 = Rs. 14887.97 … (2)

The outstanding principal ( O/P) in the beginning of the 2ndmonth is now

60000 –14887.97 = 45112.03.

Note :

(1) is called the interest part of the EMI and (2) is called as the

principal part of the EMI.

As the tenure increases the interest part reduces and the principal part

increases.

This calculation can be tabulated as follows:

O/P EMI Interest PartPrincipal

Part Month

(a) (b) (c) = (a) xi (b)-(c)

1 60000 15187.97 300 14887.97

2 45112.03 15187.97 225.56 14962.45

3 30141.02 15187.97 150.75 15037.22

4 15111.80 15187.97 75.56 15112.41

In the beginning of the 4thmonth t he outstanding principal is Rs. 15111.80

but the actual principal repayment in that month is Rs. 15112.41. This

difference is due to rounding off the values to two decimals, which leads

the borrower to pay 61 paise more!!

Example 11:Mr. Shyam Rane has bo rrowed a sum of Rs. 100000 from a

bank at 12% p.a. and is due to return it back in 5 monthly installments.

Find the EMI he has to pay and also prepare the amortization table of

repayment.

Ans: Given S= Rs. 100000; n= 5 months;

r= 12% p.a. =12

12= 1% p.m i= 0.01

Now, E=x

11(1 )nS i

i=

5100000 x 0.01

11(1 0.01)=1000

0.0485343= 20603.98munotes.in

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The amortization table is as follows:

O/P EMI Interest PartPrincipal

Part Month

(a) (b) (c) = (a) xi (b)-(c)

1 100000 20603.98 1000 19603.98

2 80396.02 20603.98 803.96 19800.02

3 60596 20603.98 605.96 19998.02

4 40597.98 20603.98 405.98 20198

5 20399.98 20603.98 204 20399.98

Check your progress

1.An overdraft of Rs. 50,000 is to be paid back in equal annual

installments in 20 years. Find the installments, if the interest is 12%

p.a. compounded annually. [(1.12)20= 9.64629]

2.A man borrows Rs. 30,000 at 6% p.a. compounded semi -annually

for 5 years. Find the periodic payments he has to make.

3.What periodic payments Mr. Narayana n has to make if he has

borrowed Rs. 1,00,000 at 12% p.a. compounded annually for 12

years? [(1.12)12= 3.896]

4.Find the future value of an immediate annuity of Rs. 1200 at 6% p.a.

compounded annually for 3 years.

5.Find the future value of an immediate annui ty of Rs. 500 at 8% p.a.

compounded p.m. for 5 years.

6.Find the accumulated value after 2 years if a sum of Rs. 1500 is

invested at the end of every year at 10% p.a. compounded quarterly.

7.Find the accumulated amount of an immediate annuity of Rs. 1000 at

9%p.a. compounded semi -annually for 4 years.

8.Find the future value of an immediate annuity of Rs. 2800 paid at

10% p.a. compounded quarterly for 2 years. Also find the interest

earned on the annuity.

9.Find the sum invested and the accumulated amount for an ordinary

annuity with periodic payment of Rs. 2500, at the rate of interest of

9% p.a. for 2 years if the period of payment is (a) yearly, (b) half -

yearly, (c) quarterly or (d) monthly.

10.Find the sum invested and the accumulated amount for an ordinary

annui ty with periodic payment of Rs. 1500, at the rate of interest of

10% p.a. for 3 years if the period of payment is (a) yearly, (b) half -

yearly, (c) quarterly or (d) monthly.

11.Mr. Banerjee wants to accumulate Rs. 5,00,000 at the end of 10

years from now. How much amount should he invest every year at

the rate of interest of 9% p.a. compounded annually?munotes.in

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12.Find the periodic payment to be made so that Rs. 25000 gets

accumulated at the end of 4 years at 6% p.a. compounded annually.

13.Find the periodic payment to be ma de so that Rs. 30,000 gets

accumulated at the end of 5 years at 8% p.a. compounded half

yearly.

14.Find the periodic payment to be made so that Rs. 2000 gets

accumulated at the end of 2 years at 12% p.a. compounded quarterly.

15.Find the rate of interest if a pe rson depositing Rs. 1000 annually for

2 years receives Rs. 2070.

16.Find the rate of interest compounded p.a. if an immediate annuity of

Rs. 50,000 amounts to Rs. 1,03,000 in 2 years.

17.Find the rate of interest compounded p.a. if an immediate annuity of

Rs. 50 00 amounts to Rs. 10400 in 2 years.

18.What is the value of the annuity at the end of 5 years, if Rs. 1000

p.m. is deposited into an account earning interest 9% p.a.

compounded monthly? What is the interest paid in this amount?

19.What is the value of the annuit y at the end of 3 years, if Rs. 500 p.m.

is deposited into an account earning interest 6% p.a. compounded

monthly? What is the interest paid in this amount?

20.Mr. Ashish Gokhale borrows Rs. 5000 from a bank at 8% compound

interest. If he makes an annual paym ent of Rs. 1500 for 4 years,

what is his remaining loan amount after 4 years?

(Hint : find the amount using compound interest formula for 4 years and

then find the accumulated amount of annuity, the difference is the

remaining amount.)

21.Find the present val ue of an immediate annuity of Rs. 10,000 for 3

years at 6% p.a. compounded annually.

22.Find the present value of an immediate annuity of Rs. 100000 for 4

years at 8% p.a. compounded half yearly.

23.Find the present value of an immediate annuity of Rs. 1600 for 2

years at 7% p.a. compounded half yearly.

24.A loan is repaid fully with interest in 5 annual installments of Rs.

15,000 at 8% p.a. Find the present value of the loan.

25.Mr. Suman borrows Rs. 50,000 from Mr. Juman and agreed to pay

Rs. 14000 annually for 4 yea rs at 10% p.a. Is this business profitable

to Mr. Juman?

(Hint: Find the PVof the annuity and compare with Rs. 50000)

26.Mr. Paradkar is interested in saving a certain sum which will amount

to Rs. 3,50,000 in 5 years. If the rate of interest is 12% p.a., ho w

much should he save yearly to achieve his target?

27.Mr. Kedar Pethkar invests Rs. 10000 per year for his daughter from

her first birthday onwards. If he receives an interest of 8.5% p.a.,

what is the amount accumulated when his daughter turns 18?munotes.in

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28.Dr. Wakan kar, a dentist has started his own dispensary. He wants to

install a machine chair which costs Rs. 3,25,000. The machine chair

is also available on monthly rent of Rs. 9000 at 9% p.a. for 3 years.

Should Dr. Wakankar buy it in cash or rent it?

29.A sum of Rs. 50,000 is required to buy a new machine in a factory.

What sinking fund should the factory accumulate at 8% p.a.

compounded annually if the machine is to be replaced after 5 years?

30.The present cost of a machine is Rs. 80,000. Find the sinking fund

the com pany has to generate so that it could buy a new machine after

10 years, whose value then would be 25% more than of today’s

price. The rate of compound interest being 12% p.a. compounded

annually.

31.Mr. Mistry has to options while buying a German wheel alignm ent

machine for his garage: (a) either buy it at Rs. 1,26,000 or (b) take it

on lease for 5 years at an annual rent of Rs. 30,000 at the rate of

interest of 12% p.a.. Assuming no scrap value for the machine which

option should Mr. Mistry exercise?

32.Regency Co-op. Hsg. Society which has 50 members require Rs.

12,60,000 at the end of 3 years from now for the society repairs. If

the rate of compound interest is 10% p.a., how much fund the society

should collect from every member to meet the necessary sum?

33.Mr. L alwaney is of 40 years now and wants to create a fund of Rs.

15,00,000 when he is 60. What sum of money should he save

annually so that at 13% p.a. he would achieve his target?

34.If a society accumulates Rs. 1000 p.a. from its 200 members for 5

years and rec eives 12% interest then find the sum accumulated at the

end of the fifth year. If the society wants Rs. 13,00,000 for society

maintenance after 5 years, is the annual fund of Rs. 1000 per

member sufficient?

35.How much amount should a factory owner invest eve ry year at 6%

p.a. for 6 years, so that he can replace a mixture -drum (machine)

costing Rs. 60,000, if the scrap value of the mixture -drum is Rs.

8,000 at the end of 6 years.

36.If a society accumulates Rs. 800 p.a. from its 100 members for 3

years and receiv es 9% interest then find the sum accumulated at the

end of the third year. If the society wants Rs. 2,50,000 for society

maintenance after 3 years, is the annual fund of Rs. 800 per member

sufficient?

37.Mr. Kanishk wants clear his loan of Rs. 10,00,000 taken at 12% p.a.

in 240 monthly installments. Find his EMI using reducing balance

method.

38.Using the reducing balance method find the EMI for the following:munotes.in

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Loan amount

(in Rs.)Rate of Interest

(in % p.a.)Period of Loan

(in years)

i) 1000 6 5

ii) 50000 6 10

iii) 8000 7 6

iv) 12000 9 10

v) 1000 9.5 10

vi) 1100000 12.5 20

39.Mr. Vilas Khopkar has taken a loan of Rs. 90,000 at 11% p.a. Find

the EMI using (a) reducing balance method and (b) Flat interest

method, if he has to return the loan in 4 years.

40.Find the EMI using reducing balance method on a sum of Rs. 36,000

at 9%, to be returned in 6 monthly installments.

41.Find the EMI using reducing balance method on a sum of Rs. 72,000

at 12%, to be returned in 12 installments.

42.Mr. Sachin Andhale has borrowed Rs. 1 0,000 from his friend at 9%

p.a. and has agreed to return the amount with interest in 4 months.

Find his EMI and also prepare the amortization table.

43.Mr. Arvind Kamble has borrowed Rs. 30,000 from his friend at 14%

p.a. If he is to return this amount in 5 monthly installments, find the

installment amount, the interest paid and prepare the amortization

table for repayment.

44.Mrs. Chaphekar has taken a loan of Rs. 1,25,000 from a bank at 12%

p.a. If the loan has to be returned in 3 years, find the EMI, Mrs.

Chaphekar has to pay. Prepare the amortization table of repayment

of loan and find the interest she has to pay.

45.A loan of Rs. 75,000 is to be returned with interest in 4 installments

at 15% p.a. Find the value of the installments.

46.A loan of Rs. 60,000 is to b e returned in 6 equal installments at 12%

p.a. Find the amount of the installments.

47.Find the sum accumulated by paying an EMI of Rs. 11,800 for 2

years at 10% p.a.

48.Find the sum accumulated by paying an EMI of Rs. 1,800 for 2 years

at 12% p.a.

49.Find the sum accumulated by paying an EMI of Rs. 12,000 for 3

years at 9% p.a.

50.Find the sum accumulated by paying an EMI of Rs. 11,000 for 8

years at 9.5% p.a.munotes.in

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Hints & Solutions to Check your progress

(1)6694 (2)3517 (3)16,144 (4)3820.32

(5)36555.65 (6)13104 (7)9380 (8)24461

(9)

Period Sum Invested Accumulated

Amount

Yearly 5000 5225

Half-yearly 10000 10695.5

Quarterly 20000 21648

Monthly 60000 65471

(10)

Period Sum Invested Accumulated

Amount

Yearly 4500 4965

Half-yearly 9000 10203

Quarterly 18000 20693

Monthly 54000 62635

(11) 32910 (12) 5715 (13) 2498.72 (14) 225

(15) 7% (16) 6% (17) 8% (18) 75424, 15424

(19) 19688 , 1688 (20) 4719 (21) 26730

(22) 673274.5 (23) 5877 (24) 59890.65 (25) 44378, Yes

(26) 97093.4 (27) 393229.95 (28) 283021.25, take it on rent

(29)12523 (30) 17698.42 (31) 108143.28 < 126000, Mr. Mistry

should use the second option. (32) 16245 (33) 18530

(34) 1270569.47, not sufficient (35) 7454.86 (36) 2,62,248; yes

(37) 11,011

(38)

Loan amount

(in Rs.)Rate of inte rest

(in % p.a.)Period of Loan

(in yrs.)EMI

(in Rs.)

i) 1000 6 5 19

ii) 50000 6 10 555

iii) 8000 7 6 136

iv) 12000 9 10 152

v) 1000 9.5 10 13

vi) 1100000 12.5 20 12498

(39) 2326, 2700 (40) 6158.48 (41) 6397.11munotes.in

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(42)

O/P EMI Intere st PartPrincipal

Part Month

(a) (b) (c) = (a) xi (b)-(c)

1 10000 2547.05 75 2472.05

2 7527.95 2547.05 56.45 2490.6

3 5037.35 2547.05 37.78 2509.27

4 2528.08 2547.05 18.96 2528.09

(43)

O/P EMI Interest PartPrincipal

Part Month

(a) (b) (c) = (a) xi (b)-(c)

1 30000 6212.23 351 5861.23

2 24138.77 6212.23 282.42 5929.81

3 18208.96 6212.23 213.04 5999.19

4 12209.77 6212.23 142.85 6069.38

5 6140.39 6212.23 71.84 6140.39

(45)19339.57 (46) 16353 (47) 3,12,673.60

(48) 48552.24 (49) 4,93,832.6 (50)15,72,727

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UNIT -III

Unit -4

CORRELATION AND REGRESSION

Unit Structure :

4.0 Objectives

4.1 Introduction

4.2 Types of Correlation

4.3 Measurement of Correlation

4.4 Rank Correlation

4.5 Regression Analysis

4.0OBJECTIVE S

To understand the relationship between two relevant characteristics of

a statistical unit.

Learn to obtain the numerical me asure of the relationship between two

variables.

Use the mathematical relationship between two variables in order to

estimate value of one variable from the other.

Use the mathematical relationship to obtain the statistical constants

line means andS.D.’s

4.1 INTRODUCTION

Inthe statistical analysis we come across the study of two or more

relevant characteristics together in terms of their interrelations or

interdependence. e.g. Interrelationship among production, sales and

profits of a company. Inter relationship among rainfall, fertilizers, yield

and profits to the farmers .

Relationship between price and demand of a commodity When we

collect the information (data) on two of such characteristics it is called

bivariate data. It is generally denoted by (X,Y) where X and Y are the

variables representing the values on the characteristics.

Following are some examples of bivariate data.

a)Income and Expenditure of workers.

b)Marks of students in the two subjects of Maths and Accounts.

c)Height of Husband and Wif e in a couple.

d)Sales and profits of a company.munotes.in

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Between these variables we can note that there exist some sort of

interrelationship or cause and effect relationship. i.e. change in the value

of one variable brings out the change in the value of other vari able also.

Such relationship is called as correlation.

Therefore, correlation analysis gives the idea about the nature and

extent of relationship between two variables in the bivariate data.

4.2 TYPES OF CORRELATION:

There are two types of correlation .

a)Positive correlation. and b) Negative correlation.

4.2.1Positive correlation: When the relationship between the variables X

andYis such that increase or decrease in Xbrings out the increase or

decrease in Yalso, i.e. there is direct relation betwe enXand Y, the

correlation is said to be positive. In particular when the ‘change in X

equals to change in Y’ the correlation is perfect and positive. e.g. Sales and

Profits have positive correlation.

4.2.2Negative correlation: When the relationship betw een the variables X

andYis such that increase or decrease in Xbrings out the decrease or

increase in Y, i.e. there is aninverse relation between Xand Y, the

correlation is said to be negative. In particular when the ‘change in X

equals to change in Y’but in opposite direction the correlation is perfect

and negative. e.g. Price and Demand have negative correlation.

4.3 MEASUREMENT OF CORRELATION

The extent of correlation can be measured by any of the following

methods:

Scatter diagrams

Karl Pearson’ s co-efficient of correlation

Spearman’s Rank correlation

4.3.1 Scatter Diagram: The Scatter diagram is a chart prepared by

plotting the values of X and Y as the points (X,Y) on the graph. The

pattern of the points is used to explain the nature of correla tion as follows.

The following figures and the explanations would make it clearer.

(i)Perfect Positive Correlation :

If the graph of the values of the variables is a

straight line with positive slope as shown in

Figure 4.1, we say there is aperfect positive

correlation between XandY. Here r =1.

(ii)Imperfect Positive Correlation :

If the graph of the values of XandYshow a band

of points from lower left corner to upper right corner

as shown in Figure 4.2, we say that ther e is an imperfectO Fig4.1 X

Y

O Fig4.2 X

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positive correlation . Here 0 < r< 1.

(iii)Perfect Negative Correlation :

If the graph of the values of the variables is a

straight line with negative slope as shown in

Figure 4.3, we say there is a perfect negative

correlation between XandY. Here r=–1.

(iv)Imperfect Nega tive Correlation :

If the graph of the values of XandYshow a band

of points from upper left corner to the lower right

corner as shown in Figure 4.4, then we sa y that

there is an imperfect negative correlation . Here –1 (v)Zero Correlation :

If the graph of the values of XandYdo not show any of the above trend

then we say that there is a zero correlation between XandY. The graph of

such type can be a straight line perpendicular to the axis, as shown in

Figure 4.5and4.6,or may be completely scattered as shown in Figure

4.7. Here r= 0.

The Figure 4.5show that the increase in the values of Yhas no effect on

the value of X, it remains the same, hence zero correlation. The Figure 4.6

show that the increase in the values of Xhas no effect on the value of Y, it

remains the same, hence zero correlation. The Figure 4.7show that the

points are completely scattered on the graph and show no particular trend,

hence there is no correlation or zero correlation between XandY.

4.3.2Karl Pearson’s co -efficient of correlation.

This co -efficient provides the numerical measure of the correlation

between the variables X and Y. It is suggested by Prof. Karl Pearson and

calculated by the formula

( , )

.x yCov x yr

Where, Cov(x,y) : Covariance between x&y

x.: Standard deviation of x&y: Standard deviation of y

Also, Cov(x,y) =1

n(x-x) (y-y) =1

nxy-xyY

O Fig4.5 XY

O Fig4.6XY

O Fig4.7 XY

O X

Fig.4.3

Y

O Fig.4.3 X

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S.D.(x) = x =2X)1x-n =2 2X1xn and

S.D.(y) =y =2)1y-Yn2 21y Yn

Remark : We can also calculate this co -efficient by using the formula

given by

2 22 22 2X) ) X

X) )1 xyx- y- -

1 1 x yx- y- XY Y

Y Yn nr

n n n n

The Pearson’s Correlation co -efficient is also called as the ‘product

mom ent correlation co -efficient’

Properties of correlation co -efficient ‘r’

The value of ‘r’ can be positive (+) or negative( -)

The value of ‘r’ always lies between –1 & +1, i.e. –1< r<+1]

Significance of ‘r’ equals to –1, +1 & 0

When ‘r’= +1; the correl ation is perfect and positive.

‘r’=-1; the correlation is perfect and negative.

andwhen there is no correlation ‘r’= 0

SOLVED EXAMPLES :

Example .1:Calculate the Karl Pearson’s correlation coefficient from the

following.

X: 12 10 20 13 15

Y: 7 14 6 12 11

Solution: Table of calculation,

And n= 5

The Pearson’s correlation coefficient r is gi ven by,

( , )

.x yCov x yr

Where,X Y XxY X2Y2

12 7 84 144 49

10 14 140 100 196

20 6 120 400 36

13 12 156 169 144

15 11 165 225 121

x =70y= 50xy=665x2=1038y2=546munotes.in

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x=x

n=70

5= 14 y=y

n=50

5=10

Cov(x,y) = Xxy-Ynx=2 2X1xn y=2 21y Yn

=665

5-14x10 =2 1038145 =2 546105

= 133 -140 =11.6 =3.40 9.2 =3.03 =-07

... Cov(x,y) =-7x=3.40 and y= 3.03

Substituting the values in the formula ofrwe get

7

3.40x3.03r =-0.68

.

... r = -0.68

Example 2:Let us calculate co -efficient of correlation between Marks of

students in the Subjects of Maths & Accounts. in a certain test conducted.

Table of calculation:

n=10

Now Pearson’s co -efficient of correlation is given by the fomula,

( , )

.x yCov x yr

Where,

x=x

n=242

10= 24.2 y=y

n=287

10=28.7Marks

InMaths

XMarks In

Accounts

Y XY X2Y2

28 30 840 784 900

25 40 1000 625 1600

32 50 1600 1024 2500

16 18 288 256 324

20 25 500 400 625

15 12 180 225 144

19 11 209 361 121

17 21 357 289 441

40 45 1800 1600 2025

30 35 1050 900 1225

x= 242y=287xy=7824x2= 6464y29905munotes.in

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Cov(x,y) = Xxy-Ynx=2 2X1xny=2 21y Yn

=7824

10-24.2x 28.7x=2 646424.210y=2 990528.710

=782.4 -694.54 =60.76 =166.81

Cov(x,y) = 87.86 ,x=7.79 andy= 12.91

... Cov(x,y) = 87.86 x= 7.79 and y= 12.91

Substituting the values in the formula ofrwe get

87.86

7.79 12.91rx =0.87

... r = 0.87

4.4RANK CORRELATION

In many practical situations, we do not have the scores on the

characteristics, but the rank s (preference order) decided by two or more

observers. Suppose, a singing competition of 10 participants is judged by

two judges A and B who rank or assign scores to the participants on the

basis of their performance. Then it is quite possible that the ran ks or scores

assigned may not be equal for all the participants. Now the difference in

the ranks or scores assigned indicates that there is a difference of openion

between the judges on deciding the ranks. The rank correlation studies the

association in th is ranking of the observations by two or more observers.

The measure of the extent of association in rank allocation by the two

judges is calculated by the co -efficient of Rank correlation ‘R’. This co -

efficient was developed by the British psychologist Ed ward Spearman in

1904.

Mathematically, Spearman’s rank correlation co -efficient is defined as,

R= 1 -2

2d

( 1)n n

Where d= rank difference and n= no of pairs.

Remarks: We can note that, the value of ‘R’ always lies between –1 and

+1

Thepositive value of ‘R’ indicates the positive correlation (association) in

the rank allocation. Whereas ,the negative value of ‘R’ indicates the

negative correlation (association) in the rank allocation.munotes.in

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SOLVED EXAMPLES:

Example 3

a)When ranks are g iven: -

Data given below read the ranks assigned by two judeges to 8 participants .

Calculate the co -efficient of Rank correlation.

Ranks by Judge Participant

No. A BRank diff

Square d2

1 5 4 (5-4)2= 1

2 6 8 4

3 7 ` 1 36

4 1 7 36

5 8 5 9

6 2 6 16

7 3 2 1

8 4 3 1

N = 8 Total 104 =d2

Spearman’s rank correlation co -efficient is given b y

R=1-2

26 d

( 1)n n

Substituting the values from the table we get,

R= 1 -26x 104

8(8 1)=-0.23

The value of correlation co -efficient is -0.23. This indicates that there is

negative association in rank allocation by the two judges A and B

b)When scores are given: -

Example 4

The data given below are the marks given by two Examiners to a

set of 10 students in a aptitude test. Calculate the Spearman’s Rank

correlation co -efficient, ‘R’

Now the Spearman’s rank correlation co-efficient is given by

R= 1 -2

26 d

( 1)n n

Substituting the values from the table we get,

R= 1 -26x5

10(10 1)

= 1-0.03

= 0.97munotes.in

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The value of correlation co -efficient is + 0. 97. This indicates that

there is positive association in assessment of two examiners, A and B.

c)Case of repeated values: -

It is quite possible that the two participants may be assigned the

same score by the judges. In such cases Rank allocation and cal culation of

rank correlation can be explained as follows.

Example : The data given below scores assigned by two judges for 10

participants in the singing competition. Calculate the Spearman’s Rank

correlation co -efficient.

Score assigned B y

JudgesRanks Rank

difference

squareParticipant

No.

A B RA RB D2

1 28 35 9 (8.5) 6 (8.5-6)2

=6.25

2 40 26 3 10(9.5) 42.25

3 35 42 5 (4.5) 3 2.25

4 25 26 10 9 (9.5) 0.25

5 28 33 8 (8.5) 7 2.25

6 35 45 4 (4.5) 2 6.25

7 50 32 1 8 49

8 48 51 2 1 1

9 32 39 6 4 4

10 30 36 7 5 4

N = 10 Total d2=117.5

Marka ByExaminer Ranks Rank

difference

squareStudent No.

A B RA RB D2

1 85 80 2 2 0

2 56 60 8 7 1

3 45 50 10 10 0

4 65 62 6 6 0

5 96 90 1 1 0

6 52 55 9 8 1

7 80 75 3 4 1

8 75 68 5 5 0

9 78 77 4 3 1

10 60 53 7 9 1

N = 10 Total 5 =d2munotes.in

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Explanation: -In the column of A and B there is repeatation of scores so

while assigning the ranks we first assign the ranks by treating them as

different values and then for rereated scores we assign the average rank.

e.g In col A the score 35 appears 2 times at number 4 and 5 in the order of

ranking so we calculate the average rank as ( 4+5)/2 = 4.5.

Hence the ranks assigned are 4.5 each. The other repeated scores can be

ranked in the same manner.

Note: In this example we can note that the ranks are in fraction e.g. 4.5,

which is logically incorrect or meaningless. Therefore in the calculation of

‘R’ we add a correction factor (C.F.) to d2calculated as follows.

Table of correction factor (C.F.)

Value

Repeate dFrequency

M m(m2-1)

35 2 2x(22-1)=6

28 2 6

26 2 6

Totalm(m2-1)=18

Now3m ). .12mC F = 18/12 =1.5

...d2= 117.5+1.5= 119

We use this value in the calculation of ‘R’

Now the Spearman’s rank correlation co -efficient is g iven by

R= 1 -2

26 d

( 1)n n

Substituting the values we get, R= 1 -26x 119

10(10 1)= 1-0.72 = 0.28

EXERCISE I

1.What is mean by correlation? Explain the types of correlation with

suitable examples.

2.What is a scatter diagram? Draw different scattered diagrams to

explain the correlation between two variables x and y.

3.State the significance of ‘r’ = +1, –1 and 0.

4.Calculate the coefficient of correlation r from the following data.

X: 18 12 16 14 10 15 17 13

Y: 9 13 20 15 11 24 26 22munotes.in

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5.The following table gives the price and demand of a certain

commodity over the period of 8 months. Calculate the Pearson’s

coefficient of correlation.

Price: 15 12 23 25 18 17 11 19

Demand 45 30 60 65 48 45 28 50

6.Following results are obtained on a certain bivariate data.

(i) n = 10 x= 75 y = 70 x2= 480

y2= 600xy = 540

(ii) n = 15 x= 60 y = 85 x2= 520

y2=1200xy = -340

Calculate the Pears on’s correlation coefficient in each case.

7.Following data are available on a certain bi -variate data :

(i)(x-x) (y-y)=120,(x-x)2= 150(y-y)2= 145

(ii)(x-x) (y-y)=-122,(x-x)2= 136(y-y)2= 148

Find the correlation coefficient.

8. Calculate the Pearson’s coefficient of correlation fr om the given

information on a bivariate series:

No of pairs: 25

Sum of x values:300

Sum of y values:375

Sum of squares of x values: 9000

Sum of squares of y values:6500

Sum of the product of x and y values:4000.

9.The ranks assigned to 8 participants by two judges are as

followes.Calculate the Spearman’s Rank correlation coefficient ‘R’.

Participant No: 1 2 3 4 5 6 7 8

Ranks by JudgeI: 5 3 4 6 1 8 7 2

JudgeII : 6 8 3 7 1 5 4 2

10.Calculate the coefficient of rank correlation from the data given below.

X: 40 33 60 59 50 55 48

Y: 70 60 85 75 72 82 69

11. Marks given by two Judges to a group of 10 participants are as

follows. Calculate the coefficient of rank correlation.

Marks by Judge

A: 52 53 42 60 45 41 37 38 25 27

Judge B: 65 68 43 38 77 48 35 30 25 50.

12. An examination of 8 applicants for a clerical post was by a bank. The

marks obtained by the applicants in the subjects of Mathematics and

Accountancy were as f ollows. Calculate the rank correlation coefficient.

Applicant: A B C D E F G Hmunotes.in

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Marks in

Maths: 15 20 28 12 40 60 20 80

Marks in

Accounts: 40 30 50 30 20 10 25 60

4.5REGRESSION ANALYSIS

As the correlation analysis studies the nature and extent of

interrelationship between the two variables X and Y, regression analysis

helps us to estimate or approximate the value of one variable when we

know the value of other var iable. Therefore we can define the

‘Regression’ as the estimation (prediction) of one variable from the other

variable when they are correlated to each other. e.g. We can estimate the

Demand of the commodity if we know it’s Price.

Why are there two regres sions?

When the variables XandYare correlated there are two possibilities,

(i)Variable Xdepends on variable y. in this case we can find the value of

x if know the value of y. This is called regression of xon.

(ii)Variabledepends on variable X. we can find the value of yif know

the value of X. This is called regression of yonx. Hence there are two

regressions,

(a)Regression of X on Y; (b) Regression of XonY.

4.5.1 Formulas on Regression equation,

Regression of XonY Regression of XonY

Assumption: X depends on Y Y depends on X

The regression equation is The regression equation is

(x-x) = b xy(y-y) (y-y) = b yx(x-x)

bxy= Re gression co -efficient of b yx= Regression co -efficient

X on Y =(x, )

( )Cov y

V yof Y on X =(x, )

(x)Cov y

V

Where,

Cov(x,y) =1

n(x-x) (y-y) =1

nxy-xy

V(x) =1

n(x-x)2and V(y) =1

n(y-y)2

V(x) =1

nx2-x2and V(y) =1

ny2-y2

Use: To find X Use: To find munotes.in

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SOLVED EXAMPLES

Example 1:

Obtain the two regression equatio ne and hence find the value of x when

y=25

Data: -

X Y X2Y2XxY

8 15 64 225 120

10 20 100 400 200

12 30 144 900 360

15 40 225 1600 600

20 45 400 2025 900

x=65y=150x2=933y2=

5150xy=

2180

And n= 5

Now the two regr ession equations are,

(x-x) = b xy(y-y) ------- x on y (i)

(y-y) = b yx(x-x)------- y on x (ii)

Where,

x=1

nx =65

5=13 and y=1

ny =150

5=30

Also,

Cov(x,y,) =1

nxy-xy V(x) =1

nx2-x2V(y) =1

ny2-y2

=2180

5-13x30 =933

5-132=5150

5-302

= 436 -390 = 186.6 -169 = 1030 –900

... Cov(x,y) = 46 V(x) = 17.6 V(y) =130

Now we find,

Regression co -efficient of X on Y Regression co -efficient of X on Y

bxy=(x, )

( )Cov y

V ybyx=(x, )

(x)Cov y

V

=46

130=46

17.6

...bxy= 0.35 and byx= 2.61

Now substituting the values of x,y,bxyand b yxin the regression

equations we get,

(x-13) = 0.35(y -30)-------x on y (i)

(y-30) =2.61(x -13)------- y on x (ii)

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Now to estimate x when y =25, we use the regression equation of x on y

...(x-13) = 0.35(25 -30)

... x = 13 -1.75 = 11.25

Remark:

From the above example we can note some points about Regression

coefficients.

Both the regression coefficients carry the same sign (+ or -)

Both the regression coefficients can not be greater than 1 in

number

(e.g. -1.25 and -1.32) is not possible.

Product of both the regression coefficients b xyand b yxmust be < 1

i.e. b xy Xbyx< 1 Here 0.35x2.61 = 0.91< 1 (Check this

always)

Example 2 :

Obtain the two regression equation sand hence find the value of y when

x=10

Data: -

X Y XxY X2Y2

12 25 300 144 625

20 18 360 400 324

8 17 136 64 289

14 13 182 196 169

16 15 240 256 225

x=70y=88xy=1218x2=1060y2=1632

And n= 5

Now the two regression equations are,

(x-x) = b xy(y-y) ------- x on y ( i)

(y-y) = b yx(x-x)------- y on x (ii)

Where,

x=1

nx =70

5= 14 and y=1

ny =88

5=17.6

Also,

Cov(x,y,) =1

nxy-xy V(x) =1

nx2-x2V(y) =1

ny2-y2

=1218

5–14x17.6 =1060

5-142=1632

5-17.62

= 243.6 -246.4 =212-196 = 326.4 –309.76

... Cov(x,y) =-2.8 V(x) = 16 V(y) = 16.64munotes.in

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Now we find,

Regression co -efficient of X on Y Regression co -efficient of X on Y

bxy=(x, )

( )Cov y

V ybyx=(x, )

(x)Cov y

V

=2.8

16.64=-2.8

16.64

...bxy=-0.168 byx= 0.175

Now substituting the values of x,y,bxyand b yxin the regression

equations we get,

(x-14) = -0.168(y -17.6) ------- x on y (i)

(y-17.6)= -0.175(x -14)------- y on x (ii)

as the two regression equations.

Now to estimate y when x =10, we use the regressi on equation of y on x

...(y-17.6) = -0.175(10 -14)

... y =17.6 +0.7 = 24.3

Example 3 :

The following data give the experience of machine operators and

their performance rating given by the number of good parts turned out per

100 pieces.

Operator: 1 2 3 4 5 6 7 8

Experience: 16 12 18 4 3 10 5 12

(in years)

Performance: 87 88 89 68 78 80 75 83

Rating

Obtain the two regression equations and estimate the permance rating of

an operator who has put 15 years in service.

Solution : We define the varia bles,

X: Experience y: Performance rating

Table of calculations:

X Y Xy x2Y2

16 87 1392 256 7569

12 88 1056 144 7744

18 89 1602 324 7921

4 68 272 16 4624

3 78 234 9 6084

10 80 800 100 6400

5 75 375 25 5625

12 83 996 144 6889

x=80y=648xy=6727x2= 1018 y2=52856munotes.in

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Now the two regression equations are,

(x-x) = b xy(y-y) ------- x on y (i)

(y-y) = b yx(x-x)------- y on x (ii)

Wher e,

x=1

nx =80

8= 10 and y=1

ny =648

8=81

Also,

Cov(x,y,) =1

nxy-xy V(x) =1

nx2-x2V(y) =1

ny2-y2

=6727

8–10x81 =1018

8-102=52856

8-812

= 840.75 -810 = 127.25 -100 = 6607 –6561

... Cov(x,y) = 30.75 V(x) = 27.25 V(y) = 46

Now we find,

Regression co -efficient of X on Y Regression co -efficient of X on Y

bxy=(x, )

( )Cov y

V ybyx=(x, )

(x)Cov y

V

=30.75

46=30.75

27.25

...bxy= 0.67 and byx= 1.13

Now substituting the values of x,y,bxyand b yxin the regression

equations we get,

(x-10) = 0.67(y -81)------- x on y (i)

(y-81)=1.13(x -10)------- y on x (ii)

as the two regression equations.

Now to estimate Performance rating (y) when Experience (x) = 15, we use

the regression equation of y on x

...(y-81) =1.13(15 -10)

...y = 81+ 5.65 = 86.65

Hence the est imated performance rating for the operator with 15 years of

experience is approximately 86.65 i.e approximately 87

4.5.2 Regression coefficients in terms of correlation coefficient.

We can also obtain the regression coefficients b xyand b yxfrom

standa rd deviations, x.,yand correlation coefficient ‘r’ using the

formulas

bxy=x

yr

and byx=y

xr

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Also consider,

bxyXbyx=x

yr

y

xr

= r2i.e. r = xxy yxb b

Hence the correlation coefficient ‘r’ is the geometric mean of the

regression coefficients, bxyandbyx

Example 5:

You are given the information about advertising expe nditure and sales:

---------------------------------------------------------------------------------------

Exp. on Advertisiment Sales (Rs. In Lakh)

(Rs. In Lakh)

----------------------------------------------------------------------------------- -----

Mean 10 90

S.D. 3 12

----------------------------------------------------------------------------------------

Coefficient of correlation between sales and expenditure on Advertisement

is 0.8 .Obtain the two regression equations.

Find the likely sales when advertisement budget is Rs. 15 Lakh.

Solution : We define the variables,

X: Expenditure on advertisement

Y: Sales achieved.

Therefore we have,

x= 10, y=90, 6x= 3, 6y = 12 and r = 0.8

Now, using the above results we can write the two regression equations as

(x-x) =x

yr

(y-y)------- x on y (i)

(y-y) =y

xr

(x-x)------- y on x (ii)

Substituting the values in the equations we get,

(x-10) =30.812(y-90)

i.e x -10 = 0.2 (y -90) ------- x on y (i)

also (y-90) =120.83(x-10)

i.e. y-90 = 3.2 (x -10) ------- y on x (ii)

Now when expenditure on advertisement (x) is 15, we can find the sales

from eqn (ii) as,

y-90 = 3.2 (15 -10)

...y = 90 + 16 = 106munotes.in

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Thus the likely sales are Rs.106 Lakh.

Example 6: Comput the two regression equations on the basis of the

following information:

X Y

Mean 40 45

Standard deviation 10 9

Karl Pearson’s coefficient of correlation between x and y = 0.50.

Also estimate the value of x when y = 48 using the appropriate equation.

Solution: We have,

x= 40, y=45,x= 10,y = 9 and r = 0.5

Now, we can write the two regression equations as

(x-x) =x

yr

(y-y)------- x on y (i)

(y-y) =y

xr

(x-x)------- y on x (ii)

Substituting the values in the equations we get,

(x-40) =100.59(y-45)

i.e x-40 = 0.55 (y -45)------- eqn of x on y (i)

and (y -45) =90.510(x-40)

i.e. y-45 = 0.45(x -40)------- eqn of y on x (ii)

Now when y is 48, we can find x from eqn (i) as,

x-40 = 0.55(48 -45)

...x = 40 +1.65 = 41.65

Example 7:

Find the marks of a student in the Subject of Mathematics who

have scored 65 marks in Accountancy Given,

Average marks in Mathematics 70

Accountancy 80

Standard Devi ation of marks in Mathematics 8

in Accountancy 10

Coefficient of correlation between the marks of Mathematics and marks of

Accountancy is 0.64.munotes.in

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Solution : We define the variables,

X: Marks in Mathematics

Y: Marks in Accountancy

Therefore w e have,

x= 70, y=80,σx= 8,σy=10 and r= 0.64

Now we want to approximate the marks in Mathematics (x), we obtain the

regression equation of x on y, which is given by

(x-x)=x

yr

(y-y)------- x on y (i)

Substituting the values we get,

(x-70) =80.6410(y-80)

i.e x-70 = 0.57 (y -80)

Therefore, when marks in Accountancy (Y) = 65

x-70 = 0.57(65 -80)

...x = 70 -2.85 = 67.15 i.e. 67 appro.

Use of regression equations to find means x,yS.D.sx ,yand

correlation coefficient ‘r’

As we have that, we can obtain t he regression equations from the

values of Means, standard deviations and correlation coefficients ‘r’, we

can get back these values from the regression equations.

Now, we can note that the regression equation is a linear equation

in two variables x and y. Therefore, the linear equation of the type

Ax+By+C = 0 or y = a+bx represents a regression equation.

e.g. 3x+5y -15 = 0 and 2x+7y+10 = 0 represent the two regression

equations.

The values of means x,ycan be obtain by solving the two equations as

the simultaneous equations.

Example 8 :

From the following regression equation, find means x,y,x,yand ‘r’

3x-2y-10 = 0, 24x -25y+145 = 0

Solution : The two regression equations are,

3x-2y-10 = 0 -------- (i)

24x-25y+145 = 0 ---(ii)

Now for xand ywe solve the two equations as the simultaneous

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Therefore, by (i) x 8 and (ii) x1, we get

24x-16y-80 = 0

24x-25y+145 = 0

-+-

9y-225 = 0 y =225

9= 25

Putting y = 25 in eqn (i), we get

3x-2(25) -10 = 0

3x–60 = 0 x =60

3= 20

Hence x= 20 and y= 25.

Now to find ‘r’ we express the equations in the form y=a+bx

So, from eqns (i) and (ii)

y =3

2x–10

2and y =24

25x+145

25

... b1=3

2= 1.5 ... b 2=24

25= 0.96

Since, b1> b 2(i.e. b 2is smaller in number irrespective of sign + or -)

... Equation (ii) is regression of y on x and b yx= 0.96

Hence eqn (i) is regression of x on y and bxy= 1/1.5 = 0.67

________ _________

Now we find, r = √bxy Xbyx i.e. r =√0.67x0.96 = + 0.84

(The sign of ‘r’ is same as the sign of regression coefficients)

Example 9:

Find the means values of x,y, and r from the two regression equations.

3x+2y -26=0 and x+y-31=0. Also find xwheny= 3.

Solution : The two regression equations are,

3x+2y -26=0 -------- (i)

6x+y -31=0 ---------- (ii)

Now for x and y we solve the two equations as the simultaneous

equations.

Therefore, by (i) x 2 and (ii) x1, we get

6x+4y -52 = 0

6x+ y -31 = 0

--+ ...y =21

3= 7

3y-21 = 0munotes.in

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Putting y = 7 in eqn (i), we get

3x+2(7) -26 = 0

3x–12 = 0 x =12

3= 4.

Hence x = 4 and y = 7.

Now to find ‘r’ we express the equations in the form y= a+bx

So, from eqns (i) and (ii)

y =-3

2x–26

2and y =-6

1x +31

1

... b1=-3

2=-1.5 ... b 2=6

1=-6

since, b1< b 2(i.e. b 1is smaller in number irrespective of sign + or -)

... Equation (i) is regression of y on x and byx=-1.5

Hence, eqn (ii) is regression of x on y and bxy=-1/6 = -0.16

Now we find, . r =bxyxbyx r = 0.16x1.5 = = -0.16

Note: The sign of ‘r’ is same as the sign of regression coefficients

Now to find 6x when 6y = 3, we use the formula,

byx=x

yr

-1.5=-0.16x3

6x

... 6x =0.48

1.5= 0.32

Hence means x= 4,y= 7, r = -0.16 and 6x = 0.32.

EXERCISES

1.What is mean by Regression? Expl ain the use of regression in the

statistical analysis.

2.Why are there two Regressions? Justify.

3.State the difference between Correlation and Regression.

4.Obtain the two regression equations from the data given bellow.

X: 7 4 6 5 8

Y: 6 5 9 8 2

Hence estimate y when x = 10.

5.The data given below are the years of experience (x) and monthly

wages (y) for a group of workers. Obtain the two regression equations

and approximate the monthly wages of a workers who have completed

15 year s of service.munotes.in

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Experience: 11 7 9 5 8 6 10

In years

Monthly wages: 10 8 6 8 9 7 11

(in ‘000Rs.)

6.Following results are obtained for a bivariate data. Obtain the two

regression equa tions and find y when x = 12

n = 15x= 130y = 220 x2= 2288y2= 5506xy = 3467

7.Marks scored by a group of 10 students in the subjects of Maths and

Stats in a class test are given below.Obtain a suitable regression

equation to find th e marks of a student in the subject of Stats who have

scored 25 marks in Maths.

Student no: 1 2 3 4 5 6 7 8 9 10

Marks 13 18 9 6 14 10 20 28 21 16

in Maths

Marks 12 25 11 7 16 12 24 25 22 20

in Stats:

8.The data given below are the price and demand for a certain

commodity over a period of 7 years. Find the regression equation of Price

on Demand and hence obtain the most likely demand for the in the year

2008 when it’s price is Rs.23.

Year: 2001 2002 2003 2004 2005 2006 2007

Price(in RS): 15 12 18 22 19 21 25

Demand 89 86 90 105 100 110 115

(100 units)

9.For a bivariate data the following results were obtained

x= 53.2 , y= 27.9 , 6x = 4.8, y =.4 and r =0.75

Obtain the two regression equations, fi nd the most probable value of x

when y =25.

10. A sample of 50 students in a school gave the following statistics

about Marks of students in Subjects of Mathematics and Science,

------------------------------------------------------------------------------ ---------

Subjects: Mathematics Science

---------------------------------------------------------------------------------------

Mean 58 79

S.D. 12 18

--------------------------------------------- -------------------------------------------munotes.in

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Coefficient of correlation between the marks in Mathematics and marks in

Science is 0.8. Obtain the two regression equations and approximate the

marks of a student in the subject of Mathematics whose score in Sci ence is

65.

11. It is known that the Advertisement promotes the Sales of the

company. The company’s previous records give the following results.

---------------------------------------------------------------------------------------

Expenditure on Advertise ment Sales

(Rs. In Lakh) (Rs. In Lakh)

----------------------------------------------------------------------------------------

Mean 15 190

S.D. 6 20

-------------------------------------------------------------------------- --------------

Coefficient of correlation between sales and expenditure on Advertisement

is 0.6. Using the regression equation find the likely sales when

advertisement budget is Rs.25 Lakh.

12. Find the values of x,y, and r from the two regression equ ations given

bellow. 3x+2y -26=0 and 6x+y -31=0. Also find 6x when y = 3.

13. Two random variables have the regression equations:

5x+7y -22=0 and 6x+2y -20=0. Find the mean values of x and y. Also

find S.D. of x when S.D. of y = 5.

14.The two regression equat ions for a certain data were y = x+5 and 16x

= 9y-94. Find values of x,yand r. Also find the S.D. of y when S.D. of x

is 2.4.

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UNIT I V

Unit -5

TIME SERIES

Unit Structure :

5.0 Objectives

5.1 Introduction

5.2 Importance of Time Series Analysis

5.3 Components of Time Series

5.4 Methods tofind Trend

5.0OBJECTIVE S

From this chapter student should learn analysis of data using

various methods. Methods involve moving average method and least

square method seasonal fluctuations can be studied by business for casting

method.

5.1INTRODUCTION

Every business venture needs to know their performance in the

past and with the help of som e predictions based on that, would like to

decide their strategy for the present By studying the past behavior of the

characteristics, the nature of variation in the value can be determined. The

values in the past can be compared with the present values o f comparisons

at different places during formulation of future plan and policies. This is

applicable to economic policy makers, mete orological department, social

scienti sts, political analysis. Forecasting thus is an important tool in

Statistical analysis .The statistical data, particularly in the field of social

science, are dynamic in nature. Agricultural and Industrial production

increase every year or due to improved medical facilities, there is decline

in the death rate over a period of time. There is increase in sales and

exports of various products over a period of years. Thus, a distinct change

(either increasing or decreasing) can be observed in the value of time -

series.

A time series is a sequence of value of a phenomenon arranged in

order of their occurrence. Mathematically it can expressed as a function,

namely y = f(t) where t represents time and y represents the corresponding

values. That is, the value y 1, y2, y3…… of a phenomenon with respect to

time periods t 1, t2, t3…. F orm a Tim e Series.munotes.in

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Forecasting tech niques facilities prediction on the basic of a data

available from the past. This data from the past is called a time series. A

set of observations, of a variable, taken at a regular (fixed and equal)

interval of time is calle d time series. A time series is a bivariate data,

with time as the independent variable and the other is the variable under

consideration. There are various forecasting method for time series which

enable us to study the variation or trends and estimat e the same for the

future.

5.2IMPORTANCE OF TIME SERIES ANALYSIS

The analysis of the data in the time series using various forecasting

model is c alled as time analysis. The importance of time series analysis is

due to the following reasons:

Understand ing the past behavior

Planning the future action

Comparative study

5.3COMPONENTS OF TIME SERIES

The fluctuation in a time series are due to one or more of the

following factors which are called “components” of time series.

(a) Secular Trend :

Thegeneral tendency of the data, either to increase, to decrease or

to remain constant is called Secular Trend. It is smooth, long term

movement of the data. The changes in the values are gradual and

continuous. An increasing demand for luxury items like r efrigerators or

colour T.V. sets reflect increasing trend. The production of steel, cement,

vehicles shows a rising trend. On the other hand, decreasing in import of

food grains is an example of decreasing trend. The nature of the trend

may be linear or curvilinear, in practice, curvilinear trend is more

common.

Trend in due to long term tendency. Hence it can be evaluation if

the time series is a available over a long duration.

(b)Seasonal Variation :

The regular, seasonal change in the time series are called

“Seasonal Variation”. It is observed that the demand for umbrellas,

raincoats reaches a peak during monsoon or the advertisement of cold

drinks, ice creams get a boom in summer. The demand for greeting cards,

sweets, increase during festival l ike Diwali, Christmas. In March, there is

maximum withdrawal of bank deposits for adjustment of income -tax

payment, so also variation tax -saving schemes shoot up during this period.

The causes, for these seasonal fluctuations, are thus change in

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component is measured to isolate these change from the trend component

and to study their effect, so that, in any business, future production can be

planned accordingly and necessary adjus tments for seasonal change can be

made

(c)Cyclical Variation :

These are changes in time series, occurring over a period which is

more than a year. They are recurring and periodic in nature. The period

may not be uniform. These fluctuations are due to ch anges in a business

cycle. There are four important phases of any business activity viz.

prosperity, recession, depression and recovery. During prosperity, the

business flourishes and the profit reaches a maximum level. Thereafter, in

recession, the pro fit decreases, reaching a minimum level during

depression. After some time period, the business again recovers

(recovery) and it is followed by period of prosperity. The variation in the

time series due to these phases in a business cycle are called “Cyc lical

Variation:.

Depression RecoveryT im e S c a le

One cycle

The knowledge of cyclic variations is important for a businessman

to plan his activity or design his policy for the phase of recession or

depression. But one should know that the factors affecting the cyclical

variations are quite irregular, difficult to identify and measure. The

cyclical variation are denoted by

(d)Irregular Variation :

The changes in the time series which can not be predicated and are

erratic in nature are called “Irregular Variation ”. Usually, these are short

term changes having signification effect on the time series during that time

interval. These are caused by unforeseen event like wars, floods, strikes,

political charges, etc. During Iran-Iraq war or recent Russian revolution ,

prices of petrol and petroleum product soared very high. In recent budget,

control on capital issued was suddenly removed. As an effect, the all

Indian -Index of share market shooted very high , creates all time records.munotes.in

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If the effect of other components of the time series is eliminated, the

remaining variation are called “Irregular or Random Variations”. No

forecast of these change can be made as they do not reflect any fixed

pattern.

MODELS FOR ANALYSIS OF TIME SERIES

The purpose of studying time s eries is to estimate or forecast the

value of the variable. As there are four components of the time series,

these are to be studied separately. There are two types of models which

are used to express the relationship of the components of the time series .

They are additive model and multiplicative model.

O = Original Time Series

T = Secular Trend

S = Seasonal Variations

C = Cyclical Variations and

I = Irregular Variations

In Additive model, it is assumed that the effect of the individual

compo nents can be added to get resultant value of the time series, that is

the components are independent of one another. The model can be

expressed as

O = T + S + C + I

In multiplicative model, it is assumed that the multiplication of the

individual effect of the components result in the time series, that is, the

components are due to different causes but they are not necessarily

independent, so that changes in any one of them can affect the other

components. This model is more commonly used. It is expres sed as

O = T × S × C × I

If we want to estimate the value in time series, we have to first

estimate the four components and them combine them to estimate the

value of the time series. The irregular variations can be found. However,

we will restrict ou rselves, to discuss method of estimating the first

components, namely Secular Trend.

5.4METHODS TO FIND TREND

There are various method to find the trend. The major methods are

as mentioned below:

I.Free Hand Curve.

II.Method of Semi –Averages.

III.Method of Moving Averages.

IV.Method of Least Squares.

we will study only the method of moving average and least squares.munotes.in

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5.4.1.Method of Moving Averages

This is a simple method in which we take the arithmetic average of

the given times series over a certain period o f time. These average move

over period and are hence called as moving averages. The time interval

for the average is taken as 3 years, 4 years or 5 years and so on. The

average are thus called as 3 yearly, 4 yearly and 5 yearly moving average.

The movi ng average sare useful in smoothing the fluctuations caused to

the variable. Obviously larger the time interval of the average more is the

smoothing. We shall study the odd yearly (3 and 5) moving average first

and then the 4 yearly moving average.

Odd Y early Moving Average

In this method the total of the value in the time series is taken for

the given time interval and is written in front of the middle value. The

average so taken is also written in front of this middle value. This average

is the trend for that middle year. The process is continued by replacing the

first value with the next value in the time series and so on till the trend for

the last middle value is calculated. Let us understand this with example:

Example 1:

Find 3 years moving aver ages and draw these on a graph paper.

Also represent the original time series on the graph.

Year 199

9200

0200

1200

22003 200

4200

5200

62007

Production

(in thousand

unit)12 15 20 18 25 32 30 40 44

Solution:

We calculate arithmetic mean of first three observations viz. 12, 15

and 20, then we delete 12 and consider the next one so that now, average

of 15, 20 and 18 is calculated and so on. These averages are placed

against the middle year of each group, viz. the year 2000, 2001 and so on.

Note moving averages are not obtained for the year 1999 and2007.

Year Production

(in thousand

unit)3 Years Total 3yrly. Moving

Average

1999 12

2000 15 12 + 15 + 20 = 47 47 / 3 = 15.6

2001 20 15 + 20 + 18 = 53 53 / 3 = 17.6

2002 18 20 + 18 + 25 = 63 63 / 3 = 21.0

2003 25 18 + 25 + 32 = 75 75 / 3 = 25.0

2004 32 25 + 32 + 30 = 87 87 / 3 = 29.0

2005 30 32 + 30 + 40 = 102 102 / 3 = 34.0

2006 40 30 + 40 + 44 = 114 114 / 3 = 38.0

2007 44munotes.in

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Example 2:

Find 5 yearly moving average for the following data.

Year 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006

Sales (in

lakhs of

Rs.)51 53 56 57 60 55 59 62 68 70

Solution:

We find the average of first five values, namely 51, 53, 56 , 57 and 60.

Then we omit the first value 51 and consider the average of next five

values, that is, 53, 56, 57, 60 and 55. This process is continued till we get

the average of the last five values 55, 59, 62, 68 and 70. The following

table is prepared.

Year Sales

(in lakhs of Rs.)5 Years Total Moving

Average

(Total / 5)

1997 51 …. ….

1998 53 …. ….

1999 56 51 + 53 + 56 + 57 +60 = 277 55.4

2000 57 53 + 56 + 57 + 60 + 55 = 281 56.2

2001 60 56 + 57 + 60 + 55 + 59 = 287 57.4

2002 55 57 + 60 + 55 + 5 9 + 62 = 293 58.6

2003 59 60 + 55 + 59 + 62 + 68 = 304 60.8

2004 62 55 + 59 + 62 + 68 + 70 = 314 62.8

2005 68 …. ….

2006 70 …. ….

Example 3:

Determine the trend of the following time series using 5 yearly moving

average s.

Solution :The time series is divided into overlapping groups of five years,

their 5 yearly total and average are calculated as shown in the following

table.

Year Export (Y) 5–yearly total (T) 5–yearly moving

average: (T/5)

1981

1982

1983

1984

1985

198678

84

80

83

86

8978+84+80+83+86 = 411

84+80+83+86+89 = 422

80+83+86+89+88 = 426411 / 5 = 82.2

422 / 5 = 84.4

85.2Year 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991

Exports

in‘000Rs 78 84 80 83 86 89 88 90 94 93 96munotes.in

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1987

1988

1989

1990

199188

90

94

93

9683+86+89+88+90 = 436

86+89+88+90+94 = 447

89+88+90+94+93 = 454

88+90+94+93+96 = 46187.2

89.4

90.8

92.2

Observation s:

I.In case of the 5 –yearly moving avera ge, the total and average for

the first two and the last two in the time series is not calculated.

Thus, the moving average of the first two and the last two years in

the series cannot be computed.

II.To find the 3 –yearly total (or 5 –yearly total) for a particular

years, you can subtract the first value from the previous year’s

total, and add the next value so as to save your time!

Even yearly moving averages

In case of even yearly moving average the method is slightly

different as here we cannot find the middle year of the four yea rs in

consideration. Here we find the total for the first four years and place it

between the second and the third year value of the variable. These totals

are again sunned into group of two, called as centered total and is placed

between the two totals. The 4 –yearly moving average is found by

dividing these centered totals by 8. Let us understa nd this method with

an example

Example 4:Calculate the 4 yearly moving averages for the following

data.

Ans: The table of calculation is show below. Student should leave one

line blank after every to place the centered total in between two years.

Years(Import)

Y4-Yearly

Total4-Yearly

Centered Total4-Yearly

Moving Averages :

1991 15 - -

1992 18 - -

1993 20 77 77 + 83 = 160 160/8 = 20

1994 24 83 83 + 90 = 173 173/8 = 21.6

1995 21 90 90 + 98 = 188 188/8 = 23.5

1996 25 98 98 + 100 = 198 198/8 = 24.8

1997 28 100 100 + 109 = 209 209/8 = 26.1

1998 26 109 - -

1999 30 - - -Year 1991 1992 1993 1994 1995 1996 1997 1998 1999

Import

in‘000Rs 15 18 20 24 21 25 28 26 30

munotes.in

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Example 5:

Find the moving average of length 4 for the following data. Represent the

given data and the moving average on a graph paper.

Year 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007

Sales (in

thousand

unit)60 69 81 86 78 93 102 107 100 109

Solution: We prepare the following table.

Year Sale (in

thousand

unit)4 Yearly Totals Centred Total Moving /

Avg. Central

= Total / 8

1998 60

1999 69

60+ 69 + 81 + 86 = 296

2000 81 296 + 314 = 610 76.25

69 + 81 + 86 + 78 = 314

2001 86 314 + 338 = 652 81.5

81 + 86 + 78 + 93 = 338

2002 78 338 + 359 = 697 87.125

86 + 78 + 93 + 102 = 359

2003 93 359 + 380 = 739 92.375

78 + 93 + 102 + 107 = 380

2004 102 380 + 402 = 782 97.75

93 + 102 + 107 + 100 = 402

2005 107 402 + 418 = 820 102.5

102 + 107 + 100 + 109 = 418

2006 100

2007 109

Note that 4 yearly total are written between the years 1999 -2000, 2000 -

01, 2001 -02 etc. and the central total are written against the years 2000,

2001, 2002 etc. so also the moving average are considered w.r.t. years;

2000, 2001 and so on. The moving averages are obtained by dividing the

certain total by 8.

The graph of the given set of values and the moving averages

against time representing the trend component are shown below. Note that

the moving averages are not obtained for the years 1998, 1999, 2006 and

2007. (i.e. first and last two extreme years).

When the values in th e time series are plotted, a rough idea about the

type of trend whether linear or curvilinear can be obtained. Then,

accordingly a linear or second degree equation can be fitted to the values.

In this chapter, we will discuss linear trend only.munotes.in

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5.4.2.LEAST SQUARES METHOD:

Let y = a + bx be the equation of the straight line trend where a, b are

constant to be determined by solving the following normal equations,

∑ y = na + b∑ x

∑ xy = a∑ x + b∑ x2

where y represents the given time series.

We de fine x from years such that ∑ x = 0. So substituting ∑ x = 0 in the

normal equation and simplifying, we get

b =

x2xyand a =ny

Using the given set of values of the time series, a, b can be calculated and

the s traight line trend can be determined as y = a + bx. This gives the

minimum sum of squares line deviations between the original data and the

estimated trend values. The method provides estimates of trend values for

all the years. The method has mathemati cal basis and so element of

personal bias is not introduced in the calculation. As it is based on all the

values, if any values are added, all the calculations are to be done again.

Odd number of years in the time series

When the number of years in t he given time series is add, for the

middle year we assume the value of x = 0. For the years above the middle

year the value given to x are …, -2,-1 while those after the middle year

are values 1,2, … and so on.

Even number of years in the time series

When the number of years in the time series is even, then for the

upper half the value of x are assumed as…., -5,-3,-1. For the lower half

years, the values of x are assumed as 1, 3, 5, …. And so on.

Example 6:

Fit a straight line trend for the follo wing data giving the annual profits (in

lakhs of Rs.) of a company. Estimate the profit for the year 1999.

Years 1992 1993 1994 1995 1996 1997 1998

Profit 30 34 38 36 39 40 44

Solution : Let y = a + bx be the straight line trend.

The number of years i s seven, which is sold. Thus, the value of x is taken

as 0 for the middle years 1995, for upper three years as -3,-2,-1 and for

lower three years as 1, 2, 3.munotes.in

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The table of computation is as shown below:

Years Profit (y) x xy x2Trend Value:

Yt= a+ bx

1992

1993

1994

1995

1996

1997

199830

34

38

36

39

40

44-3

-2

-1

0

1

2

3-90

-68

-38

0

39

80

1329

4

1

0

1

4

931.41

33.37

35.33

37.29

39.25

41.21

43.17

Total∑y = 261∑x = 0∑xy = 55∑x2= 28

From the table : n = 7, ∑xy = 55, ∑x2= 28,∑y = 261

There fore b =

x2xy=2855= 1.96 and a =ny=7261= 37.29

Thus, the straight line trend is y = 37.29 + 1.96x.

The trend values in the table for the resp ective years are calculated by

substituting the corresponding value of x in the above trend line equation.

Forthe trend value for 1992: x = -3:

y1992= 37.29 + 1.96 (-3)=37.29 -5.88 = 31.41

Similarly, all the remaining trend values are calculate d.

(A short -cut method in case of odd number of years to find the remaining

trend values once we calculate the first one, is to add the value of b to the

first trend value to get the second trend value, then to the second trend

value to get the third one and so on. This is because the difference in the

values of x is 1.)

To estimate the profit for the years 1999 in the trend line equation, we

substitute the prospective value of x, if the table was extended to 1999. i.e.

we put x = 4, the next value afte r x = 3 for the year 1998.

y1999= 37.29 + 1.96 (4) = 45.13

There fore the estimated profit for the year 1999 is Rs. 45.13 lakhs.

Example 7:

Fit straight line trend by the method of lease squares for the following data

represent ing production in thousand units. Plot the data and the trend line

on a graph paper. Hence or otherwise estimate the trend for the years

2007 .munotes.in

## Page 95

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Year 1999 2000 2001 2002 2003 2004 2005

Production (in

thousand unit) 14 15 17 16 17 20 23

Solution :

Here, the total number of years is 7, an odd number. So we take the center

as 1986 the middle -most year and define x as year 2002. The values of x

will be -3,-2,-1, 0, 1, 2, 3.

Prepare the following table to calculate the required summations. Note

that the trend values can be written in the table only after calculation of a

and b.

Year Production

(y)x x2x y Trend

Values

1999 14 -3 9 -42 13.47

2000 15 -2 4 -30 14.79

2001 17 -1 1 -17 16.11

2002 16 0 0 0 17.43

2003 17 1 1 17 18.75

2004 20 2 4 40 20.07

2005 23 3 3 69 21.39

122 28 37

Here, n = 7, ∑ y = 122, ∑ x2= 28,∑ x y = 37

Now, a and b are calculated as follows:

a = 43.17 4286.177122

ny

b = 32.1 3214.12837

2

xxy

So, the equation is used to find trend values.

y = a + b x

i.e. y = 17.43 + 1.32x

The equation is used to find trend va lues.

For the year 1999, x = -3, substituting the value od x, we get,

y = 17.43 + 1.32 ( -3) = 17.43 –3.96 = 13.47

to find the remaining trend values we can make use of the property of a

straight line that as all the values of x are equidistant with diff erent of one

unit ( -3,-2,-1,----and so on), the estimated trend value will also be

equidistant with a difference of b unit.

In this case as b = 1.32, the remaining trend values for x = -2,-1, 0, ---

etc. are obtained by adding b = 1.32 to the previo us values. So, the trend

values are 13.47, 14.79, 16.11, 17.43, 18.75, 20.07 and 21.39.munotes.in

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Now to estimate trend for the year 2007, x = 5, substituting in the equation

y = 17.43 + 1.32x

= 17.43 + 1.32 (5) = 24.03

So, the estimated trend value for the year 2007 is 24,030 unit.

For graph of time series, all point sare plotted. But for the graph of trend

line, any two trend values can be plotted and the line joining these points

represents the straight line trend.

0510152025

1999 2000 2001 2002 2003 2004 2005 2006 2007

Yea rsTimeSeries

Time

S eries

Trend L ine

For the trend line, the trend values 17.43 and 21.39 for the years

2002 and 2005 are plotted and then a straight line joining these two points

is drawn and is extended on both the sides.

The estimate of trend for the year 2007 can also be obtained from

the graph by drawing a perpendicular for the year 2007, from x -axis which

meet the trend line at point P. From P, a perpendicular on y -axis gives the

required ternd estimate as 24.

Now, to find straight line trend, when number of years is even, consider

the follo wing example.

Example 8:

Fit a straight line trend to the following time –series, representing

sales in lakhs of Rs. of a company, for the year 1998 to 2005. Plot the

given data well as the trend line on a graph paper. Hence or otherwise

estimate tre nd for the year 2006.

Year 1998 1999 2000 2001 2002 2003 2004 2005

Sales

(Lakhs of Rs.) 31 33 30 34 38 40 45 49munotes.in

## Page 97

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Solution:

Here the number of years = 8, an even number, so we define

x =5.05. 2001 year,so that the values of x are -7,-3,-1, 1, 3, 5 and 7, to

get∑ x = 0.

Prepare the following table to obtain the summations ∑ x2,∑ y, ∑ x y.

Year Sales

(in Lakhs of

Rs.)x x2x y Trend

Values

1998 31 -7 49 -217 28.33

1999 33 -5 25 -165 30.95

2000 30 -3 9 -90 33.57

2001 34 -1 1 -34 36.19

2002 38 1 1 38 38.81

2003 40 3 9 120 41.43

2004 45 5 25 225 44.05

2005 49 7 49 343 46.67

300 168 220

Here, n = 8, ∑ y = 300, ∑ x2= 168,∑ x y = 220

Now, a and b are calculated as follows:

a = 5.378300

ny

b = 31.1168220

2

xxy

So, the equation of the straight line trend is y = a + b x

i.e. y = 37.5 + 1.31 x

To obtain the trend values, first calculate y for x = -7, for the year 1998

y = 37.5 + 1.31 ( -7)

= 37.5 –9.17 = 28.33

To find the successive trend values, go on addition 2b = 2 × 1.31 = 2.62,

to the preceding values as in this case the different between x values is of

2 units.

So, the estimated values of trend for x = -5,-3,-1, 1, 3, 5, 7 and 7 are

30.95, 33. 57, 36.19, 38.81, 41.43, 44.05 and 46.67 respectively.

Write down these values in the table.

Hence the estimated trend value for the year 2006 is 49.29 (in lakhs of

Rs.).munotes.in

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Now, for the graph of trend line, note that only two trend values 30.95 and

46.67 w.r.t. years 1999 and 2005 are considered as point. The line joining

these two points represents trend line.

0102030405060

1998 1999 2000 2001 2002 2003 2004 2005 2006

Yea rsTimeSeries

Time S eries

To estimated the trend for the year 2006, drawn a perpendicular

from x -axis at this point meeting the line in P. then from P, draw another

perpendicular on y -axis which gives estimate of trend as 49.

Example 9:

Fit a straight line trend to the following data. Draw the graph of

the actual time series and the trend line. Estimate the sales for the year

2007.

Year 1998 1999 2000 2001 2002 2003 2004 2005

Sales

in‘000Rs 120 124 126 130 128 132 138 137

Solution : lety = a + bx be the straight line trend.

The number of years in the given time series is eight, which is an even

number. The upper four years are assigned the values of x as 1, 2, 3, and

7. Note that ere the difference between the values of x is 2, but the sum is

zero.

Now, the table of computation is completed as shown below:

Years Profit (y) X Xy X2Trend Value:

Yt= a + bx

1998

1999

2000

2001

2002

2003120

124

126

130

128

132-7

-5

-3

-1

1

3-840

-620

-378

-130

128

39649

25

9

1

1

9120.84

123.28

125.72

128.16

130.06

133.04munotes.in

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2004

2005138

1375

7390

35925

49135.48

137.92

Total∑y = 1035∑x = 0∑xy = 205∑x2= 168

From the table : n = 8, ∑xy = 205, ∑x2= 168,∑y = 1035

b = 22.1168205

2

xxyand a = 38.12981035

ny

Thus, the straight line trend is y = 129.38 + 1.22x.

The trend values in the table for the respective years are calculated by

substituting the corresponding value of x in the above trend line equation.

For the trend value for 1998: x = -7:

y1998= 129.38 + 1.22 ( -7)= 129.38 -8.54 = 120.84

Similarly, all t he remaining trend values are calculated.

(A short -cut method in case of even number of years to find the remaining

trend values once we calculate the first one, is to add twice the value of b

to the first trend value to get the second trend value, then t o the second

trend value to get the third one and so on. This is because the difference

in the values of x is 2. In this example we add 2 x 1.22 = 2.44)

Estimation:

To estimate the profit for the years 2007 in the trend line equation, we

substitute the prospective value of x, if the table was extended to 2007. i.e.

we put x =11, the next value after x = 9 for the year 2006 and x = 7 for

2005.

y2007= 129.38 + 1.22 (11) = 142.8

There fore the estimated profit for the year 2007 is Rs. 1,42,800.

Now w e draw the graph of actual time series by plotting the sales

against the corresponding year, the period is taken on the X -axis and the

sales on the Y -axis. The points are joined by straight lines. To draw the

trend line it is enough to plot any two poin t (usually we take the first and

the last trend value) and join it by straight line.

To estimate the trend value for the year 2007, we draw a line

parallel to Y -axis from the period 2007 till it meet the trend line at a point

say A. From this point we d raw a line parallel to the X -axis till it meet the

Y-axis at point say B. This point is our estimate value of sales for the year

2007. The graph and its estimate value (graphically) is shown below:munotes.in

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From the graph, the estima ted value of the sales for the year 2007

is 142 i.e. Rs 1,42,000 (approximately)

Example 10:

Fit a straight line trend to the following data. Draw the graph of

the actual time series and the trend line. Estimate the import for the year

1998.

Solution : Here again the period of years is 6 i.e. even. Proceeding

similarly as in the above problem, the table of calculation and the

estimation is as follows:

Years Import

(y)x xy x2Trend Value:

Yt= a + bx

1991

1992

1993

1994

1995

199640

44

48

50

46

52-5

-3

-1

1

3

5-200

-132

-48

50

138

26025

9

1

1

9

2541.82

43.76

45.7

47.64

49.58

51.52

Total∑y = 280∑x = 0∑xy = 68∑x2= 70

From the table : n = 6, ∑xy = 68, ∑x2= 70,∑y = 280

Their four b = 97.07068

2

xxyand 67.468280

ny

Thus, the straight line trend is y = 46.67 + 0.97x.Year 1991 1992 1993 1994 1995 1996

Import

in‘000Rs 40 44 48 50 46 52

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All the remaining trend value s are calculated as described in the above

problem.

Estimation:

To estimate the import for the year 1998, we put x = 9 in the tried line

equation. There fore y1997= 46.67 + 0.97 (9) = 55.4

There fore the imports for the year 1997is Rs. 55,400.

The gr aph of the actual time series and the trend values along with the

graphical estimation is an shown below:

From graph the estimated import are Rs. 55,000.

Example 11:

Fit a straight line trend to the following time –series, r epresenting

sales in lakhs of Rs. of a company, for the year 1998 to 2005. Plot the

given data well as the trend line on a graph paper. Hence or otherwise

estimate trend for the year 2006.

Year 1998 1999 2000 2001 2002 2003 2004 2005

Sales

(Lakhs of Rs.) 31 33 30 34 38 40 45 49

Solution:

Here the number of years = 8, an even number, so we define

x =5.05. 2001 year, so that the val ues of x are -7,-3,-1, 1, 3, 5 and 7, to

get∑ x = 0.

Prepare the following table to obtain the summations ∑ x2,∑ y, ∑ x y.munotes.in

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Year Sales

(in Lakhs of

Rs.)x x2x y Trend

Values

1998 31 -7 49 -217 28.33

1999 33 -5 25 -165 30.95

2000 30 -3 9 -90 33.57

2001 34 -1 1 -34 36.19

2002 38 1 1 38 38.81

2003 40 3 9 120 41.43

2004 45 5 25 225 44.05

2005 49 7 49 343 46.67

300 168 220

Here, n = 8, ∑ y = 300, ∑ x2= 168,∑ x y = 220

Now, a and b are calculated as follows:

a = 5.378300

ny

b = 31.1168220

2

xxy

So, the equation of the straight line trend is y = a + b x

i.e. y = 37.5 + 1.31 x

To obtain the trend values, first calculate y for x = -7, for the year 1998

y = 37.5 + 1.31 ( -7)

= 37.5 –9.17 = 28.33

To fi nd the successive trend values, go on addition 2b = 2 × 1.31 = 2.62,

to the preceding values as in this case the different between x values is of

2 units.

So, the estimated values of trend for x = -5,-3,-1, 1, 3, 5, 7 and 7 are

30.95, 33. 57, 36.1 9, 38.81, 41.43, 44.05 and 46.67 respectively.

Write down these values in the table.

Hence the estimated trend value for the year 2006 is 49.29 (in lakhs of

Rs.).

Now, for the graph of trend line, note that only two trend values 30.95 and

46.67 w.r.t. y ears 1999 and 2005 are considered as point. The line joining

these two points represents trend line.munotes.in

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0102030405060

1998 1999 2000 2001 2002 2003 2004 2005 2006

Yea rsTimeSeries

Time S eries

To estimate the trend for the year 2006, drawn a perpendicular

from x -axis at this point meeting the line in P. then from P, draw another

perpendicular on y -axis which gives estimate of trend as 49.

MEASURESMENT OF OTHER COMPONENTS

We have studied four method of estimation of Secular Trend. The

following procedure is applied to separate the remaining components of

the time s eries.

Using seasonal indices (s), the seasonal variations in a time series

can be measured. By removing the trend and the seasonal factors, a

combination of cyclical and irregular fluctuations is obtained.

If we assume, multiplicative model, represen ted by the equation

O = T × S × C × I

Then, to depersonalize the data, the original time series (O) divided

by the seasonal indices (S), which can be express as,

ICTSICST

SO

If it is further divided by trend values (T), then we have

ICTICT

Thus a combination of cyclical and irregular variation can be

obtained. Irregular fluctuations, because of their nature, can not be

eliminated completely, but these can be minimized by taking short term

averages and then the estimate of cyclical variation can be obtained.

METHODS TO ESTIMATE SEASONAL FLUCTUATIONS

We have seen methods to separate the trend component of Time

Series. Now, let us see, how to separate the seasonal component of it.

Methods of Seasonal Indexmunotes.in

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It is used to finds the effect of seasonal variations in a Time Series. The

steps are as follows:

i.Find the totals for each season, as well as the grand total, say G.

ii.Find the arithmetic means of these total, and the grand total by

dividin g the values added.

iii.Find seasonal indices, representing the seasonal component for

each season, using the formula

Seasonal Index =ge GrandAveraSeasonal Averagefor 100

Where, Grand Average =ofValues TotalNoG

.

Example 12:

Find the seasonal component of the time s eries, using method of seasonal

indices.

Seasonal /

YearsI II IV Grand

2003 33 37 32 31

2004 35 40 36 35

2005 34 38 34 32

2006 36 41 35 36

2007 34 39 35 32

Solution:

I II III IV Grand

Total

Average

Seasonal

Index172

34.4 (172 /

5)

25.351004.34

= 97.59195

39

25.3510039

= 110.64172

34.4

25.351004.34

= 97.59166

33.2

25.351002.33

= 94.18705 (G)

35.25(G/20)

The time series can be deseasonalised by removing the effect of seasonal

component from it. It is done using the formula.

Deseasonalised Value =dex SeasonalInlue OriginalVa 100

BUSINESS FORECASTING:

In this chapter, few methods of analyzing the past data and

predicting the future values are already discussed. Analysis of time series

an important ro le in Business Forecasting. One of the aspects of it

estimating future trend values. Now -a-days, any business or industry is

governed by factors like supply of raw material, distribution network,

availability of land, labour and capital and facilitates l ike regular supply of

power, coal, water, etc. a business has to sustain intricate government

regulations, status, everchanging tastes and fashions, the latest technology,

cut throat competition by other manufacturers and many other.munotes.in

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While making a forec ast, combined effect of above factors should

be considered. Scientific method are used to analyse the past business

condition. The study reveals the pattern followed by the business in the

past. It also bring out the relationship and interdependence of different

industries which helps in interpretation of changes in the right perspective.

The analysis gives an idea about the components of the time series and

their movement in the past. Various indices such as index of production,

prices, bank deposits, money rates, foreign exchange position etc. can

provide information about short and long term variations, the general

trend, the ups downs in a business.

The study of the past data and the comparison of the estimated and

actual values helps in pinpoint ing the areas of shortcoming which can be

overcome. For successful business forescasting co -ordination of all

departments such as production, sales, marketing is sine -qua-nin, which

result in achieving ultimate corporate goals.

There are different theor ies of Business Forecasting such as

i.Time lag or Sequence Theory

ii.Action and Reaction Theory

iii.Cross Cut Analysis Theory

iv.Specific Historical Analogy Theory

Of these, Time lag or Sequence Theory is most important. It is

based on the fact that there is a tim e lag between the effect of changes at

different stages but there is a sequence followed by these effect e.g. In

80’s, the invention of silicon ships brought fourth and fifth generation

computers in use. The computers were introduced in various fields such

as front -line and back house banking, airlines and railways reservation,

new communication technique, home appliances like washing machine

etc. this, in turn, increase the demand for qualified personnel in electronic

filed to manufacture, handle and mai ntain these sophisticated machine. It

has result in mad rush for admission to various branches of electronics and

computer engineering in the recent past.

By applying any one of the these forecasting theories, business

forecasting can be made. It shoul d be noted that while collecting the data

for analysis, utmost care has to be taken so as to increase the reliability of

estimates. The information should be collected by export investigators,

over a long period of time. Otherwise, it may lead to wrong co nclusions.

EXERCISE

1.What is a time series ? Describe the various components of a time

series with suitable example.

2.What are seasonal variation ? Explain briefly with example.

3.Describe the secular trend component of a time series,munotes.in

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4.What are the method of determining trend in a time series?

5.Compare method of moving average and least squares of estimating

trend component.

6.Find the trend values using the method of semi -averages for the

following data expressing production in thousand unit of a company

for 7 years.

7.Explain the method to calculate 3 yearly and 4 yearly moving

averages.

8.What are the merits and demerits of the method of moving average?

9.Explain the simple average method to find the seasonal indices of a

time series

10.Calculate trend by consider ing three yearly moving average for the

following time series of price indices for the years 2000 -2007. Also

plot on the graph the trend values.

Year 2000 2001 2002 2003 2004 2005 2006 2007

Price Index 111 115 116 118 119 120 122 124

11.Determine the tr end for the following data using 3 yearly moving

averages. Plot the graph of actual time series and the trend values.

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997

Sales

in‘000Rs 24 28 30 33 34 36 35 40 44

12.Determine the trend for th e following data using 3 yearly moving

averages. Plot the graph of actual time series and the trend values.

Year 1977 1978 1979 1980 1981 1982 1983 1984

Sales

in‘000Rs 46 54 52 56 58 62 59 63

13.Determine the trend for the following data us ing 3 yearly moving

averages. Plot the graph of actual time series and the trend values.

14.

Year 1979 1980 1981 1982 1983 1984 1985 1986

Profit in

lakhs of Rs 98 100 97 101 107 110 102 105

15.Determine the tren d for the following data using 5 yearly moving

averages. Plot the graph of actual time series and the trend values.

Year 1980 1982 1984 1986 1988 1990 1992 1994 1996 1998 2000

Values 34 37 35 38 37 40 43 42 48 50 52munotes.in

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16.Determine the trend for the following data giving the pro duction of

steel in million tons, using 5 yearly moving averages. Plot the graph of

actual time series and the trend values.

Year 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982

Production 28 30.5 32 36.8 38 36 39.4 40.6 42 45 43.5

17.Find five -yearly moving average for the following data which

represents production in thousand unit of a small scale industry. Plot the

given data as well as the moving average on a graph paper.

Year 1980 1981 1982 1983 1984 1985 1986 1987 1988 19891990

Production 110 104 78 105 109 120 115 110 115 122 130

Ans. The trend values are 101.2, 103.2, 105.4, 111.8, 113.8, 116.4 and

118.4 for the years 1982 to 1988.

18.Find the trend component of the following time series of production in

thousand k ilogram during 1971 -1980. Plot the moving average and the

original time on a graph paper.

Year 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980

Production 12 15 18 17 16 20 23 22 24 25

Ans. The trend values are 16, 17.125, 18.375, 19.625, 21.25, 22. 875 for

the years 1973 to 1978.

19.Fit a straight line trend to the following data representing import in

million Rs. of a certain company. Also find an estimate for the year 2008.

Ans. The straight line trend is y = 49 -x. the trend values are 52, 51, 50,

49, 48, 47 and 46 respectively and the estimate trend for the year 2006 is

44 million Rs.

20.The production of a certain brand of television sets in thousand unit is

given below. Fit a straight line trend to the data. Plot the given data and

the trend line on graph find an estimate for the year 2004.

Ans. The straight line trend is y = 947.71 + 33.43 x. the trend values are

846.42, 879.85, 913.28, 946.71, 1013.57 and 1047. The estimate for the

year 2004 is 1080.43 thermal million.

21.The straight line trend by the method of least squares for the following

data which represents the expenditure in lakhs od Rs. on advertisement ofYear 2000 2001 2002 2003 2004 2005 2006

Import 48 50 58 52 45 41 49

Year 1997 1998 1999 2000 2001 2002 2003

production 865 882 910 925 965 1000 1080munotes.in

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a certain company. Also find an estimate for the year 2005. Plot the given

data and the trend line on a graph paper.

Ans. The trend i s y = 40.13 +2.9x. the trend values are 19.83, 25.62,

31.43, 37.23, 43.03, 48.83, 54.63 and 60.43, 2005 is 66.23.

22.Use the method of least squares to find straight line trend for the

following time series of production in thousand units 1981 –1988. Al so

estimate trend for the year 2003.

Ans. The straight line trend is y = 91.5 + 1.167 x. the trend values are

83.331, 85.665, 87.999, 90.333, 92.667, 95.001, 97.335 and 99 .669. the

estimate of trend, for the year 2003 is 102.003

23.Calculate seasonal indices for the following data:

Year I II III IV

2003 55 53 57 51

2004 56 55 60 53

2005 57 56 61 54

Ans. 100.59, 98.2, 106.57, 94.61

24.Determine the trend for the fo llowing data giving the production of

wheat in thousand tons from the years 1980 to 1990, using the 5 -yearly

moving averages. Plot the graph of actual time series and the trend

values.

Year 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990

Producti on 13.5 14.7 17 16.2 18.1 20.4 22 21.2 24 25 26.6

25.Determine the trend for the following data giving the income (in

million dollars) from the export of a product from the year 1988 to

1999. Use the 4 -yearly moving average method and plot the graph of

actual time series and trend values.

Year 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999

Income 340 360 385 470 430 444 452 473 490 534 541 576

26.Using the 4 -yearly moving average method find the trend for the

following data.Year 1997 1998 1999 2000 2001 2002 2003 2004

Expenditure 21 24 32 40 38 49 57 60

Year 1995 1996 1997 1998 1999 2000 2001 2002

Production 80 90 92 83 94 99 92 102

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Year 1967 1968 1969 1970 1971 1972 1973 1974 1975 1976 1977

Value 102 100 103 105 104 109 112 115 113 119 117

27.Determine the trend for the following data giving the sales (in ’00 Rs.)

of a product per week for 20 weeks. Use appropriate moving average

method.

Week

Sales

Week

Sales1

22

11

292

26

12

343

28

13

364

25

14

355

30

15

356

35

16

397

39

17

438

36

18

489

30

19

5210

32

20

49

28.An online marketing company works 5 -days a week. The day -to-day

total sales (in ‘000 Rs) of their product for 4 weeks are given below.

Using a proper moving average method find the trend values.

Days

Sales

Days

Sales1

12

11

352

16

12

323

20

13

324

17

14

385

18

15

366

20

16

357

26

17

348

25

18

389

27

19

4010

30

20

41

29.Fit a straight line trend to the following data. Draw the graph of the

actual time series and the trend line. Estimate the sales for the years

2000.

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999

Sales in

‘000 Rs 45 47 49 48 54 58 53 59 62 60 64

30.Fit a straight line trend to the following data. Draw the graph of the

actual time series and the trend line. Estimate the sales for the years

2001.

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998

Profit in

‘000 Rs 76 79 82 84 81 84 89 92 88 90

31.Fit a straight line trend to the following data. Draw the graph of the

actual time series and the trend line. Estimate th e sales for the years

2007.

Year 1998 1999 2001 2002 2003 2004 2005 2006

Profit in

‘000 Rs 116 124 143 135 138 146 142 152munotes.in

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32.Fit a straight line trend to the following data giving the number of

casualties (in hundred) of motorcyclists without hel met. Estimate the

number for the year 1999.

Year 1992 1993 1994 1995 1996 1997 1998

No of

casualties 12 14.2 15.2 16 18.8 19.6 22.1

33.Fit a straight line trend to the following data. Draw the graph of the

actual time series and the trend line. Estimate the import for the years

2002

Year 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000

Import

in’000 Rs. 55 52 50 53 54 56 58 60 57 59

34.Fit a straight line trend to the following data giving the price of crude

oil per barrel in USD. Dr aw the graph of the actual time series and the

trend line. Estimate the sales for the year 2003.

Year 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001

Price per

barrel 98 102 104.5 108 105 109 112 118 115 120

35.Apply the method of le ast squar es to find the number of student

attending the library in the month of May of the academic year 2005 –

2006 from the following data.

Month Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr

Students

105 120 160 225 180 115 124 138 176 230 180

36.Assuming that the trend is absent, find the seasonal indices for the

following data and also find the deseasonalized values.

Quarters I II III IV

1977

1978

1979

198010

12

16

2412

15

18

2614

18

20

2816

22

24

34

37.Calculate seasonal indices for the follow ing data:

Year I II III IV

2003 55 53 57 51

2004 56 55 60 53

2005 57 56 61 54

Ans. 100.59, 98.2, 106.57, 94.61

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UNIT -IV

Unit -6

INDEX NUMBERS

Unit Structure :

6.0 Objectives

6.1 Introductions

6.2 Importance of Index Numbers

6.3 Price Index Numbers

6.4 Cost of Living Index Number or Consumer Price Index Number

6.5 Useof Cost of Living Index Numbers

6.6 Real Income

6.7 Demerits of Index Numbers

6.0OBJECTIVES

Tounderstand about the importance of Index Numbers.

Tounderstand different types of Index Numbers and their

computations.

Tounderstand about Real Income andCost of Living Index Numbers .

To understand the problems in constructing Index Numbers.

To know the merits and demerits of Index Numbers.

6.1INTRODUCTION

Every variable undergoes some changes over a period of time or in

different regions or due to some factors affecting it. These changes are

needed to be measure d. In the last chapter we have seen how a time series

helps in estimating the value of a variable in future. But the magnitude of

the changes or variations of a variable, if known, are useful for many more

reasons. For example, if the changes in prices of various household

commodities are known, one can plan for a proper budget for them in

advance. If a share broker is aware of the magnitude of fluctuations in the

price of a particular share or about the trend of the market he can plan his

course of action of buying or selling his shares. Thus, we can feel that

there is a need of such a measure to describe the changes in prices, sales,

profits, imports, exports etc, which are useful fr oma common man to a

business organization.

Index number is an important statistical relative tool to measure the

changes in a variable or group of variables with respect to time,

geographical conditions and other characteristics of the variable(s). Index

number is a relative measure, as it is independent of th e units of the

variable(s) taken in to consideration. This is the advantage of index

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are absolute measures, i.e.they are expressed in units, while index

numbers are percentage values which are independent of the units of the

variable(s). In calculating an index number, a base period is considered for

comparison and the changes in a variable are measured using various

methods.

Though index numbers were initially used for measuring th e

changes in prices of certain variables, now it is used in almost every field

of physical sciences, social sciences, government departments, economic

bodies and business organizations. The gross national product (GNP), per

capita income, cost of living in dex, production index, consumption,

profit/loss etc every variable in economics uses this as a tool to measure

the variations. Thus, the fluctuations, small or big, in the economy are

measure dby index numbers. Hence it is called as a barometer of

economic s.

6.2IMPORTANCE OF INDEX NUMBERS

The important characteristics of Index numbers are as follows:

(1)It is a relative measure : As discussed earlier index numbers are

independent of the units of the variable(s), hence it a special kind of

average which can be used to compare different types of data expressed in

different units at different points of time.

(2)Economical Barometer : A barometer is an instrument which measure s

the atmospheric pressure. As the index numbers measure all the ups and

downs in the econ omy they are hence called as the economic barometers.

(3)To generalize the characteristics of a group : Many a time it is

difficult to measure the changes in a variable in complete sense. For

example, it is not possible to directly measure the changes in a bu siness

activity in a country. But instead if we measure the changes in the factors

affecting the business activity, we can generalize it to the complete

activity. Similarly the industrial production or the agricultural output

cannot be measured directly.

(4)To forecast trends : Index numbers prove to be very useful in

identifying trends in a variable over a period of time and hence are used to

forecast the future trends.

(5)To facilitate decision making : Future estimations are always used for

long term and short term planning and formulating a policy for the future

by government and private organizations. Price Index numbers provide the

requisite for such policy decisions in economics.

(6)To measure the purchasing power of money and useful in deflating :

Index numb ers help in deciding the actual purchasing power of money.

We often hear from our elders saying that “ In our times the salary was

just Rs. 100 a month and you are paid Rs. 10,000, still you are not happy!”munotes.in

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The answer is simple (because of index numbers!) that the money value of

Rs. 100, 30 years before and now is drastically different. Calculation of

real income using index numbers is an important tool to measure the

actual income of an individual. This is called as deflation.

There are different types of index numbers based on their requirement

like, price index, quantity index, value index etc. The price index is again

classified as single price index and composite price index.

6.3PRICE INDEX NUMBERS

The price index numbers are classified as shown in the following diagram:

Notations:

P0: Price in Base Year Q0: Quantity in Base Year

P1: Price in Current Year Q1: Quantity in Current Year

Thesuffix ‘0’stands for the base year and the suffix ‘1’stands for the

current year .

6.3.1 Simple (U nweighted) Price Index N umber By Aggregative

Method

In this method we define the price index number as the ratio of sum of

prices in current year to sum of prices in base year and express it in

percentage. i.e.multiply the quotient by 100.

Symbolical ly, … (1)

Steps for computation:

1.The total of all base year prices is calculated and denoted by0P.

2.The total of all current year pri ces is calculated and denoted by

1P.

3.Using the above form ula, simple price index number is computed.

Example 1

Forthe following data, construct the price index number by simple

aggregative method:

I=1

0P

P

x 100munotes.in

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Price inCommodity Unit1985 1986

A Kg 10 12

B Kg 4 7

C Litre 6 7

D Litre 8 10

Solution: Follo wing the steps for computing the index number, we find

the totals of the 3rdand 4thcolumns as shown below:

Price inCommodity Unit1985( P0) 1986(P1)

A Kg 10 12

B Kg 4 7

C Litre 6 7

D Litre 8 10

I=1

0P

P

x100 =36

28x 100 = 128.57

Meaning of the value of I:

I= 128.57 means that the prices in 1986, as compared with that in 1985

have increased by 28.57 %.

6.3.2 Simple (Unweighted) Price Index Number by Average of

Price Relatives Meth od

In this method the price index is calculated for every commodity

and its arithmetic mean is taken. i.e.the sum of all price relative is divided

by the total number of commodities.

Symbolically, if there are ncommodities in to consideration, then

the simple price index number of the group is calculated by the formula:

… (2)

Steps for computation

1.The price relatives for each commodity are calculated by the formula:

1

0x 100P

P.

2.The total of these price relatives is calcul ated and denoted as:

1

0x 100P

P

.

3.The arithmetic mean of the price realtives using the above formula no.

(2) gives the required price index number.I=1

01x 100P

n P

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Example 2

Construct the simple price index number for the following data

using average of price relatives method:

Price inCommodity Unit1997 1998

Rice Kg 10 13

Wheat Kg 6 8

Milk Litre 8 10

Oil Litre 15 18

Solution : In this method we have to find price relatives for every

commodity and then total these price relatives. Following the steps for

computing as mentioned above, we introduce first, the column of price

relatives. The table of computation is as follows:

Price in

Commodity Unit1997( P0)1998( P1)1

0x 100P

P

Rice Kg 10 13 130

Wheat Kg 6 8 133.33

Milk Litre 8 10 125

Oil Litre 15 18 120

Total: 508.33

Now, n= 4 and the total of price relatives is 508.33

1

01x 100PIn P

=508.33

4= 127.08

The prices in 1998 have increased by 27 % as compared with in 1997.

Remar k:

1.The simple aggregative method is calculated without taking into

consideration the units of individual items in the group. This may give

a misleading index number.

2.This problem is overcome in the average of price relatives method, as

the individual price relatives are computed first and then their average

is taken.

3.Both the methods are unreliable as they give equal weightage to all

items in consideration which is not true practically.

6.3.3 Weighted Index Numbers by Aggregative Method

In this method w eights assigned to various items are considered in

the calculations. The products of the prices with the corresponding

weights are computed; their totals are divided and expressed in

percentages.

Symbolically, if Wdenotes the weights assigned and P0,P1have

their usual meaning, then the weighted index number using aggregative

method is given by the formula:munotes.in

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… (3)

Steps to find weighted index number using aggregative method

1.The columns of P1WandP0Ware introduced.

2.The totals of thes e columns are computed.

3.The formula no. (3) is used for computing the required index number.

Example 3

From the following data, construct the weighted price index number:

Commodity A B C D

Price in 1982 6 10 4 18

Price in 1983 9 18 6 26

Weight 35 30 20 15

Solution : Following the st eps mentioned above, the table of computations

is as follows:

Using the totals from the table, we have

Weighted Index Number I=1

0PW

PW

x100 =1365

860x 100 = 158.72

Remark :

There are different formulae based on what to be taken as the weight while

calculating the weighted index numbers. Based on the choice of the weight

we are going to study here three types of weighted index numbers: (1)

Laspeyre’s Index Number, (2) Paasche’s Index Number and (3) Fisher’s

Index Numb er.

(1)Laspeyre’s Index Number :

In this method, Laspeyre assumed the base quantity ( Q0) as the

weight in constructing the index number. Symbolically, P0,P1and Q0

having their usual meaning, the Laspeyre’s index number denoted by ILisCommodit

yWeight

(W)Price

in 1982

(P0)P0WPrice in

1983

(P1)P1W

A 35 6 210 9 315

B 30 10 300 18 540

C 20 4 80 6 120

D 15 18 270 26 390

Total - -0PW =

860- 1PW = 1365I=1

0PW

PW

x 100

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given by the formu la:

… (4)

Steps to compute IL:

1. The columns of the products P0Q0andP1Q0are introduced.

2. The totals of these columns are computed.

3. Using the above formula no. (4), ILis computed.

Example 4

From the data given below, construct the Laspeyre’s index number:

1965 1966CommodityPrice Quantity Price

A 5 12 7

B 7 12 9

C 10 15 15

D 18 5 20

Solution : Introducing the columns of the products P0Q0and P1Q0, the

table of computation is completed as shown below:

1965 1966

Commodity Price

(P0)Quantity

(Q0)Price

(P1)P0Q0 P1Q0

A 5 12 7 60 84

B 7 12 9 84 108

C 10 15 15 150 225

D 18 5 20 90 100

Total - - - 0 0P Q = 3841 0PQ = 517

Using the totals from the table and substituting in the formula no. (4), we

have

1 0

0 0517x 100384LPQIP Q x 100 = 134.64

(2)Paasche’s Index Number :

In this method, Paasch assumed the current year quantity ( Q1) as the

weight for constructing the index number. Symbolically, P0,P1andQ1

having their usual meaning, the Paasche’s index number denoted by IPis

given by the formula:

… (5)

The steps for computing IPare similar to that of IL.1 0

0 0LPQIP Qx 100

1 1

0 1PPQIP Qx 100munotes.in

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Example 5

From the data given below, construct the Paasche’s index number:

1985 1986CommodityPrice Price Quantity

A 5 8 10

B 10 14 20

C 6 9 25

D 8 10 10

Solution : Introducing the columns of the products0 1P Qand1 1PQ, the table

of computations is completed as shown below:

1985 1986

Commodity Price

(P0)Price

(P1)Quantity

(Q1)P0Q1 P1Q1

A 5 8 10 50 80

B 10 14 20 200 280

C 6 9 25 150 225

D 8 10 10 80 100

Total - - -P0Q1

= 480P1Q1=

685

Using the totals from the table and substituting in th e formula no. (5), we

have:1 1

0 1PPQIP Qx 100 =685

480x 100 = 142.71

(3)Fisher’s Index Number :

Fisher developed his own method by using the formulae of Laspeyre and

Paasche. He defined the index number as the geometric m ean of ILandIP.

Symbolically, the Fisher’s Index number denoted as IFis given by the

formula: IF= xL PI I =1 0 1 1

0 0 0 1xPQ PQ

P Q P Q

x 100 .. (6)

Note :

1.The multiple 100 is outside the square root sign .

2.While computing products of t he terms, care should be taken to

multiply corresponding numbers properly.

Example 6

From the following data given below, construct the (i) Laspeyre’s index

number, (ii) Paasche’s index number and hence (iii) Fisher’s index

number.

1975 1976ItemPrice Quantity Price Quantity

A 4 12 6 16

B 2 16 3 20

C 8 9 11 14munotes.in

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Solution: Introducing four columns of the products of P0Q0,P0Q1,P1Q0

andP1Q1, the table of computations is completes as shown below:

From the table, we have P0Q0= 152, P0Q1= 216, P1Q0= 219 and

P1Q1= 310

IL=1 0

0 0PQ

P Q

x 100 =219

152x 100 = 144.08

IP=1 1

0 1PQ

P Q

x 100 =310

216x 100 = 143.52

IF= xL PI I =144.08 x 143.52 = 143.8

Remark :

1.Laspeyre’s index number though popular has a drawback that it does

not consider the change in consumption over a period. (as it does not

take into account the current quantity).

2.Paasch e’s index number overcomes this by assigning the current year

quantity as weight.

3.Fisher’s index number being the geometric mean of both these index

numbers, it considers both the quantities. Hence it is called as the ideal

index number.

Example 7

From th e following data given below, construct the Kelly’s index number:

Base Year Current YearItemPrice Quantity Price Quantity

A 18 20 24 22

B 9 10 13 16

C 10 15 12 19

D 6 13 8 15

E 32 14 38 18

Solution: Introducing the columns of Q=0 1

2Q Q,P0QandP1Q, the table

of computations is completed as shown blow:Item P0 Q0 P1 Q1 P0Q0P0Q1P1Q0P1Q1

A 4 12 6 16 48 64 72 96

B 2 16 3 20 32 40 48 60

C 8 9 11 14 72 112 99 154

Total 152 216 219 310

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Item Q0 Q1 Q P0 P0Q P1 P1Q

A 20 22 21 18 378 24 504

B 10 16 13 9 117 13 169

C 15 19 17 10 170 12 204

D 13 15 14 6 84 8 112

E 14 18 16 32 512 38 608

Total 1261 -- 1597

From the table, we have P0Q= 1261 and P1Q= 1597

IK=1

0PQ

P Q

x 100 =1597

1261x 100 = 126.65

6.3.4 Weighted Index Numbers using ave rage of price relatives

method

This is similar to what we have seen in subsection 7.3.2. Here the

individual price relatives are computed first. These are multiplied with the

corresponding weights. The ratio of the sum of the products and the total

value o f the weight is defined to be the weighted index number.

Symbolically, if Wdenotes the weights and Idenote the price relatives

then the weighted index number is given by the formula:IW

W

… (8)

One of the important weighted index num ber is the cost of living index

number , also known as the consumer price index (CPI) number .

6.4 COST OF LIVING INDEX NUMBER OR

CONSUMER PRICE INDEX NUMBER

There are two methods for constructing this index number:

(1) Aggregative expenditure method and (2) Family Budget Method

(1)In aggregative expenditure method we construct the index number by

taking the base year quantity as the weight. In fact this index number is

nothing but the Laspeyre’s index number.

(2)In family budget method, value weights are com puted for each item in

the group and the index number is computed using the formula:

IW

W

, where I=1

0P

Px 100 and W=P0Q0 … (9)munotes.in

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Example 8

A survey of families in a city revealed the following inf ormation:

Item Food Clothing FuelHouse

RentMisc.

%

Expenditure30 20 15 20 15

Price in 1987 320 140 100 250 300

Price in 1988 400 150 125 250 320

What is the cost of living index number for 1988 as compared to that of

1987?

Solution : Here % expendi ture is taken as the weight ( W). The table of

computations is as shown below:

Item P0 P1I=1

0P

Px 100%

Expenditure

(W)IW

Food 320 400 125 30 3750

Clothing 140 150 107.14 20 2142.8

Fuel 100 125 125 15 1875

House Rent 250 250 100 20 2000

Miscellaneous 300 320 106.67 15 1600.05

Total W=100 11367.85

From the table, we have W= 100 and IW= 11376.85

cost of living index number =IW

W

=11367.85

100= 113.68

6.5USE OF COST OF LIVING INDEX NUMBERS

1.These index numbers reflect the effect of rise and fall in the economy

or change in prices over the standard of living of the people.

2.These index number s help in determining the purchasing power of

money which is the reciprocal of the cost of living index number.

3.It is used in deflation. i.e.determining the actual income of an

individual. Hence it also used by the management of government or

private orga nizations to formulate their policies regarding the wages,

allowance to their employees.

6.6 REAL INCOME

As discussed earlier in this chapter, index numbers are very useful

in finding the real income of an individual or a group of them, which

facilitat es the different managements to decide their wage policies. Themunotes.in

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process of measuring the actual income vis -a-vis the changes in prices is

called as deflation .

The formula for computing the real income is as follows:

Real Income of a year =Money Income for the year

Price Index of that yearx 100

Example 12

Calculate the real income for the following data:

Year 1990 1991 1992 1993 1994 1995

Income in

Rs.800 1050 1200 1600 2500 2800

Price

Index100 105 115 125 130 140

Solution : The real income is calculated by the formula:

real income =Money Income for the year

Price Index of that yearx 100

The table of computation of real income’s is completed as shown below:

Year Income in Rs.Price

IndexReal Income

1990 800 100 800

1991 1050 1051050

105x100 = 1000

1992 1200 1151200

115x 100 = 1043

1993 1600 1251600

125x 100 = 1280

1994 2500 1302500

130x 100 = 1923

1995 2800 1402800

140x 100= 2000

6.7DEMERITS OF INDEX NUMBERS

(1)There are numero us types and methods of constructing index

numbers. If an appropriate method is not applied it may lead to wrong

conclusions.

(2)The sample selection may not be representative of the complete series

of items.

(3)The base period selection also is personalized and hence may be

biased.

(4)Index number is a quantitative measure and does not take into

account the qualitative aspect of the items.

(5)Index numbers are approximations of the changes, they may not

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Check Your Progress

1.Define Index Numbers.

2.Write a shor t note on the importance of Index Numbers.

3.“Index Numbers are the Economical barometers”. Discuss this

statement with examples.

4.Discuss the steps to construct Index Numbers.

5.What are the problems in constructing an Index Number?

6.Define Cost of Living Index Number and explain its importance.

7.What do you mean by (i) Chain Based Index Number and (ii) Fixed

Base Index Number? Distinguish between the two.

8.Define (i) Laspeyre’s Index Number, (ii) Paasche’s Index Number

and (iii) Fisher’s Index Number. What is the difference between the

three? Which amongst them is called as the ideal Index Number?

Why?

9.What are the demerits of Index Numbers?

10.From the following data, construct the price index number by simple

aggregative method:

Price inCommodity Unit1990 1991

A Kg 14 18

B Kg 6 9

C Litre 5 8

D Litre 12 20

Ans: 148.65

11.From the following data, construct the price index number for 1995,

by simple aggregative method, with 1994 as the base:

Price inCommodity Unit1994 1995

Rice Kg 8 10

Wheat Kg 5 6.5

Oil Litre 10 13

Eggs Dozen 4 6

Ans: 131.48

12.From the following data, construct the price index number for 1986,

by average of price relatives method:

Price inCommodity Unit1985 1986

Banana Dozen 4 5

Rice Kg 5 6

Milk Litre 3 4.5

Slice Bread One Pac ket 3 4

Ans: 132.08munotes.in

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13.From the following data, construct the price index number, by method

of average of price relatives:

Price inCommodity Unit1988 1990

A Kg 6 7.5

B Kg 4 7

C Kg 10 14

D Litre 8 12

E Litre 12 18

Ans: 148

14.From the following data, construct the price index number for 1998,

by (i) simple aggregative method and (ii) simple average of price relatives

method, with 1995 as the base:

Price inCommodity Unit1995 1998

Rice Kg 12 14

Wheat Kg 8 10

Jowar Kg 7 9

Pulses Kg 10 13

Ans: (i) 124.32, (ii) 125.06

15.From the following data, construct the weighted price index number:

Commodity A B C D

Price in 1985 10 18 36 8

Price in 1986 12 24 40 10

Weight 40 25 15 20

Ans: 121.29

16.From the following data, construct the index number using (i) simple

average of price relatives and (ii) weighted average of price relatives:

Ans: (i) 130.13, (ii) 128.82Price inCommodity Weight1988 1990

Rice 4 8 10

Wheat 2 6 8

Pulses 3 8 11

Oil 5 12 15munotes.in

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17.From the data given below, construc t the Laspeyre’s index number:

1975 1976CommodityPrice Quantity Price

A 5 10 8

B 6 15 7.5

C 2 20 3

D 10 14 12

Ans: 131.40

18.From the data given below, construct the Paasche’s index number:

1980 1985CommodityPrice Price Quantity

A 4 7 10

B 14 22 16

C 5 7 30

D 8 10 21

Ans: 144.67

19.From the following data given below, construct the (i) Laspeyre’s

index number, (ii) Paasche’s index number and hence (iii) Fisher’s index

number.

1980 1990CommodityPrice Quantity Price Quantity

A 6 15 9 21

B 4 18 7.5 25

C 2 32 8 45

D 7 20 11 29

Ans: (i) 203.83, (ii) 203.37, (iii) 203.60

20.From the following data given below, construct the (i) Laspeyre’s

index number, (ii) Paasche’s index number and (iii) Fisher’s index

number.

Base Year Current Yea rCommodityPrice Quantity Price Quantity

Cement 140 200 167 254

Steel 60 150 95 200

Coal 74 118 86 110

Limestone 35 50 46 60

Ans: (i) 103.98, (ii) 127.4, (iii) 115.09munotes.in

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21.From the following data given below, construct the Fisher’s index

number:

Base Year Current YearCommodit yPrice Quantity Price Quantity

A 2 8 4 14

B 6 14 7 20

C 8.5 10 12 15

D 14 8 19 12

E 22 60 38 85

Ans: 131.37, 120.15

22.From the following data, construct the aggregative price index

numbers by taking the average price of the t hree years as base.

Commodity Price in 1980 Price in 1981 Price in 1982

A 10 12 16

B 16 19 25

C 5 7 10

Ans: 81.58, 100, 134.21

23.From the following data, construct the price index number by taking

theprice in 1978 as the base price using aggregative m ethod:

CommodityPrice in

1978Price in1979Price in

1980

A 16 18 24

B 4 6 7.5

C 11 15 19

D 20 28 30

Ans: 131.37, 120.15

24.From the following data, construct the price index number by taking

the price in 1998 as the base price:

Commodity WeightPrice in

1998Price in

1999Price in

2000

A 3 12 15 20

B 1 8 9 11

C 4 16 20 25

D 2 15 18 22munotes.in

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25.From the following data, construct (i) IL, (ii) IP, (iii) IF

1969 1970CommodityPrice Quantity Price Quantity

Rice 2 10 3 12

Wheat 1.5 8 1.9 10

Jowa r 1 6 1.2 10

Bajra 1.2 5 1.6 8

Pulses 4 14 6 20

Ans: 144.4, 144.56, 144.28

26.Construct the cost of living index number for 1980 using the Family

Budget Method:

Ans: 192.95

27.Construct the cost of living index number for the following data with

base year as 1989.

Ans: for 199 0: 114.49, for 1991 : 132.20

28.A survey of families in a city revealed the following information:

Item Food Clothing FuelHouse

RentMisc.

%

Expenditure30 20 15 20 15

Price in

1987320 140 100 250 300

Price in

1988400 150 125 250 320

What is the cost of living index number for 1 988 as compared to that of

1987 ? Ans: 113.65Price inItem Quantity1975 1980

A 10 5 7

B 5 8 11

C 7 12 14.5

D 4 6 10

E 1 250 600

Price inItem Weight1989 1990 1991

Food 4 45 50 60

Clothing 2 30 33 38

Fuel 1 10 12 13

House Rent 3 40 42 45

Miscellaneous 1 5 8 10munotes.in

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29.Construct the consumer price index number for the following

industrial data:

Ans: 167.30

30.Calculate the real income for the following data:

Year 1988 1989 1990 1991 1992 1993

Income in

Rs.500 550 700 780 900 1150

Price

Index100 110 115 130 140 155

31.The employees of Australian Steel lt d. have presented the following

data in support of their contention that they are entitled to a wage

adjustment. Dollar amounts shown represent the average weekly take

home pay of the group:

Year 1973 1974 1975 1976

Pay in $ 260.50 263.80 274 282.50

Index 126.8 129.5 136.2 141.1

Compute the real wages based on the take home pay and the price indices

given. Also compute the amount of pay needed in 1976 to provide buying

power equal to that enjoyed in 1973.

32.Calculate the real income for the following d ata:

Year 1977 1978 1979 1980 1981 1982

Income in

Rs.250 300 350 500 750 1000

Price

Index100 105 110 120 125 140Item Weight Price Index

Industrial Production 30 180

Exports 15 145

Imports 10 150

Transpor tation 5 170

Other activity 5 190

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33.The per capita income and the corresponding cost of living index

numbers are given below. Find the per capita real income:

Year 1962 1963 1964 1965 1966 1967

per capita

income220 240 280 315 335 390

cost of

living I.N.100 110 115 135 150 160

34.The following data gives the salaries (in ’00 Rs.) of the employees of

Hindusthan Constructions Ltd with the cost of living index number . Find

the real income and suggest how much allowance should be paid to them

to maintain the same standard of living.

Year 1990 1991 1992 1993 1994 1995

Income 12 14 17 20 24 28

Price

Index100 120 135 155 180 225

35.The income of Mr. Bhushan Damle in 1 999 was Rs. 8,000 per month.

If he gets an increment of Rs. 1,200 in 2000 and the price index being 115

with base as 1999, can you conclude that Mr. Damle has got an increment

which will maintain his standard of living as compared with the previous

year?

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UNIT -V

Unit -7

PROBABILITY DISTRIBUTIONS

Unit Structure :

7.1 Introduction

7.2 Mathematical Expectation and Variance

7.3 Binomial Distribution

7.4 Normal Distribution

7.1INTRODUCTION

In the previous chapter we have seen that outcomes of an

experime nt can be expressed in numbers. In an experiment of throwing a

die, the possible outcomes are expressed as 1, 2, 3, 4, 5, or 6. In

experiments like tossing a coin or picking a card or drawing a ball from a

bag, the outcomes are numbers. But they can be ass igned values like in

tossing a coin, the outcome of heads can be assigned value 0 and that of a

tails can be assigned value 1. Thus, in some way all points in a sample

space can be assigned numerical values.

A relation which assigns every outcome of an e xperiment to a real

number is called as a random variable also called as stochastic variable.

We can also say thus, that a random variable assures the probability for

every outcome of an experiment.

Example 1:

In the above example of throwing a die, the random variable say X

takes values {1, 2, 3, 4, 5, 6}.

In an experiment of tossing two coins, we can assume the random

variable as the number of heads (or tails) in an outcome. The sample space

isS= {HH, HT, TH, TT}. If number of heads denotes the val ue of the

random variable ( X), then the first outcome has 2 heads, second and third

has 1 heads and fourth outcome has no heads. Thus, X= {2, 1, 1, 0}. We

will have a different random variable ( Y) if we take the number of tails as

the counter for the rand om variable. In that case Y= {0, 1, 1, 2}.

Discrete random variable:

A random variable which takes discrete (distinct) values or to say

in mathematical words as the variable which takes finite or countably

infinite values is called as discrete random var iable .munotes.in

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Example 2:

The scores on a die, the number of calls received at a call centre,

number of letters typed by a secretary, number of strikes in a factory etc

are examples of discrete random variables.

Probability Distribution:

We know that with every value of the random variable ( X) there is

a probability ( P(X)) assigned for that particular outcome. The set of all

values of the random variable along with their corresponding probabilities

is called as the probability distribution of the random variable .

Example 3:

In an experiment of tossing three coins, if the random variable X

denotes the number of heads in every outcome then the probability

distribution table of Xis as shown below:

X 0 1 2 3

P(X) 1/8 3/8 3/8 1/8

In an experiment of throwing two dice, if the random variable

represents the sum of the scores on the upper faces of both the dice then its

probability distribution table is as shown below:

X Events P(X)

2 (1, 1) 1/36

3 (1, 2), (2, 1) 2/36 = 1/18

4 (1, 3), (2, 2), (3, 1) 3/36 = 1/12

5 (1, 4), (2, 3), (3, 2), (4, 1) 4/36 = 1/9

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 = 1/6

8 (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) 5/36

9 (3, 6), (4, 5), (5, 4), (6, 3) 4/36 = 1/9

10 (4, 6), (5, 5), (6, 4) 3/36 = 1/12

11 (5, 6), (6, 5) 2/36 = 1/18

12 (6,6) 1/36

The events are explicitly written to count the probability. It is not

expected to write the events every time. One should also observe that the

total of the column of probabiliti es is 1, which we already know from the

previous chapter that ( )iP A = 1 for all events Ai.

The function P(X) is called the probability function ofX.

The probability distribution of a random variable is called as

discrete probability d istribution and the corresponding probability

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7.2MATHEMATICAL EXPECTATION and VARIANCE

Let Xbe a random variable taking values x1,x2,…xnwith

corresponding probabilities1p,2p, …,nprespectively. The Expectation

ofx, denoted as E(x) is given by the formula:

1 1 2 2 ( ) ....n n i i E x p x p x p x p x , whereip= 1

In cases of games of chance if the amount received b y a player if

he wins a game is a, the amount he looses if he does not win the game is b

and the probability of winning the game is pthen the expectation is given

by the formula: ap bp, where 1p p.The negative sign b ecause of the

loss the person suffers.

The expectation of a random variable is called as its mean . The

mean of a random variable is denoted by the Greek letter (pronounce as

‘mu’ ofmusic ).

Laws of Expectation

IfXandYare two r andom variables then

1.E(X)0.

2.The expected value of their sum X + Y is given by : E(XY) =E(X)

E(Y).

3.Similarly, E(aXbY) =aE(X)bE(Y).

4.The expected value of their product XYis given by: E(XY) =E(X).E(Y)

Variance of a random variable

Variance of a random variable Xdenoted by V(X) is the square of

the standard deviation of Xand is calculated by the formula: V(X) =E(X2)

–[E(X)]2

Example 4:

A die is thrown at random. What is the expectation and variance of

the number on it?

Ans: When a die is thrown the possible outcomes are X = {1, 2, 3, 4, 5,

6} with each having a probability of 1/6. This can be tabulated as follows:

xi 1 2 3 4 5 6

pi 1/6 1/6 1/6 1/6 1/6 1/6

E(X) =1 1 2 2 ....n n p x p x p x = (1 + 2 + 3 + 4 + 5 + 6) x 1/6 = 21/6 = 3.5

Now, E(X2) = 1 x 1/6 + 22x 1/6 + 32x 1/6 + 42x 1/6 + 52x 1/6 + 62x 1/6

= (1 + 4 + 9 + 25 + 36) x 1/6 = 75/6 = 12.5

V(X) =E(X2)–[E(X)]2= 12.5 –(3.5)2= 12.5 –12.25 = 0.25

Example 5:

Three coins are tossed. What is the expectation of a heads

occurring?

Ans: When three coins are tossed the sample space is as follows:

S= {HHH ,HHT ,HTH ,THH ,HTT,THT,TTH,TTT}munotes.in

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The outcomes for a head are: 0 heads, 1 heads, 2 heads and 3 heads.

The corresponding probabilities are shown below:

xi(heads) 0 1 2 3

pi 1/8 3/8 3/8 1/8

E(no.of heads ) =E(x) =i ip x =1 3 3 10 x 1 x 2 x 3 x8 8 8 8 =12

8

E(x) = 1.5

Example 6:

The probability that it will rain on a day is 0.6. Find the

expectation of umbrella seller who earns a profit of Rs. 3050 if it rains and

a loss of Rs. 250 if it does n ot rain on a day.

Ans: Given p= 0.6p= 1–0.6 = 0.4, a= 3050 and b= 250

the mathematical expectation = 3050 x 0.6 –250 x 0.4 = 1830 –

100 = 1730

Exercise

(1)Define ( a) random vari able , ( b) discrete random variable, ( c)

probability mass function

(2)Define ( a) mean and ( b) of a random variable.

(3)Prepare a frequency distribution table of tossing 3 coins.

(4)Prepare a frequency distribution table of throwing two dice.

(5)If it rains, a dealer i n umbrella can earn Rs. 300 per day and if it does

not rain he can lose Rs.80 per day. What is the expectation if the

probability of a rainy day is 0.57?

(6)A person plays a game where he earns Rs. X2if he gets Xon a die.

Find the mean and variance of his earning.

(7)In a game, a person Xthrows a coin three times. He is paid Rs. 400 if

he gets a heads all three times. The entry fee for the game is Rs. 80.

What is the mathematical expectation of X?

(8)A shop owner earns a profit of Rs. 1000 per day. On a holiday he

suffers a loss of Rs. 300 per day. If the probability that a day is a

holiday is 0.14, find his mathematical expectation.

(9)A die is thrown at random. What is the expectation of the number on

it?

(10)If two dice are thrown, what is the expectation of the sum of the

sample points?

(11)Ifndice are thrown, what is the expectation of the sum of the sample

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(Hint: In above two problems use laws of expectation)

(12)Find the mean and variance when a coin is tossed two times.

(13)If three coins are tossed, what is the ex pectation and variance of the

number of tails?

(14)A person draws 2 balls from a bag containing 6 white and 5 black

balls. He is paid Rs. 22, if he draws both balls of white color and Rs.

11 if he draws one of each color. Find his expectation.

(15)A person draws 3 balls from a bag containing 3 white, 4 red and 5

black balls. He is offered Rs. 10, Rs. 5 and Rs. 2 if he draws three

balls of same color, 2 balls of same color and 1 ball of each color

respectively. Find his expectation.

(16)The probability there is at least one error in accounts statement

prepared by A is 0.25, by B is 0.35 and by C is 0.4. If A, B and C

prepared 10, 16 and 20 statements respectively, find the expected

number of error free statements.

(17)Three Mathematics teachers X, Y and Z were given 120, 200 and 150

papers of an examination to assess respectively. The probability that

there is totaling mistake in a paper by A, B and C is 0.2, 0.55 and

0.25 respectively. Find the expected number of ( a) papers in which

error is possible and ( b) error free paper s.

(18)IfXandYare two independent discrete random variables such that

E(X) = 12, E(Y) = 20, then find the expectation of Z= 2X+ 3Y.

(19)In a lottery game there are 10 tickets. 3 tickets have a prize of Rs. 2,

2 tickets have a prize of Rs. 5 and 1 ticket has a prize of Rs. 10. The

remaining tickets are blank. Find the expectation of a player winning

a prize.

(20)A newspaper agent earns Rs.200 a day if there is some breaking news

in it and loses Rs. 20 a day if there is no breaking news. The

probability that there is a breaking news in the paper is 0.45. Find the

expectation of his earnings.

(21)A gambler draws a card from a pack of 52 cards. He earns Rs. 104 if

it is an ace, Rs. 52 if it is a King or a Queen, Rs. 26 if it is a Jack and

loses Rs. 13 if it is any other c ard. Find his expectation.

(22)Find the mean and variance for the following probability distribution:

X -5 0 5 10 15

P(X) 1/3 1/9 1/3 1/9 1/9

(23)Find the mean and variance for the following probability

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X 0 2 4 6 8 10

P(X) 0.05 0.15 0.02 0.28 0.3 0.2

(24)The probability distribution of daily demand of cell phones in a

mobile gallery is given below. Find the mean and variance.

Demand 5 10 15 20

Probability 0.4 0.22 0.28 0.10

(25)The probability distribution of number of divorce cases withdrawn

per da y from a court is given below. Find the mean and variance.

No. of

Cases0 1 2 3 4 5

Probability 0.2 0.26 0.12 0.22 0.08 0.12

(26)At the famous ‘Jumbo Vada’ shop near Dadar station in Mumbai,

the probability distribution of arrival of number of customers pe r minute is

given below. Find the expectation of customers per minute.

No of

Customers0 2 4 6 8

Probability 0.24 0.22 0.26 0.18 0.1

(27)Mr. Chinmay has bought a new motorcycle from ‘Swastik

Agency’. The agency offers after sales service contract for Rs. 60 0 for

four years. From a market survey the probability distribution of the

expenses on service in four years for the same brand is known and is given

below. Should Mr. Chinmay pay for the service contract?

Expenses 200 400 600 800 1000 1200

Probability 0.34 0.22 0.1 0.08 0.22 0.04

7.3BINOMIAL DISTRIBUTION

In many situations we see that a same experiment is repeated

number of times. Also the probabilities of the outcomes are fixed

irrespective of the previous trials. Let us consider the example of tossi ng a

coin three times. Here one experiment of tossing a coin is done three times

and the probability of getting a head or getting a tail is always the same

i.e.1/2. Let us call the event of getting a heads as a ‘success’ and getting a

tails as a ‘failure’ and the corresponding probabilities as pandqwhere p +

q= 1 (or q =1–p). Consider an outcome ‘HTT’. With our notations just

introduced, there is one success and 2 failures in this outcome. The

probability of this outcome is obtained by multiplying t he individual

probabilities. So we have the probability of HTT as1

2x1

2x1

2=1

8orpxq

xq=pq2. But if the order is not important then the probability of getting

onesuccess and two failures is 3 x pq2, as there are three such outcomes

vizHTT, THT and TTH. It is easy to do so if the number of trials is less.munotes.in

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When the number of trial is more and the probabilities of outcomes are

fixed either a success or a failure, Be rnoulli characterized such probability

distributions and called them as Binomial Distributions .

Characteristics of Binomial Distributions

1.The experiment consists of n(a finite number) number of trials.

2.The outcome of every trial is either a success or a failure and is

independent of the previous trial(s).

3.The probabilities of success (and hence failures) remains constant for

every trial.

4.Ifpdenotes the probability of success, q= 1 –p denotes the

probability of failure and rdenote the number of succes ses in all n

trials then the probability of rnumber of successes is given by the

formula: P(X = r ) =n r n r

rC p q. This is the probability mass function

ofr. Here r= 0, 1, 2, 3, ……., n.

5.The probabilities given in the formula are nothing but the terms in the

binomial expansion of ( p + q )n. Hence the name given to this

distribution is Binomial Distribution.

6.Mean of a Binomial Distribution is given by =np.

7.Variance of a Binomial Distribution is given by npq. Hence its

standard deviation is npq .

Example 7:

The mean of a binomial distribution is 12 and standard deviation is

3. Calculate n,pandq.

Ans: Given np= 12, npq = 3npq= 9

12 x q= 9q= 9/12 = ¾

p= 1–q= 1–¾ = ¼

nx ¼ = 12 n= 12 x 4 = 48

Thus, n= 48, p= ¼ and q= ¾

Example 8:

The mean and variance of a binomial distribution are 14 and 9.

Comment on this statement.

Ans: From the given information, np= 14 and npq = 9npq=

81

14 x q= 81q= 81/14 = 5.78

We know that pr obability of any outcome is never greater than 1.

Thus, the information given is inconsistent.

Example 9

The probability that a youth exercises every day is 0.6. Find the

probability that out of 5 youths selected ( i) none of them do exercise, ( ii)

atleast one exercises.munotes.in

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Ans: Given p= 0.6q= 1–p =1–0.6 = 0.4 and n= 5

(i) To find probability that none of them exercise means r= 0

We know that: P(X=r) =n r n r

rC p q

P(X= 0) =5 0 5 0

0(0.6) (0.4)C= (0.4)5= 0.01024

(ii) To find probability of atleast one does exercise. This is a

complementary event of no one does exercise, whose probability we have

already found.

P(atleast one does exercise) = 1 –P(X= 0) = 1 –0.01024 = 0.98976

Example 10

The average rainfall in a 30 days’ month is 50%. Find the

probability that ( i) the first four days of a given week will be fine and the

remaining will be wet, ( ii) rain will fall on just three days of a week.

Ans: Given p= 50% = 0.5 q= 0.5

(i) To find probability that the first four days of a given week will be fine

and the remaining will be wet. Here the days of rainfall is fixed.

probability for rainfall in remaining days of a week = (0.5)3(0.5)4

= 0.0078

(ii) To find the probability that rain will fall on just three days of a week.

Here which three days is specified.

P(X =3) =7 3 4

3(0.5) (0.5)C =7 x 6 x 5

3 x 2x 0.0078 = 0.27

Example 12

The food inspect or along with his colleagues comes for an

inspection of a drug. The number of faulty drug tablets found in every

sample pack of 16 tablets is given below. The following table shows the

distribution of 160 tablets.

No of faulty

tablets0 1 2 3 4 5

No of sa mples 26 22 30 35 36 11

(i) Fit a binomial distribution and find the expected frequencies, if the

chance of the machine being defective is ½

(ii) Find the mean and standard deviation of the fitted distribution.

Ans: (i) Given p= ½q= ½ , n= 5 and N= 160

We will now find the probabilities of the number of faulty tablets

by finding the binomial expansion of ( p + q )5.

Now, ( p + q )5=5 4 3 2 2 3 4 51 5 10 10 5 1p p q p q p q pq q

Substituting values of pandq. we have

(p + q )5=1 5 10 10 5 1

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Now, the numbers1 5 10 10 5 1, , , , ,32 32 32 32 32 32are the corresponding probabilities

of 0, 1, 2, 3, 4 and 5 no. of faulty tablets.

The expected frequencies are calculated by multiplying each probability

with N= 160

Thus, the observed frequencies w ith the expected frequencies are tabulated

as shown below:

No of faulty

tablets0 1 2 3 4 5

Observed

Frequency26 22 30 35 36 11

Expected

Frequency5 25 50 50 25 5

(ii) Mean = np= 5 x ½ = 2.5

standard deviation = npq =1 15 x x2 2= 1.12

Exercise

(1)Ifn= 10 and p= ¼, find the mean and variance of the binomial

distribution.

(2)The mean of a binomial distribution is 40 and standard deviation is 6.

Calculate n,pandq.

(3)The mean of a binomial distribution is 6 and its st andard deviation is

. Find n,pandq.

(4)The mean of a binomial distribution is 3 and the variance is 1.2, find

nandp.

(5)The mean of a binomial distribution is 5 and variance is 5/2. Find

P(X= 4).

(6)The mean and variance of a binomial distributio n are 3 and 6/5. Find

P(X >3).

(7)Ifn= 8 and p= 2/3, find the mean and standard deviation of the

binomial distribution.

(8)Comment on “The mean and variance of a binomial distribution are 4

and 6”.

(9)Comment on “The mean and standard deviation of a binomial

distribution are 6 and 4”.

(10)A bag contains 10 white and 5 black balls. If 4 balls are selected at

random, find the probability that ( i) 3 are black balls, ( ii) atleast one

black ball is selected.

(11)Out of the total passengers travelling by BEST buses 40% do not

have the exact money for their tickets. If a conductor gives tickets tomunotes.in

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15 passengers, find the probability that less than 5 passengers will

pay exact money.

(12)In a government department an officer is not in his chair 12 out of 30

days in a month. If the se nior officer investigates his office 6 times,

what is the probability that he is not in his chair 3 times.

(13)The probability that internet users will buy a product from an online

marketing advertisement is 0.35. If 100 users login the site, find the

probabil ity that15 or more than that would actually buy the product?

(14)The EGS of central government guarantees employment to every 4

out of 6 unemployed persons. If 100 people are selected at random

from a village, find the probability that 10 of them have received

employment.

(15)The probability that a student from distance education passes is 0.6.

Find the probability that out of 6 students selected ( i) no student

passes, ( ii) atleast one student passes.

(16)The probability that an evening college student will graduate is 0.4.

Determine the probability that out of 5 students ( i) none, ( ii) one, ( iii)

atleast one student will graduate.

(17)The overall passing percentage of students enrolled in a distance

education institute is 35%. Find the probability that out 6 students 3

students pass.

(18)The probability that a micro wave oven is found to be defective is

0.2. If 6 microwave ovens are selected, find the probability that all

are defective.

(19)The probability of a defective bolt is 1/10. Find the mean and

variance of defective bolts o ut of a total of 400 bolts.

(20)5 out of 25 items are found to be defective. If 4 items are selected

find the probability distribution of the defective items.

(21)6 out of 50 items in a lot are found to be defective. Find the

probability of the following if 4 item s are selected: ( i) one defective

item, ( ii) 3 defective items, ( iii) at most 3 defective items.

(22)6% of articles in a given bundle are defective. Find the probability

that in a sample of 5 articles, none is defective.

(23)The probability that a long range missi le hits a target is 0.8. If 4

missiles are shot, find the probability that ( i) exactly two will hit the

target, ( ii) atleast two will hit the target.

(24)The probability of a man hitting a target is ¼. If h fires 7 times, what

is the probability that he hits t he target atleast twice?

(25)The average rainfall in a 30 days’ month is 40%. Find the probability

that ( i) the first four days of a given week will be fine and the

remaining will be wet, ( ii) rain will fall on just three days of a week.

(26)Out of 1000 families w ith 4 children each, how many would you

expect to have ( i) 2 boys and 2 girls, ( ii) atleast one boy, ( iii) no girlmunotes.in

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(27)If on an average rain falls on 10 days in every 30 days, find the

probability that ( i) the first three days are fine and the remaining are

wet, (ii) the rain will fall on just three days of a week

(28)A box contains 60 Alphanso mangoes of which 8 are normal

mangoes. 8 mangoes are selected by the food inspector. Find the

probability that ( i) 6 mangoes are normal type, ( ii) atleast one is an

Alphanso type mango.

(29)The incidence of a certain disease is such that on an average 20% of

workers suffer from it. If 10 workers are selected at random, find the

probability that ( i) exactly 2 workers suffer from the disease, ( ii) not

more than 2 workers suffer fro m the disease, ( iii) not more than 2

workers suffer from the disease.

(30)An unbiased die is tossed 3 times. Find the probability of obtaining

(i) no six, ( ii) all sixes.

(31)An unbiased die is tossed 4 times. Find the probability of obtaining

(i) atleast one six, (ii) 4 sixes.

(32)Four coins are tossed simultaneously. What is the probability of

getting ( i) 4 heads, ( ii) 2 heads, 2 tails, ( iii) atleast one heads

(33)Five coins are tossed 3200 times. Find the frequency distribution of

heads and tails. Also find the mean and variance.

(34)8 coins are tossed 256 times. Find the expected frequencies of

success. Also find the mean and variance of the fitted values.

(35)The screws produced by a certain machine were checked by

examining the number of defectives in a sample of 12. The foll owing

table shows the distribution of 128 samples.

(36)

No of

defectives0 1 2 3 4 5 6 7

No of samples 7 6 19 35 30 23 7 1

(i) Fit a binomial distribution and find the expected frequencies, if the

chance of the machine being defective is 0.5

(ii) Find the m ean and standard deviation of the fitted distribution.

7.4NORMAL DISTRIBUTION

The Binomial distribution is a discrete probability distribution as

the random variable considered is a discrete. But for many practical

problems related to sales volume, he ight, weight of an individual or a

product, length of an item, strength, resistance, life of an electrical

instrument the random variable is of continuous type

Normal distribution is the most commonly used continuous

probability distribution. The probabi lity density function of a normal

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P(X) =21

2 1

2X

e

for–Hereis the mean and is the standard deviation of the

distri bution.

Characteristics of a Normal curve

Fig7.1

1.The curve of the normal curve is bell shaped.

2.The curve is symmetrical about the central vertical line corresponding

to the mean of the distribution.

3.The peak of the curve is obtain ed at X=.

4.All the three measures of central tendencies mean, median and mode

coincide for a normal curve.

5.There are two tails of the curve which extend infinitely in both

positive and negative X -axis and never touch the axis.

6.The area under the normal curve is unity i.e.1. This is because the

area represents the probabilities of the variable and the sum of all

probabilities as we know is 1. Due to symmetry, the area to the left of

the mean is exactly 50% i.e.0.5 and similarly the area to the right of

the mean is also 0.5. Thus, for a normal curve, P(X) =P(X).

7.The standard deviation determines the spread of the distribution

around the m ean. As shown in Fig 9.2, if the value of is small then

the curve will be narrow and if the value of is large the curve will

become wider indicating the deviations of data around the mean.

8.The area covered b etween is 68.26% of the total area under the

normal curve.

9.The area covered between 2is 95.44% of the total area under the

normal curve.

10.The area covered between 3is 99.74% of the total area under the

normal curve.

Fig7.2

Area under the Normal Curve

IfXis a random normal variable, then the formula to calculate the

probability function uses methods of integration which are quite

cumbersome. This lead to a procedure of defining a new variable z=

X

, called as the standard normal variate (S.N.V.). It is observed that

the mean of this SNV is zero and its standard deviation is unity. Now the

problem of finding the probabilities for X=X1, is equated with finding the

area under the standard normal curve at z=z1. The area between z =0 andmunotes.in

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z=z1is read from the standard normal tables. This table is provided in the

appendix.

Now let us see some examples regarding how to read the standard

normal tables:

Example 13:

To find the area under the curve when z= 1.35

Ans: We look at the first column of the table for the number

1.3 and then move horizontally till the column of 0.05

which corresponds to our 1.35 (1.3 + 0.05). The value

in this cell is 0.4115. Thus, the area under the curve for

z= 1.35 is 0.4115. This also means that z= 1.35 represents 41.15%

of the total area.

Fig 7.3

Example 14:

To find area under the curve for z> 1.62

Ans: Repeating the previous step we find the value of

1.6 + 0.02 from th e table, this is 0.4474.

Now, we want to find area for z> 1.62. What we

Have got is the area between z= 0 and z= 1.62. The

total area to the right of z= 0 is 0.5. Thus the required

area is 0.5 –0.4474 = 0.0526

Fig 7.4

Example 15:

To find area un der the normal curve between z>–1.25

Ans: We know that the normal curve is symmetric about the

mean. So we repeat the same steps as described above

to find the area between z= 0 and z= 1.25. From the table this area is

0.3944. Now, the area represe nting z>–1.25 is the area between 0 to –

1.25 plus the remaining 50% area i.e0.5

Thus, the required area is 0.3944 + 0.5 = 0.8944

Fig 7.5

Example 16

To find area between –1.2z2

Ans: The required ar ea is split into two areas:

–1.2z0 and 0 z2

The first area as we have already seen how to calculate

is 0.3849 and the second area is 0.4772

Thus, the re quired area is a sum of these two areas

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Now let us solve some actual problems using our knowledge of

reading the standard normal table values.

Fig 7.6

Example 17:

The heights of 1000 students in a college are normally distributed

with m ean 160 cm and standard deviation 12 cm. How many students will

be there such that ( i), their heights are greater than 165 cm ( ii) their

heights are between 150 cm and 170 cm, ( iii) heights are less than 145cm.

Ans: Given: N= 1000, = 160,=12

(i)X= 165

z=X

=165 160

12=0.42

P(X> 165) = P(z> 0.42)

The area under the normal curve represented by

z= 0 and z =0.42 is 0.1628

The probability of students with heights greater than 165 cm is 16.28%

and the number of students are 1000 x 0.1628 = 162.8 163

(ii)X1= 150 and X2= 170

1

1Xz

=150 160

12=–0.83

and2

2Xz

=170 160

12= 0.83

The required probability is P(X1XX2)

=P(–0.83z0.83)

=P(–0.83z0) + P(0z0.83)

=P(0z0.83) + P(0z0.83)

= 2P(0z0.83)

From th e table we have the required probability as 2 x 0.2967 = 0.5934

Thus , the number of students whose heights are between 150cm

and 170 cm is

1000 x 0.5934 = 593.4 593.

(iii)X= 145

Xz

=145 160

12=–1.25

The area represented by z–1.25,

Now, the area between z =0 and z= 1.25 is 0.3944

the required area is 0.5 –0.3944 = 0.1056

the number of students with heights less than 145cm are 106.

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Example 18:

If 2.28% of teachers in a State have salary less than Rs. 4,500 and

30.85% teachers have salary greater than Rs. 7,000. Find the average

salary and standard deviation.

Ans: (In such kind of problems we do not find the area but we find the

value of z corresponding to the area known. In simple words, we look

inside the table and find the corresponding zvalue from the first column

and first row.)

Since 2.28% of teachers have salary less than 4500, the remaining

47.72% lie to the right of X= 4500. The area to the right of X= 4500 is

0.5–0.028 = 0.4772. Now look in the table, to find z-value for this is –2.

(As it is to the left of the mean)

4500z

=–2 4500 –=–2 … (1)

Since 30.85% teachers have salary more than Rs. 7000, the area to

the right of X= 7000 is 0.3085. So, the remaining area to the left of X=

7000 is 0.5 –0.3085 = 0.1915

Thez-value corresponding to th is area is 0.5 from the table

7000z

= 0.5 7000 –= 0.5 ... (2)

Solving (1) and (2) we have = 1000 and hence = 6500. Thus, the

mean salary is Rs. 6,500 and standard deviation is Rs. 1000

Exercise

(1)Define ( i)standard normal curve, ( ii) Standard normal variate.

(2)State the properties of standard normal curve.

(3)Using the table of areas under a norm al curve, find the probabilities

of: (a)P(0z1.2), ( b)P(0z2.3), ( c)P(–1z0), (d)P(–1z

1), (e)P(–1.5z0.2), ( f)P(z-1.6), ( g)P(z2).

(4)If a random variate X is normally distributed with mean 45 and

standard deviatio n 12, find the probabilities of the following: ( a)X

30, (b)X55, (c)X60, (d) 35X50

(5)IfXis a normal random variable with mean 14 and standard

deviation 6, find the probabilities of: ( a)X10, ( b)X16, ( c)X

18, (d) 10X20.

(6)If X 1and X 2are two random variates with mea ns 30, 25 and standard

deviations 16, 12 respectively, find P(60Y80), where Y= 3X 1+

2X2.

(7)The heights of 600 students in a college are normally distributed with

mean 154 cm and standard deviation 16 cm. How many students will

be there such that ( i) their heights are between 140 cm and 160 cm,munotes.in

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(ii) their heights are greater than 175 cm, ( iii) heights are less than

135 cm.

(8)The average life of a cartridge is 7 days with standard deviation 1.5

days. If the life e xpectancy of a cartridge shows a normal

distribution, what is the probability that a cartridge functions for

more than 10 days?

(9)The weights of 1000 chocolates of a brand are normally distributed

with mean weight as 15 mg and standard deviation 4mg. Find th e

number of chocolates with weights greater than 20mg.

(10)The marks of 1200 students in a College show a normal distribution

with mean 56 and standard deviation 6. Estimate the number of

students with marks ( i) less than 50, ( ii) greater than 60, ( iii) betwee n

45 and 65.

(11)The mean height of soldiers is 68.22 inches with variance 10.8. If the

heights show a normal distribution, find the number of soldiers out of

a regiment of 1000 whose height is greater than 6 feet.

(12)The marks obtained by students are normally d istributed with mean

65 and variance 25. What is the probability of students getting marks

more than 75?

(13)To pass a physical test for Air force, the height of a cadet should be

at most 162cm. Out of 1400 cadets appeared for the physical test

how many could not clear it, if their average height was 160 cm with

standard deviation 9cm. Assume that the heights of the cadets show a

normal distribution.

(14)A manufacturer knows from his experience that the resistance of

resistors he produces is normally distributed wi th mean 100 ohms

and standard deviation 2 ohms. What is the percentage of resistors

having resistance between 98 and 102 ohms?

(15)The average life of a battery is 150 minutes with standard deviation

14 min. If the average life of battery shows a normal distri bution,

find the probability that a battery works for more than 180 min.

(16)The mean and standard deviation of average monthly salaries of 6000

people are Rs. 20,000 and Rs. 3,500 respectively. Assuming that the

data shows a normal distribution, find the ( i)number of people with

salaries greater than Rs. 25,000, ( ii) salaries between Rs. 15,000 and

Rs. 20, 000 and ( iii) less than Rs. 10,000.

(17)If 1.88% of teachers in a State have salary less than Rs. 3,500 and

4.85% teachers have salary greater than Rs. 8,000. F ind the average

salary and standard deviation.

(18)In an examination marks obtained by students in Mathematics,

Statistics and Economics are normally distributed with average

marks 51, 53 and 46 and standard deviation 15, 12 and 16

respectively. Find the proba bility that the total marks are ( i) 180 and

above, ( ii) less than 90.munotes.in

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(19)The aggregate of students in a FYBSc class of a College is normally

distributed with mean 425 marks and standard deviation 36 marks. A

student is said to pass if he scores 40% of the tot al of 700 marks.

Find the number of students ( i) who have passed, ( ii) whose

aggregate is less than 325 marks.

(20)The average diameter of a needle is measure as 0.1 mm with a

standard deviation of 0.005mm. If 1600 samples are selected, find

the number of nee dles with diameter ( i) less than 0.08mm, ( ii) greater

than 0.2mm and ( iii) between 0.6mm and 0.3mm

(21)The daily sales of a firm are normally distributed with mean Rs. 8000

and standard deviation Rs. 100. What is the probability that on a

certain day sales wil l be less than Rs. 8200 and what is the

percentage of days with sales between Rs. 8050 and Rs. 8250?

(22)1000 light bulbs are installed in new factory, show a normal

distribution with mean life of 120 days and standard deviation of 20

days. How many bulbs will expire in less than 90 days?

(23)Record kept by goods inwards department of a large factory show

that average number of lorries arriving each week is 248. It is known

that the distribution is normal with standard deviation26. If this

pattern of arrival contin ues, what is the percentage of weeks expected

to have the number of arrivals ( i) less than 230 per week, ( ii) more

than 280 per week.

(24)The income distribution workers in a certain factory were found to be

normal with mean Rs. 500 and standard deviation Rs. 50. There were

228 persons with income above Rs. 600. How many workers were

there in all?

(25)The marks obtained by students in a subject showed a normal

distribution with mean 65 and standard deviation 14. If there were

142 students with marks less than 50, f ind the total number of

students.

(26)For a normal distribution 30% items are below 45, 8% are above 64.

Find the mean and variance.

(27)In a normal distribution 10% item are under 35 and 89% under 63.

Find the mean and standard deviation.

(28)In a class, 30% students have marks less than 40, 33% have marks

between 40 and 50 and the remaining have marks above 50. If the

data is normally distributed, find the average marks and standard

deviation.

(29)A STD booth owner has an average balance of Rs. 230 and standard

deviation of Rs. 30. Assuming that the balance with the owner

behave normally, find the proportion of the balance being ( i) less

than Rs. 180 and ( ii) between Rs. 200 and Rs. 250.

(30)In an examination in a college 46% students secured a pass class and

10% students sec ured first class. If the minimum marks for pass classmunotes.in

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and first class are 35 and 60 respectively. Find the average marks

obtained by students.

(31)The life of army shoes is normally distributed with mean 8 months

and standard deviation 2 months. If 5000 pairs of shoes are issued ,

how many pairs would be expected to need replacement after 12

months?

(32)The marks obtained by students in an examination are normally

distributed with mean of 70 and standard deviation 6. If the top 5%

students get grade A and the botto m 25% get grade F, what marks is

the lowest A and the highest F?

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QUESTION PAPER PATTERN

MARKS: -100 TIME: -3 HRS

N.B :

1)All questions are compulsory

2)All question carry equal marks

3)Figures to the right indicate marks to a sub -question.

4)Graphs paper will be supplied on request.

5)Use of non -programmable calculator is allowed.

SECTION -I

Q.1 Attempt any four of the following

(a)5 marks (b)5 marks (c)5 marks

(d)5 marks (e)5 marks 20 marks

Q.2 Attempt any four of the following

(a)5 marks (b)5 marks (c)5 marks

(d)5 marks (e)5 marks 20 marks

SECTION -II

Q.3 Attempt any four of the following

(a)5 marks (b)5 marks (c)5 marks

(d)5 marks (e)5 marks 20 marks

Q.4 Attempt any four of the following

(a)5 marks (b)5 marks (c)5 marks

(d)5 marks (e)5 marks 20 marks

Q.5 Attempt any four of the following

(a)5 marks (b)5 marks (c)5 marks

(d)5 marks (e)5 marks 20 marks

munotes.in