CONTENTS
Unit No. Title Page No.
SEMESTER - I
1. Shares and Mutual Funds 01
2. Permutations and Combinations 16
3. Linear Programming Problems 47
4. Measures of Central Tendency 59
5. Measures of Dispersion 85
6. Elementary Probability Theory 99
7. Statistical Decision Theory 117
munotes.in
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UNIT I
Unit -1
SHARES AND MUTUAL FUNDS
Unit Structure :
1.0 Objectives
1.1 Introduction
1.1.1 Bonus Shares
1.1.2 Splitting of shares
1.2 Mutual Funds
1.3 Systematic Investment Plan (SIP)
1.0OBJECTIVES:
After going through this chapter you will be able to:
Define a share, face value, market value, dividend, equity shares
preferential shares, bonus shares.
Understand the concept of Mutual fund.
Calculate Net Income after considering entry load, dividend,
change in Net Asset Value (N.A.V) and exit lo ad.
Understand the Systematic Investment Plan (S.I.P).
1.1INTRODUCTION
In day -to-day life we hear about shares, share market etc. Here
we will see, exactly what these terms deal with .
When a group of persons plan to establish a company, they form
a company under the companies Act 1956. Now this company is an
established company .The people who establish this company are
called promoters of the company . These promoters can now raise a
certain amount of cap ital to start (run) the company .They divide this
required amount into small parts called shares .
A share is the smallest unit of capital of a company. Stock is the
American term for share. Usually a share is of value Rs. 100 / -or
Rs.50/ -or Rs. 10 / -or Rs. 5/ -or Rs. 2/ -or Rs./ -1 . This value is called
the face value of the share. These shares are sold to the public. (usually
face value is Rs. 10/ -, unless otherwise specified ) .This sale is called
theInitial Public Offer (IPO) of the company.
The company issues sha re certificates to the persons from whom
it accepts the money to raise the capital. Persons who have paid money
to form the capital are called share holders. Now -a-days the shares are
not in the form of paper, but in the electronic dematerialised (De mat)
form, hence the allotment of shares is done directly in the demat
account,without a certificate.munotes.in
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Face value or n ominal value orPar value is the value printed on
the share certificate. Since shares exist in electronic demat form, we
can say that the face value is the value stated in the I.P.O. subscription
form.
The shareholders enjoy the profits (if any) of the company, after
providing for the taxes and other reserve funds . This is called as
dividend .
Types of shares ; Mainly the sh ares are of two types i)Preference
shares andii)Equity shares or common shares or ordinary shares .
i) Preference shares : These shares have a priority over the equity
shares. From the profits made by a company , a dividend at a fixed rate
is paid to them first, before distributing any profit amount t o the equity
shareholders. Also , if and when the company is closed down then
while returning of the capital, these shareholders get a preference.
Again, preference shares are mainly of two types:
a)Cumulative Preference shares: In case of loss or inadequate
profit , The preference shareholders are not paid their fixed rate of
dividend , then the dividend is accumulated in the subsequent years
to these shareholders & is paid preferentially whenever possi ble .
b)Non-cumulative Preference shares: As in the case of cumulative
preference shares, here the unpaid dividends do not accumulate.
ii) Equity shares : These are the shares for whom the dividend and
the return of capital is paid after paying the p reference shareholders.
In case of equity shares , the rate of dividend is not fixed and it is
decided by the Board of Directors .
Share Market
Shareholders are allowed to buy or sell shares like
commodities. Selling or buying a share for a price higher than its face
value is legal. The share prices are allowed to be subject to the market
forces of demand and supply and thus the prices at which shares are
traded can be above or below the face value.
The place at which the share s are bought and sold is called a
share market or stock Exchange and the price at which a share is traded
is called its Market Price (MP) or the Market value. If the market price
of a share is same as its face value, then the share is said to be traded at
Par.
If M.P. is greater tha n face value of a share , then the share is
said to be available at a premium or above par andis called premium
share or above par share.
If M.P. is lower th an face value of a share , then the s hare is
said to be available at a discount or below par & the share is called a
discount share or below par share .munotes.in
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The purchase and sale of shares can take place through
brokerage firms and depositary P articipant (DP). e.g. Sharekhan .com,
Kotak Securities ltd , ICICI direct .com etc. They charge a commission
on the purchase and sale of shares, which is called as a brokerage. The
brokerage is charged as a percentage of the M.P. of the share.
Normally it is below 1%.
1.1.1Bonus Shares
Sometimes, when a company's free reserves are high, it may
choose to capitalize a part of it by converting it into shares. This is
done by issuing bonus shares to existing shareholders. These bonus
shares are issued free of cost. The ratio of bonus sh ares to the existing
shares is fixed.
Getting bonus shares increases the number of shares of
shareholders. But since this applies to all the shareholders in a fixed
ratio, hence the percentage of a shareholder's ownership of the
company remain s same as before.
Now, we will study some examples based on the above concepts :
Example 1
Mr. Prashant invested Rs. 75,375/ -to purchase equit y shares of a
company at market price of Rs. 250 / -through a brokerage firm,
charging 0.5% brokerage. The face value of a share is Rs. 10/ -. How
many shares did Mr. Prashant purchase?
Solution : Brokerage per share = 250 x0.5
100=1.25
cost of purchasing one share = 250+1.25 =251.25
Number of shares purchased =75375
251 .25= 300
Example 2
Mr. Sandeep received Rs. 4,30,272 / -after selling shares of a company
at market price of Rs. 720 / -through Sharekhan Ltd., with brokerage
0.4%.Find the number of shares he sold.
Solution : Brokerage per share = 720 x 0.4= 2.88
100
selling price of a share = 720 -2.88 = 717.12
Number of shares sold =430272
717.12=600
Example 3
Ashus Beauty World ' has issued 60,000 shares of par value ofRs.
10/-each. The company declared a total dividend of Rs. 72,000 / -.
Find the rate of dividend paid by the compa ny.munotes.in
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Solution : Face value of 60,000 shares = 60,000 x 10 = 6,00,000
Rate of Dividend =Total Dividend
Face value of 60,000 sharex100
=72000
600000x 100 = 12
The rate of dividend pa id by the company is 12%
Example 4
The capital of ABC Company consists of Rs. 15 lakh sin 6 %
cumulative preference shares of Rs. 100 each and Rs. 30 lakh sin
equity shares of Rs.10/ -each. The dividends on cumulative preference
shares for earlier year was not paid . This year , the company has t o
distribute profit of Rs . 3 lakh after keeping 20 % as reserve fund. Find
the percentage rate of dividend paid to the equity shareholders.
Solution: Reserve fund =20
100x 300000 = Rs. 60,000/ -
Profit to be distrib uted= 3,00,000 -60,000 = 2,40,000
Annual dividend for 6 % cumulative preference shareholders
=6
100x 1500000 = 90,000
This needs to be paid for 2 years (last year & current year ) as the
preference shares are cumulative & last year's dividend was not paid .
Total Dividend paid to Preference shareholders
= 2 x 90000= 1,80,000/ -
Now , dividend to be distri buted to the equity shareholders
= 2,40,000 -1,80,000 = Rs. 60,000/ -
Rate of dividend =60,000
30,00,000x 100 = 2
The rate of dividend to the equity shareholders is 2 %
Example 5
Mr. Dinesh boug ht some shares of a company which had a face value
of Rs.100 / -. The company declared a dividend of 15 % but Mr.
Dinesh's rate of return on investment was only 12% . At what market
price did he purchase the shares ? There was no brokerage involved.
Solution:
Dividend on oneshare =Rate of Dividend
100x face value of one share
=15x 100 = Rs. 15/ -
Rate of Return on investment =Dividend on one share
purchase price of 1 sharex 100munotes.in
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12 =15
purchase of 1 sharex 100
purchase pric e of 1 share =15
12x 100 = 125
Example 6: Comparison of two stocks
Mr. Subu invested Rs. 20,000 / -in Rs. 100/ -shares of company A at
the rate of Rs. 125 /-per share . He received 10 % dividend on these
shares. Mr. Subu also invested Rs. 24,000/ -in Rs. 10/ -shares of
company B at Rs.12/ -per share and he received 15 % dividend. Which
investment is more beneficial?
Solution : Company A
Rate of re turn =Dividend on one share
purchase price of 1 sharex 100 =10
125x 100 =8%
125
Company B
Rate of return =1.5
12x 100 = 12.5 %
Investment in company B is more profitable .
Example 7
Ms. Ashma Mehta bought 300 shares of a company of face value Rs.
100 / -each at a market price of Rs. 240 / -each . After receiving a
dividend at 8 % , she sold the shares at Rs . 256 / -each. Find her rate
of return on investment. There was no brokerage involved.
Solution : Difference in the market price = 256 -240= 16
Dividend on 1 share =Rate of dividend
100x face value of 1 share
=8
100x 100= 8
Rate of Return on Investment
=(Price change) (Dividend on 1 share)
purchase price of 1 share=16 8
240x 100
=2400
240= 10
The rate of return on investment was 10 % .
1.1.2Splitting of shares:
Sometimes companies split the face value of a share & break it
up into sm aller units . For e.g. a Rs. 100 / -share can be split into 10
shares each of face value Rs. 10 / -or a Rs. 10/ -share can be split into
two shares of face value Rs. 5/ -each . Usually this does not affect a
shareholder's wealth . However , it can mak e selling of a part of the
holdings easier.munotes.in
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Example 8
Mr. Joshi purchased 30 shares of Rs. 10/ -each of Medi computers Ltd.
on20th Jan . 2007, at Rs. 36/ -pershare. On 3rd April 2007, the
company decided t split their shares from the face value of Rs. 1 0/-per
share to Rs. 2/ -pershare. On 4th April 2007, the market value of each
share was Rs. 8/ -pershare. Find Mr. Joshi's gain or loss, if he was to
sell the shares on 4th April 2007? (Nobrokerage was involved in the
transaction) .
Solution : On 20 th Jan 2007
Purchase cost of 30 shares = 30 x 36 = 1080 /-
On 3rd April 2007, each Rs. 10/ -share became 5 shares of Rs. 2/ -
each.
No. of shares = 30 x 5 =150
On4thApril 2007, market value of 150 shares was @ Rs. 8 each
= 150 x 8 = 1200
His gain = 1200 -1080 = 120/ -
Example 9
Rahul purchased 500 shares of Rs. 100 of company A at Rs. 700 / -.
After 2 months, he received a di vidend of 25 % . After 6 months , he
also got one bonus share for every 4 share sheld. After 5 months, he
sold all his shares at Rs. 610/ -each. The brokerage was 2% on both ,
purchases & sales.Find his percentage return on the investment.
Solution :For purchase:
Face value = Rs. 100 / -, No. of shares = 500, market price = Rs.700/ -
Dividend = 25 %,brokerage = 2%
Purchase price of one share = 700 +2
100x700= 714
Total purchase = 500 x 714 = Rs.3,57,000/ -
Dividend =25
100of 100 i.e. Rs. 25 / -per share
Total dividend = 500 x25 = Rs. 12,500 / -
Now, bonus shares are 1 for every 4 shares .
No. of bonus shares =1
4x 500 = 125
Total No. of shares = 500 +125 =625
For sales,
No. of shares = 625, market price = 610 , Brokerage 2%
Sale price of one share = 610 -2% of 610 = 597.8
Total sale value = sale price of one share x No. of shares
= 597.8 x 625 = Rs. 3, 73,625/ -
Net profit = sale value + Dividend -purchase value
= 3,73,625 + 12500 -3,57,000munotes.in
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= Rs . 29,125/ -.
% gain =29,125
3,57,000x 100 = 8.16
= 8.16
EXERCISE :
1) Mr. Am ar invested Rs 1,20,480/ -to buy equity shares of a company
at market price of Rs . 480 / -at 0.4 % brokerage. Find the N o. of
shares he purchased.
Ans:250
2)Aditi invested Rs. 19,890 / -to purchase shares of a company with
face value of Rs. 10/-each , at market price of Rs. 130/ -. She received
dividend of 20 % as well Afterwards , she sold these shares at market
price of Rs. 180/ -.She had to pay brokerag e of 2 % for both purchase
and sales of shares. Find her net profit.
Ans:No. of shares =150 , sales = 26460 , Divi dend = 300 , purchase =
19,890, profit= 6870
3) Amol wants to invest some amount in company A or company B ,
by purchasing equity shares o f face value of Rs. 10 / -each , with
market price of R. 360/ -and Rs. 470/ -respectively . The companies
are expected to declare dividends at 20 % and 45% respectively .
Advise him on the choice of shares of company.
Ans: company B is a better choice .
4) Find the percentage gain or loss if 200 shares of face value Rs. 10/ -
were purchased at Rs . 350/ -each and sold later at Rs. 352 / -, the
brokerage being 0.5 % on each of the transaction .
Ans:-0.43 % i.e. a loss of
43 %
5) Find the number of shares if the total dividend at 8% on the shares
with face value Rs.10/ -was Rs. 240. Ans :-
300
1.2MUTUAL FUNDS
In the previous unit shares, we have studied how one can
transact in shares. Now, we will study what are the mutual funds and
how they function.
An investor can invest money directly in shares or he can invest
his money through mutual funds . Mutual funds are managed by large
financial se rvices with a professional team of fund Managers &
research experts.
Mutual fund is a pool of money , drawn from investors .The
amount collected is invested in dif ferent portfolios of securities , by themunotes.in
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fund managers and the profits (returns), proportional to the investment ,
are passed back to the investors.
At a given time, the total value is divided by the total number
of units to get the value of a single unit a given time . This is called Net
Asset Value (NAV).
NAV = Net Assets of the scheme
Total No. of unit s outstanding
or NAV = Total Assets -liabilitie s____
Total No. of units outstanding
There are mixed or hybrid funds which invest in both debt and
equity. The offer docume nts give the guidelines / constraints under
which the fund managers would operate. e.g. investment in equity 80
% to 100 %, investment in money markets 0 % to 20 % etc.
In India , the mutual funds are governed by SEBI ( Securities
and Exchange Board of I ndia ) .There are different companies , called
the ' Fund Houses ' (like SBI or Reliance or HDFC) which float
different mutua l funds. Each such fund is called a 'scheme', e .g. HDFC
has a scheme ' HDFC Tax saver ' etc.
Like IPO of a compan y's share , a mutual fund scheme starts
by having a N.F.O. (New Fund Offer) . Investors can invest by
purchasing Units of the mutual funds .Usually a unit is of Rs. 10/ -. A
share is the smallest unit of a company's capital , whereas in mutual
funds , e ven a fraction of a unit can be purchased after the N.F. O.
Let us study the following example to understand this concept :
Example 10
A mu tual fund 's scheme shows the following on 01/01/2007
Total value of securities
(Equity , Bonds etc.)Rs. 1500 crores
Cash Rs . 100 crores
Liabilities Rs . 200 crores
Total No. of units outstanding Rs. 100 crores
NAV = Rs. 1500 crores + Rs. 100 crores -Rs. 200 crores
100 crores
= Rs. 1400 crores = Rs . 14 per unit .
100 crores
The NAV of a mutu al Fund scheme is calculated anddisclosed to the
publc for evey work ing day . The NAV change s daily. Investors try to
invest when NAV is low and sell the units and get profits when the
NAV is high .munotes.in
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Most mutual fund schemes are not traded at stock marke t.
Thus, investor purchases as well as sells the units to AMC i.e. Asset
manage ment company, This sale is called redemption of units.
Basically funds are of two types : -
1) close ended funds 2) open ended funds .
1)Close ended mutual funds : -These are offered with a fixed date of
maturity and can be purchased fr om mutual fund companies during a
specific period . The investor can get the amount after expiry date of
the fund . If an investor wants to exit before the maturity date , he can
sell the units on the stock exchange at a discount or through a buy -
back option by the fund .
2)Open ended funds : These have no fixed date of maturity and the
units can be sold or repurchased at any time .The no. of units & its
capital changes daily .
Entry load & Exit load : Some mutual fund schemes collect a
charge when investors purchase or redeem units . These are usually
percentage of NAV . The charge levied while purchasing a unit is
called the entry load & the charge collected on redemption is called
exit load .
Usually , the debt funds have not loads . When there are no charges
while purchasing or selling of units , these funds are called No Load
Funds .
Mutual Funds can be broadly categorised into two types : 'Dividend '
funds which offer a dividend and 'Growth ' funds which do not offer a
divid end .
In mutual unds , the dividend given has no direct relation to the
profit earned . The mutual fund invests the money in different shares
that may or may not give a dividend at different times & different rates
. The fund manager may at any arbi trary point , decide to give a part
of the unit s' value back to the investors . This is called dividend .
For a growth fund , the NAV does not com edown due to dividends .
It moves up or down purely on the basis of the gains or losses of the
securiti es that the fund has invested in.
For a growth fund , the gains per unit are purely from the difference
between the redemption price and the purchase price i.e. the total gain
is purely the capital gain . For a dividend fund , the total gain is the
addition of the capital gain & the dividend .
Capital gain = Amount received aft er redemption -Amount invested .
Rate of Return = Change in NAV + Dividend x 100
NAV at the beginning of the period
(This is for a given period) .munotes.in
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Annualised rate of Return = Rate of Return x 365
n
where n is the number of days .
Some important Terms :
i) Assets : -It refers to market value of investment of M.F. in
government securities , bonds etc. , its receivable s , accrued income &
other assets .
ii) liabilities : -It includes all expenses like accrued expenses , payables
and other liabilities for the M.F. scheme .
iii) Net Assets : -Total Assets -liabilities
iv) The valuation Date is the date on which NAV is calculated .
Example 11
Mr. Deore invested Rs. 25,000/ -to purchase 2,500 uits of ICICI MF -
B plan on 4th April 2007 . He decided to sell t he units on 14th Nov.
2007 at NAV of Rs. 16.4 / -. The exit load was 2.5 % . Find his profit
(Calculations are upto 2 decimal points )
Solution :
No. of units =2500 , purchase cost = Rs. 25,000/ -
NAV on the date of sale = RS. 16.4/ -, exit load = 2.5%= of 16.4 =
0.41
selling price of 1 unit = 16.4 -0.41 = 15.99
sale value = 2500 x 15.99
= Rs. 39,975/ -
Profit = 39,975 -25 ,000
= Rs. 14,975 .
Example 12
Ragini invested Rs. 94,070/ -in mutual Fund when NAV was Rs. 460 / -
with entry load of 2.25 % . She received a dividend of Rs. 5/ -per unit .
She, later sold all units of fund with an exit load of 0.5 % . If h er gain
was Rs. 1654/ -, find NAV at which she sold the units .
(Calculations are upto 2 decimal points)
Solution : purchase price of one unit = 460 + 2.25% of 460
= 460 +10.35 = 470.35
No. of units purchased = 94,070 = 200
470.35
Total dividend = 200 x5 = 1000
Gain = Profit + Divi dend
1654 = Profit + 1000
Profit = 1654 -1000= 654munotes.in
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While selling l et NAV of one unit be y
sale price of one unit = NAV -exit load
= y-0.5% of y
= 0.995 y
sale price of 200 units = 200 x 0.995 y= 199 y
Also , profit = Total sale -Total pur chase
654 = 199y -94,070
199y = 654 + 94,070
199y = 94724
y= 476
NAV at which she so ld units = Rs. 476/ -.
Example 13
If a mutual fund had NAV of Rs. 28 / -at the beginning of the year and
Rs. 38/ -at the end of the year , find the absolute change and the
percentage change in NAV during the ye ar .
Solution : NAV at the beginning = Rs. 28/ -
NAV at the end = Rs. 38/ -
Absolute change inNAV = in 38 -28 = Rs. 10/ -
% change = Absolute change x 100 = 10x100 = 35.71 %
NAV at the beginning 28
Example 14
If NAV was Rs. 72/ -at the end of the year , with 12.5 % increase
during the year , find NAV at the beginning of the year .
Solution : Let 'x' be the NAV at the beginning of the year .
Absolute chang e in NAV = 12.5 % of x = 12.5 x x = 0.125 x
100
NAV at the end of the year = x + 0.125 x = 1.125 x
1.125 x = 72
x =72
1.125
= 64
NAV's initial value was Rs. 64 / -.
Example 15
Rohit purchased some units in open end equity fund at R s. 16/ -. The
fund distributed interim dividend of Rs. 5/ -per unit , and the NAV of
the fund at the end of the year was Rs. 25/ -. Find the total percentage
return . (Calculations are upto 2 decimal points)
Solution : Total gain = change in NAV + Dividen d
= (25 -16) + 5
= 9+5
= 14
Total percentage gain = Total gain x 100
NAV at the begin ningmunotes.in
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=14x 100 = 87.5 %
16
Example 16
Mr.Hosur purchased some units in open -end fund at Rs. 30/ -and its
NAV after 18 months was Rs. 45/ -. Find the annualised change in
NAV as a percentage .
Solution : change in NAV for 18 months = 45 -30 = Rs. 15/ -
annualised change = change in NAV x 12 x 100
NAV at beginni ng No. of months
=15x12x 100
30 18
= 33.33 %
Check your progress :
1) Mr. Kamble purchased 586. 909 units of Kot akcash plus retail
Growth on 1st June 2007 when the NAV was RS. 20.4461. Its NAV as
on 3rdDecember, 2007 was Rs. 21.1960/ -.The fund has neither entry
load nor an exit load. Find the amount invested on 1st June 2007 and
the value of Mr. Kamble's inves tment on 3rdDecember 2007 .
Ans .12,000 , 12440.12 .
2) Ms . Kannan purchased 113.151 units of 'FT India Prima Plus' on
9th April 2007 and redeemed all the units on 7th Aug 2007 when the
NAV was Rs. 35.5573 . The entry load was 2.25 % and the exit lo ad
was 1 % . If she gained Rs. 483.11 , find the NAV on 9th April 2007
.(Calculations are upto 2 decimal points)
Ans . 30.2514
3) Mr. Pandit invested Rs. 10,000/ -in Birla Sunlife Equity Fund -
Divjdend plan ' on 10/07/2007 , when the NAV was Rs. 78.04 ,and
redeemed all the units on 12/11/2007 when the NAV was Rs. 84.54 . In
the meanwhile , on 31/08/2007 , she had received a dividend @ Rs. 10
per unit . Find her total gain and the rat e of retu rn considering loads as
follows:
The entry load was 2 .25 % and the exit load was 0.5 % The number of
units were calculated correct upto 3 decimal places.
Ans . Total gain = Rs . 1794.46 ,Rate of return =
17.94%
4) Given the following information , calculate NAV of the mutual
fund : -
No. of units =15000
Market value of investments in Govt . securities = Rs. 20 lakhs
Market value of investments in corporate Bonds = Rs. 25 lakhs
Other Assets of the fund = Rs. 15 lakhs
Liabilities of the fund = 6 lakhs
Ans . Rs. 360/ -.
5)Mumtaz purchased 1200 units of TATA BIG Bond -G Rs. 12,000
/-on 14th April 2007 . She sold her units on 9th Dec 2007 at NAV ofmunotes.in
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Rs. 15.36/ -. The short term gain tax (STGT) was 10% of the profit .
Find her net profit . (Calculations upto 2 decimal points )
Ans . profit = 6432 , STGT = 643.2 , Net profit = 5788.8
(profit -STGT) .
1.3SYSTEMATIC INVESTMENT PLAN (SIP)
In SIP an investor invests a fixed amount (e.g. say 1000/ -) every
month on a fixed date (e.g. 1st of every month ) . In general the
minimum amount is Rs. 1000/ -per month , in diversified equity
schemes . It can be even Rs. 500/ -as well in ELSS schemes . If this is
done for many months , then each time units are purchased at a
different NAV . Over a period of few months, an investor gets the
benefit of a phenomenon called 'Rupee cost Averaging' .
Rupee -cost-averaging : -If NAV increases , the no. of units
decreases & if NAV decreases , the no. of units purchased increases .
Thus on the whole , it lowers the average cost of units because
indirectly ,the investor buys more units when NAV prices are low &
he buys less units , when NAV prices are high . It is called Rupee -cost-
Averaging .
Consider the following example : -
Mr. Shaikh keeps Rs. 5000/ -on 3rd of every month for 4 months as
follows : -(Calculations are correct to 2 points of decimal)
Month Amount (in Rs.) NAV No. of units he gets
1 5000 109.48 5000/109.48=45.67
2 5000 112.36 5000/112.36=44.50
3 5000 108.14 5000/108.14 =46.24
4 5000 105.62 5000/105.62=47.34
Total 20,000 183.75
Avg price of units = 20,000 / 183.75 = 108.84
If Mr. Shaikh would have invested the entire amount of Rs. 20 ,000/ -
0n 3rd of first month only , with NAV Rs. 109.48/ -, the no. of units
purchased would have been 20,000/ 109 .48 = 182.68
Thus he gained more units and average price of units also was
Rs.108.84 instead of Rs.109.48 which was NAV on 3rd of the first
month
If SIP is followed for a long period of time , it can create wealth to
meet a person's future needs like housing , higher education etc .
Now , we will study the following examples to understand SIP .munotes.in
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Example 17
Mr. Pat il invested in a SIP of a M.F. , a fixed sum of Rs. 10,000/ -on
5th of every month , for 4 months . The NAV on these dates were Rs.
34.26 , 46.12 , 39.34 and 41.85 . The entry load was 2.25 % through
out the period . Find the average price , including the entry load , using
the Rupee -cost-Averaging method .How does it compare with the
Arithmetic mean of the prices ? (Calculations are correct to 4 digits
decimal)
Solution :
Month NAV Entry load =
2.25%Total price No.ofunits =1000 /
Total price
1 34.26 00 0.7708 35.0308 285.4627
2 46.12 00 1.0377 47.1577 212.0544
3 39.34 00 0.8851 40.2252 248.6006
4 41.85 00 0.9141 42.7916 233.6906
TOTAL 165.2053 979.8083
By using Rupee -cost-Averaging method : -
Avg Price = Total amount
Total No. of units
=40,000 = 40 .8243
979.8083
A.M. of price = Total price =165.2053 = 41.3013
4 4
Avg. pric e using Rupee -cost-Averaging method is less than
A.M. of prices .
Example 18
Mr. Desai invested Rs. 5000/ -on 1st of every month for 5 months in a
SIP of a M.F. with NAV's as 48.15 , 52.83 ,41.28,35.44 & 32.65
respectively . There was no entry load cha rged . Find the average price
, Mr. Desai paid using the Rupee -cost-Averaging method . After 6
months ,he sold all his units , when NAV was Rs. 51.64 with
contingent deferred sales charge (CDSC) as 2.25 % . Find his net gain.
(Calculations are correct to 2 digits decimal)
Solution : consider the following table : -
Month Amount (in Rs.) NAV No. of units
1 5000 48.15 5000/48.15=103.84
2 5000 52.83 5000/52.83=94.64
3 5000 41.28 5000/41.28=121.12
4 5000 35.44 5000/35.44=141.08
5 5000 32.65 5000/32.65=153.14
TOTAL 25000 613.82
Avg . price of units = 25000 =40.73
613.82 0munotes.in
15
For selling :
selling price of one unit = 51.64 -2.25% of 51.64 = 50.48
Total sales = 50.48 x 613.82= 30,991.77
Net gain = 30,991.77 -25,000 = Rs. 599 1.77 .
Check your progress :
1) Mr. Thomas started a SIP in 'HDFC long term advantage Fund ' . On
the 10th July , Aug and Sept 2007 he invested Rs. 1000/ -each at the
NAVs Rs. 44.100 , Rs. 43.761/ -,s. 45.455 respectively . The entry load
was 2.25% . Find his average acquisition cost per unit upto 3 decimal
places . (Calculations are up to 3 decimal points) .
Ans. Rs. 45.427/ -.
2)Maneesha d Rs. 20,000/ -on 2nd of every month for 5 onths in a
SIP of a mutual fund , with NAVs as Rs. 53.12 , Rs. 56.2 6 , Rs. 48.86
,Rs.50.44 and Rs. 54.62 respectively . The entry load was 2.25 %
throughout this period .Find average price , including the entry load ,
using the Rupee -cost -Averaging method and compare it with
Arithmetic mean of prices .
(Calculate up to 2 decimal points )
Ans . 53.70 , 53.84 .
munotes.in
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Unit -2
PERMUTATIONS AND COMBINATIONS
Unit Structures :
2.1 Introduction
2.2 Prerequisites and Terminology
2.3 Permutation
2.4 Formula to compute permutations of ndifferent objects
2.5 Permutations of nobjects not all different
2.6 Combination
2.1INTRODUCTION
We often hear people saying that ‘probably it will rain today’; ‘it is
likely that India will win the match against Australia on this pitch’; ‘the
chance of passing the CET is 10% only’ and so on. The words ‘probably’,
‘by chance’ or ‘likely ’ are statistical terms but are very commonly used by
all of us. It is very natural that people are interested to know about the
possibility that something happens. People interested in sports are eager to
know, the possibility of their team to win a game; political activists want
to be sure their chances of winning an election, meteorological department
would like to know about the weather, an economists may want to know
the chance that sales will increase if the price of a commodity is decreased
etc. Also it becomes necessary sometimes to know in how many different
ways a particular event may happen.
All these calculations can be broadly classified into two types
namely permutations andcombinations .
2.2PREREQUISITES AND TERMINOLOGY
Before going i nto the detailed study of these two methods, we shall
discuss some of the prerequisites which are useful in understanding the
concepts and formulae related to them.
Factorial:
The factorial of a natural number nis defined as the product of all
numbers fr om 1to n. It is denoted as n!
For example :
Factorial of 3 i.e.3! = 3 x 2 x 1 = 6 ( This is read as 3 factorial equal to 6 )munotes.in
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Similarly, 5! = 5 x 4 x 3 x 2 x 1 = 120
In general, n! =n(n–1)(n–2)(n–3)……3 x 2 x 1
This formula gives a recursive relation :n! =n(n–1)!
For example :
6! = 6 x 5 x 4 x3 x 2 x 1 = 6 x 5!
Remark : We define 0! = 1
Fundamental Principal of Counting :
If there are mways doing one thing and nways of doing another thing
then the total number of ways of doing both the things t ogether is mn.
2.3PERMUTATION
Apermutation ofnobjects is an arrangement of some of these (or
all) objects in a definite order.
The order in which the arrangement is done is important in
permutations.
Example 1:In how many different ways can thr ee friends Mitesh, Ritesh
and Paresh stand for a group photograph?
Ans:Let us denote these friends by their first alphabet M, R and P. The
three friends can be arranged as shown below:
M R P M P R R M P R P M P M R P RM
The number of ways is 6.
One should not be satisfied with this answer. The question that
should come to our mind is how did we arrange them so? Well, if we
observe the arrangement again, it can be seen that first M’s place was
fixed and the remaining two were arranged. This step was repeated again
for R and P. In terms of permutation what we did was arranging 3 objects
amongst themselves. The next question should be what if there are 10
friends? Can you write down their different arrangements explicitly as
above?
2.4FORMULA TO COMPU TE PERMUTATIONS OF N
DIFFERENT OBJECTS
nPr:The number of ways of arranging robjects out of nobjects is
denoted bynPrand is calculated by the formula:nPr=!
( )!n
n r
For e.g:4
24! 4 x 3 x 2 x 112(4 2)! 2 x 1P munotes.in
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Remark : The number of way s of arranging all nobjects is thus
! !
( )! 0!n
nn nPn n =n!
Now if we go back to the first example, where we had to arrange 3 friends,
then the number of ways of their arrangement using the above formula is
3! 6,which is the same answer what we had got by arranging them
explicitly.
Example 2:In how many ways can 6 people be photographed, if only 4
can be seated at a time?
Ans:Here n= 6 and r= 4.
no. of ways of arranging 4 out of 6 people =6
46!
(6 4)!P=
6 x 5 x 4 x 3 x 2 x 1
2 x 1= 720
Example 3:If3 542n nP P then find the value of n.
Ans:5!
5 !n nPnand3!
3 !n nPn
Since3 542n nP P
! !425 ! 3 !n n
n n
3 !425 !n
n
i.e.
3 ( 4)( 5)!425 !n n n
n
(n–3)(n–4) = 42
n2–7n+ 12 = 42
n2–7n–30 = 0
(n–10)(n+ 3) = 0
n= 10, –3
Asn≠ –3n= 10
Example 4:Show that1
1 .n n
r r P n P
Ans: Consider RH.S. =1
1 .n
r n P
=n. ( 1)!
[ 1) ( 1 ]!n
n r
= ( 1)!
1 1 !n n
n r
=!
!n
n r=n
rP= L.H.S.
Example 5:In how many different ways can a 4 digit number be formed
from the numbers 1, 2, 3, …, 9, with no digit being repeated?
Ans: Since no repetition is allowed, the number of ways of forming a four
digit number from the given 9 d igits ismunotes.in
19
9
4P=9!
9 4 !=9 x 8 x 7 x 6 x 5!
5!= 3024.
Example 6: In how many different ways can the letters of the word
“MATHS” be arranged if no letter is to be repeated in the same word?
Ans: The word “MATH S” consists of four letter M, A, T, H and S.
The required number of ways is arranging 4 objects all at a time.
Thus, the no. of different ways = 4! = 24.
Example 7:Eight candidates are to appear for an interview in a company.
In how many ways can the H R manager schedule the candidates for their
interview? What if two of the eight candidates are not be interviewed?
Ans: The number of ways of arranging 8 candidates is 8! = 40320
If two out the eight are not be arranged, it means to arrange the remaining
six out of 8.
This can be done in8
6Pways.
the number of ways of scheduling the candidates now =8!
8 6 !
=8!
2!=40320
2= 20160
Example 8:Find the number of ways in which 3 books on Economics, 4
books on Mathematics and 2 books on Law can be arranged on a book
shelf so that books of the same subject are together. Find the number of
arrangements if books are to be arranged at random.
Ans: We can consider this problem of arranging three blocks E(books on
Economics), M(books on Mathematics) and L(books on Law), which can
be done in 3! ways.
Now,
3 books in Economics can be arranged amongst themselves in 3! ways,
4 books in Mathematics can be ar ranged amongst themselves in 4! ways,
and 2 books in Law can be arranged amongst themselves in 2! Ways
Thus, by the fundamental principle of counting,
the total number of arrangements = 3! x 3! x 4! x 2! = 1728 ways.
If there is no condition on the posit ion of any book of any subject we may
consider this of arranging 3 + 4 + 2 = 9 books, which can be done in 9!
Ways
i.e. 362880 ways.munotes.in
20
Example 9:In how many ways can 3 boys and 4girls be seated for a
group photograph if (i) no two boys sit together, (ii) no two girls sit
together, (iii)all boys sit together
Ans: (i)no two boys sit together
4 girls can be arranged in 4! = 24 number of ways
In each such arrangement there are five places (marked as X) where boys
can be seated so that no two boys sit togethe r as shown below:
X G X G X G X G X
Now, 3 boys can be seated in these 5 places in5
3Pways = 60 ways
Thus, the total number of arrangements = 24 x 60 = 1440.
(ii)no two girls sit together :
3 boys can be arranged in 3!number of ways
In each such arrangement there are four places (marked as X) where girls
can be seated so that no two girls sit together as shown below:
X B X B X B X
Now, 4 girls can be seated in these 4 places in4
4Pways = 4! ways
Thus, the total number of arrangements = 3! x 4! =144.
(iii)all boys sit together
All boys can be considered as one block and the remaining 4 girls as four
blocks. The required arrangement is of these 1 + 4 = 5 blocks, which can
be done in 5! ways .
The 3 boys can be arranged amongst themselves in 3! ways .
Thus the total number of ways in which 3 boys and 4 girls can be seated
for a group photograph such that all boys are seated together is = 5! x 3! =
720.
Example 10:In how many different ways can the letters of the word
“JOGESHWARI” be arranged such that (i) there is no restriction, (ii) the
word starts with ‘A’, (iii) the word ends with ‘W’, (iv) the word begins
with ‘A’ and ends with ‘W’, (v) the vowels are together, (v i)the letters W,
A, R are neve r together.
Ans: The given word “JOGESHWARI” consists of 10 distinct letters, of
which 4 are vowels (A, E, I, O) and 6 are consonants ( G, H, J, R, S, W)
(i)If no restriction is there then the total number of arrangements = 8! =
40320
(ii)The word starts with ‘A’
Since the first place is fixed for the 10 letter word, it remains to arrange
the remaining 9 letters.
Thus, the total number of ways = 9! = 362880
(iii)The word ends with ‘W’munotes.in
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This is similar to the above problem
(iv)the word begins with ‘A’and ends with ‘W’
Since two letters AandWare fixed, it remains to arrange the remaining 8
letters which can be done in 8! = 40320 ways.
(v)the vowels are together
4 vowels can be considered as one block and the remaining 6 consonants
as 6 blocks.
Thus, the pro blem now reduces to arranging these 7 blocks, which can be
done in 7! = 5040 ways.
The 4 vowels can be arranged amongst themselves in 4! = 24 ways.
Thus by fundamental principle of counting,
the total number of ways = 5040 x 24 = 120960.
(vi)the letters W, A, R are never together
The required number of ways = total number of all arrangements –number
of ways of arrangements where W, A, R are together.
Now, the total number of ways of arranging 10 letters = 10!
No of ways of arranging the letters W, A, R t ogether = 8! x 3! (from the
above case (v))
Thus the required number of ways = 10! –(8! x 3!) = 3628800 –241920
= 3386880.
2.5Permutations of nobjects not all different
Consider a permutation of 2 white and 4 black hats of the same
type. The n the permutation W 1W2B1B2B3B4is same as W 2W1B1B2B3B4as
we cannot differentiate between the two white hats. Thus, in permutations
where certain numbers of objects are similar, we have to remove the
duplications of same permutations.
Letn1objects be of one kind, n2objects be of second kind,…, nk
objects be of kthkind then the number of distinct permutations of all
objects taken together i.e. n=n1+n2+ … + nkis given by:
1 2 3!
! ! !...... !kn
n n n n
Example 11: A College Librarian Mrs. Akalpita orders 15 books of
which 5 books are of Business Law, 5 books of Business Mathematics, 3
books on Principles of Management and 2 books on F.H.S. In how many
different ways can she arrange them on the shelf?
Ans: Given n= 15, n1= 5,n2= 5,n3= 3 and n4= 2
The number of ways of arranging the 20 books on the shelf
=15!
5!5!3!2!= 7567560munotes.in
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Example 12:In how many different ways can the letters of the word
“MALAYALAM” be arranged horizontally?
Ans: In the given word “MALAYALAM” has n= 9 let ters of which M, A
and L are repeating.
Here ‘M’ is repeated 2 times, n1= 2
‘A’ is repeated 4 times, n2= 4
‘L’ is repeated 2 times, n3= 2
the number of permutations =9!
2!4!2!= 378 0.
Example 13:Find the number of permutations of the letters of the word
“TOMMORROW” such that (i) no two M’s are together, (ii) all the O’s
are not together.
Ans: The given word “TOMMORROW” has 9 letters of which ‘O’, ‘M’
and ‘R’ are repeated.
Now, ‘ O’ is repeated 3 times, n1= 3
‘M’ is repeated 2 times, n2= 2
‘R’ is repeated 2 times, n3= 2
number of permutations of letters of the word “TOMMORROW”
=9!
3!2!2!= 15120 … (1)
(i)no two M ’sare togeth er
We assume the two M’s as one block and the remaining 7 letters as
remaining 7 blocks.
The number of ways of arranging these 7+1 = 8 blocks is 8!, in which the
letters RandOare repeated 2 and 3 times respectively.
Hence, the number of permutations i n which both the M’s are together is
=8!
3!2!= 3360. … (2)
From (1) and (2), the number of distinct permutations in which no two M’s
are together = 15120 –3360 = 11760.
(ii)all the O’s are not together
We consider the thr eeO’s as one block and the remaining 6 letters as one
block.
The number of ways of arranging the 6 + 1 = 7 blocks is 7!, in which the
letters MandRare repeated 2 times each.
Hence, the number of permutations in which all the O’s are together is
=7!
2!2!= 1260 … (3)munotes.in
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From (1) and (3), the number of distinct permutations in which all three
O’s are not together = 15120 –1260 = 13860
Example 14:Find the number of permutations of the letters of the word
“VOWEL” such that the vowels occupy the oddplaces in the arrangement.
Ans: The given word “VOWEL” has 2 vowels ‘E’, ‘O’ and 3 consonants.
There are two odd places viz. 1st, 3rdand 4th.
1st___ 3rd____ 5th
Two vowels can be placed in three places in3
2Pways. For each such way
the 3 consonants can be placed in the remaining three places in 3! ways.
By Fundamental principal of counting, the total number of arrangements is
given by3
2Px 3! = 6 x 6 = 36 ways.
Exerci se
1.Find the number of ways in which 4 boys can be seated for a group
photograph.
2.In how many ways can 3 boys and 2 girls be seated for a photograph.
3.In how many ways can 7 books be arranged on a book shelf?
4.Find the number of ways of making 3 people sit o n 3 chairs?
5.Find the number of ways of making 5 people sit on 3 chairs?
6.In how many ways can 3 boys and 2 girls be seated for a photograph
in two lines with the first line of boys and second line of girls?
7.20 students of a class are seated in three lines for a group photograph.
The first line has 8 chairs; second line has 3 chairs and third 5 chairs.
In how many ways can this be done?
8.In how many ways can 6 people be selected for 3 posts in a company?
9.In how many ways can 10 students be selected for Stude nt’s Council
which consists of only 4 members?
10.Find the number of ways in which a 5 digit number be formed from
the numbers 1, 2, 3, ……, 9 if ( i) no digit is repeated, ( ii) repetition is
allowed.
11.Find the number of ways of forming a 4 digit even number fro m the
digits 1, 2, 3, ……, 9 if no digit is to be repeated?
12.In how many ways can 3 -digit odd number be formed from the digits
1, 2, 3, ……, 9 if ( i) no digit is repeated, ( ii) repetition is allowed.
13.Find the number of ways of arranging the letters of the w ord ( i)
MASK, (ii)MOTHER, ( iii) RATION, ( iv) YES, ( v) BHARTI, ( vi)
GREATFUL .
14.Find the value of nfrom the following:
i.5 36n nP P
ii.6 456n nP P
iii.7 512n nP P
15.Show that1
1n
rP
= (n +1)n
rP.munotes.in
24
16.Find the number of ways in which 2 books on Mathematics, 3 books
on Law and 2 books on Economics can be arranged on a shelf so that
books of the same subject are together. Also find the number of
arrangements if books are to be arranged at random.
17.Find the number of ways in which 4 books on Physics, 3 books on
Chemistry and 2 books on Biology can be arranged on a shelf so that
(i) books of the same subject are together; ( ii) No two books on
Biology are together ; (iii)No two books on Chemi stry are together and
(iv) No two books on Physics are together.
18.In how many ways can 4 boys and two girls be seated if ( i) no two
girls sit together, ( ii) no two boys sit together and ( iii) both the girls sit
together?
19.In how many ways can 3 Africans and 3 Americans be seated so that
(i) atleast two Americans always sit together and ( ii) exactly two
Africans sit together?
20.In how many distinct ways can the letters of the word
“CHEMISTRY” be arranged such that ( i) there is no restriction; ( ii)
the word begin s with a vowel; ( ii) the word begins and ends with a
vowel; ( iii) the vowels are together and ( iv)the letters T,RandYare
never together.
21.In how many distinct ways can the letters of the word “ ANDHERI ”
be arranged such that ( i) the word begins with A; (ii) the word begins
with Aand ends with R; (iii) the word begins with vowel and ends
with vowel and ( iv) the letters N,RandIare never together?
22.A College Librarian Mrs. Parita orders 12 books of which 4 are of
Maths, 3 are of English Literature, 3 boo ks on Sociology and 4 books
on Philosophy. In how many different ways can the books be
arranged?
23.In how many distinct ways can the letters of the word “DISMISS” ?
24.In how many distinct ways can the letters of the word “STATISTICS”
such that ( i) no two I’s are together; ( ii) no two T’s are together ?
25.Find the number of permutations of the letters of the word “VIRAR”
such that the vowels occupy the even places in the arrangement.
2.6COMBINATION
A combination of nobjects is an arrangement of some of these (or
all) objects where the order of arrangement is not considered.
For e.g. : A combination considers the arrangements ‘ab’and‘ba’,
of two letters ‘a’ and ‘ b’as the same.
n
rC:The selection (or rejection) of robjects out of nobjects is denoted
byn
rCand is calculated by the formula:n
rC=!
!( )!n
r n r
For example:5
2C=5!
2!(5 2)!=5!
2!3!=5 x 4 x 3 x 2 x 1
2 x 3 x 2 x 1= 10munotes.in
25
We know that the factorial formula can be written recursively. Hence
we can write the numerator n! asn(n–1)(n–2)….( n–r+ 1)( n–r)! Due
to this recursive formula, the calculation ofn
rCbecomes easier as show
below:
Now,n
rC=( 1)( 2)....( 1)( )!
!( )!n n n n r n r
r n r
=( 1)( 2)....( 1)
!n n n n r
r
For example:10
4C=10 x 9 x 8 x 7
4 x 3 x 2 x 1= 210. This method simplifies and
speeds up the calculations
Results (without proof) :
1.0nC=n
nC= 1
2.n n
r n rC C
3.1n n
r r C C =1n
rC
Exercise
Evaluate the following:
(1)8
1C (2)15
0C (3)6
6C (4)8
3C (5)8
5C
(6)8
2C (7)5
3C (8)6
3C (9)6
4C (10)6
2C
Example 15:In how many ways can 2stude nts be selected for a student’s
committee out of 7 students?
Ans:Here n= 7 and r=2
the no. of ways of selecting 5 out of 7 students =7
2C=7 x 6
2 x 1= 21.
Example 16:A question paper h as 8 questions and only 5 questions are to
be attempted. I how many ways a student can select any 5 questions?
Ans: Here n= 8,r= 5
the required number of ways =8
5C=8
3C=8 x 7 x 6
3 x 2 x 1= 56
Example 17:In how many ways can 2 boys and 2 girls be selected from a
group of 6 boys and 5 girls?
Ans:Here there are two calculations to be made which are
interdependent.
(i) 2 boys out of 6 boys can be selected in6
2Cways and
(ii) 2 girls out of 5 girls can be selected in5
2Cways.
The total number of ways of such a selection =6 5
2 2xC C =6 x 5
2x5 x 4
2
=150 .
Example 18:A book shelf has 10 books of which 6 are of Accounts and
remaining of Management. In how many ways can a person select 3 books
on Accounts and 1 book on Management?munotes.in
26
Ans: There are 6 books of Accounts and 10 –6 = 4 books on
Management.
3 books out of 6 books on Accounts can be selected in6
3C= 20 ways.
1 book out of 4 books on Management can be selected in4
1C= 4 ways.
By fundamental principal of counting,
the total number of ways is 20 x 4 = 80.
Example 19:A Committee of 6 people is to be formed from a Staff of 4
Managers, 6 Officers and 2 peons. Find the number of distinct committees
in which there are: ( i) 2 persons from each category; ( ii) no peons; ( iii)
exactly 2 Managers and ( iv) atleast 2 Manager s.
Ans:
(i) 2 persons from each category can be selected in
4 6 2
2 2 2x xC C C = 90 ways.
(ii) No peons to be selected means the selection of 6 people is to be done
from remaining 4 + 6 = 10 people.
This can be done in10
6C=10
4C=10 x 9 x 8 x 7
4 x 3 x 2 x 1= 210 ways.
(iii) Exactly 2 Managers out of 4 can be selected in4
2C= 6 ways.
The remaining 4 persons in the committee are to be selected from the
remaining 6 + 2 = 8 persons , which can be done in8
4C=8 x 7 x 6 x 5
4 x 3 x 2 x 1= 70
ways.
The total number of ways = 6 x 70 = 420
(iv) Atleast two managers can be selected in the following ways:
Managers Others
Selection I 2 out of 4 4 out of 8
Selection II 3 out of 4 3 out of 8
Selection III 4 out of 4 2 out of 8
Selection I can be done in4 8
2 4xC C =420 ways
Selection II can be done in4 8
3 3xC C = 192 ways
Selection II can be done in4 8
4 2xC C = 28 ways
Thus, the total number of ways = 420 + 192 + 28 = 640.
Exercise
1.In how many ways can 6 books be selected out of 10 books?
2.In how many ways can 4 boys be selected out of 7 boys?
3.A question paper contains two sections. Section I consists of 4
questions of which two are to be attempted and Section II consists of
5 questions of which 3 are to be attempted. Find the number of ways
of attempting the questions in the paper.
4.There are 6 bulbs of which 3 are defective are to be put in two sockets
in a room. Find the number of ways in which the room is lighted.
5.There are 8 books on History and 4 books on Geography. In how
many ways can 4 books on History and 3 books on Geography be
selected?munotes.in
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6.A box contains 4 white and 5 black balls. How many selections of 2
balls can be made so that ( i) one ball of each color is selected; ( ii)
both balls are white; ( iii) no white ball is selected.
7.A box contains 6 blue, 4 green and 2 white balls. How many
selections of three balls can be made so that ( i)one ball of each color
is selected; ( ii) atleast one white ball is selected and ( iii) no white ball
is selected.
8.10 candidates appear for an interview. The selection committee has
time to interview only 6 candidates. Find the number of ways of
selecting t he candidates.
9.10 candidates appear for an interview. 2 candidates are disqualified as
they did not bring necessary documents and the selection panel has
time to interview only 5 people. In how many ways can this be done?
10.A committee consisting of 4 men an d 3 women is to be formed from 7
men and 6 women. Find the number of selections in which ( i) a
particular man is selected; ( ii) a particular women is not selected; ( iii)
a particular man is selected and a particular is rejected.
11.A College has 8 professors and 5 lecturers. In how many ways can a
committee of 5 teachers be formed such that it consists of ( i) two
professors, ( ii)two lecturers, ( iii) atleast two professors, ( iv) atleast
two lecturers, ( v)atleast two professors and lecturers and ( vi) no
lecturer .
12.A Cultural Committee of 8 persons from 6 men and 5 women is to be
constituted in a Society. Find the number of distinct committees if it
should consist of ( i) atleast 3 women, ( ii) exactly 2 women and ( iii)
atmost 4 men.
13.A question paper has three sectio ns with each consisting of 4
questions. A student has to attempt a total of 5 questions with atleast
one from each section. Find the number of ways in which this can be
done.
14.A case is under discussion in front of 5 judges. Find the number of
ways in which the judgment is given with a majority.
15.The Mumbai Royal Cricket Club has 22 players of which 3 are wicket
keepers, 6 are fast bowlers, 2 are spinners and 3 all rounder’s. In how
many ways can a team be formed if it should include ( i) one wicket
keeper, 2 fast bowlers and one all rounder; ( ii) 2 fast bowlers, 1
spinner and 1 wicket keeper; ( iii) atleast one spinner, ( iv) atleast one
spinner and one all rounder.
munotes.in
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UNIT II
Unit -3
LINEAR PROGRAMMING PROBLEM S
Unit Structure :
3.0 Objective
3.1 Introduction
3.2 Meaning of Linear Programming Problem
3.3 Formulation of Linear Programming Problem
3.4 Sketching of Graphs
3.4.1 Graph of a linear equation
3.4.2 Graphs of Linear Inequalities
3.5 Solution of L.P.P. by graphical method
3.0OBJECTIVE
From this chapter students should learn Introduction, meaning
of linear programming problem, formulation of linear programming
problem, some examples on formulation, Sketchi ng of graphs of linear
equation and linear inequalities, and solving of linear programming
problem graphically.
3.1INTRODUCTION
Planning is the heart and the soul of any project, be it a
business empire or a simple task required to be done by an indivi dual.
However, we will be discussing planning w.r.t. production houses
here.
Every organization, uses labour, machine, money, materials,
time etc. These are called resources. As one cannot have an unlimited
supply of resources, there is always an upp er limit on these resources.
Therefore, the management has to plan carefully and systematically to
use these resources, so as to get the maximum profit at a minimum
cost. This is Basic Principle of running any business successfully.
Such a planning is c alled “Programming the strategies”. This is done
by writing a management problem as a mathematical model and then
solve it scientifically.
Thus, a programming problem consists of Business problems
where one faces several limitations causing restrictio n. One has to
remain within the frame –work of these restriction and optimize
(maximize or minimize, as the case may be) his goals. The strategies
of doing itsuccessfully, is called solving a programming problem.munotes.in
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We will be learning Linear Programmin g in the chapter, which
is the most widely used technique in Production Planning.
3.2MEANING OF LINEAR PROGRAMMING
PROBLEM
As the name suggest, a Linear Programming problem is the
problem of maximizin gor minimizing a linear function, subject to
linea r constraints.
Consider a general Programming Problem with a certain goal.
Obviously, there are restrictions also.
i.If the restrictions , when express edmathematically, are in the form
of Linear inequalities, the programming problem is called a Linear
Programming problem (L.P.P)
The restrictions are also called constraints.
ii.If the constraints do not have more than two variables, the L.P.P.
can be solved graphically.
iii.The goals when written mathematically are called objective
function.
We will be solving only those linear programming problems
having only one objective to be achieved at a time. We will be
required to optimize (i.e. maximize or minimize) this objective
function.
For example: if the variable is the profit, then we would like to
maximize i t, but if the variable is the cost or expenditure, we would
naturally wish to minimize it. Hence, solving a Linear Programming
Problem means, optimizing the given objective function within the
given constraints.
Our first job will be to transform the ma nagement problems
into appropriate mathematical modules.
3.3FORMULATION OF LINEAR PROGRAMMING
PROBLEM
The easiest way to learn this concept is with the help of
examples. We begin with the following:
EXAMPLE 1:
Amanufacturer produces two types of t oys for children, Flutes
and drums, each of which must be processed through two machines A
and B. The maximum availability of machine A and B per day are 12
and 18 hours respectively. The manufacturing a Flutes requires 4
hours in machine A and 3 hours i n machine B, whereas a drum require
2 hours of machine A and 6 hours of machine B, if the profit for Flute
is Rs. 20 and per drum is Rs. 50, formulate the problem to maximize
the profit.munotes.in
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Solution: Let us suppose that, the manufacture produces x
Flutes and y drums per day. Then tabulating the given data as fallows,
we observe that, if x Flutes and y drums are manufactured per day, he
will require 4x + 2y hours of machine A and 3x + 6y hours of machine
B.
Machine Flutes (x) Drum (y) Maximum ava ilable
A
B4
32
612
18
Since the availability of the machine A and B are not more than
12 and 18 hours respectively, we must have 4x + 2y 12 and 3x + 6y
18.
Note that he may not be able to complete a fluteor a drum in a
day therefore, x or y need not be integers, but x and y can never be
negative, since nobody can m anufacture a negative number of
production .Hence, the condition 0 ,0y x willhave to be taken
Thus, the problem will be written mathematically as
4x + 2y12
3x + 6y18 and
x0,y0
These inequalities are the constrains on the problem.
Now to find the objective function.
Since the profit per Flute and drum is Rs. 20 and Rs. 50
respectively, the objective function would be
Profit z = 20x + 50y, which is to be maximized
Thus Maxmise z = 20x + 50y
Subject to 4x + 2y12
3x + 6y18
x0, y0
EXAMPLE 2:
Three different kinds of food A,B and C are to be considered to
form a weekly diet. The minimum weekly requirements for fats,
carbohydrates and proteins are 12, 30 and 20 units respectively. One
Kg. of food A has 2, 16 and 4 units respectively of these ingredients
and one Kg. of food B has 6, 4 and 3 units respectively whereas one
Kg. of food C has 1, 5and 7 kgs of these ingr edients. If the cost per kg.
of food A is Rs. 75, per kg. and that of food B is Rs. 80 and per kg. of
food C is Rs. 60, construct the problem to minimize the cost.
Solution: If x kg. of food A, y kg. of food B and z kg. of food
C are to be considered for weekly diet, then the data can be represented
by the following tabular form.munotes.in
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A
(x)B
(y)C
(z)Minimum
requirements
Fats
Carbohydrates
Proteins2
16
46
4
31
5
712
30
20
The constraints can be written as
2x + 6y + z12
16x + 4y + 5z 30
4x+ 3y + 7 z20,
x0, y0
Since the cost per kg. of the food A, B and C are given to be Rs
75, Rs. 80 Rs. 60 respectively, the objective function would be:
Cost C = 75x + 80y + 60z, which is to be minimized under the
given constraints.
Thus Minimize C = 75x + 80y + 60z
Subject to 2x + 6y + z12
16x + 4y + 5z 30
4x + 3y + 7z 20,
x0, y0
EXAMPLE 3:
Two types of food packets A and B are available. Each contain
vitamins N 1and N 2. A person need 4 de cigrams of N 1and 12
decigrams of N 2per day. Food packet A contain 2 decigram of vitamin
N1and 4 decigram of vitamin N 2. Food packet B contain 1 decigrams
of vitamin N 1and 4 decigrams of vitamin N 2. Food packed A and B
cost Rs. 15 and Rs. 10 respecti vely. Formulate L.P.P. which will
minim izethe cost.
Solution: Let x = no. of packet of food A. y= no. of packet of food B.
Food packet A cost Rs. 15 and food packet B cost Rs. 10.
Objective function i .e. cost function
Z = 15x + 10y subject to
Side constraints
Food packed A Food packet B Requirement ( )
Vitamin N 1
Vitamin N 22
41
44
12
xand y are no. of packet of food A and B respectively.
x0, y0
Mathematical form
Min Z = 15x + 10y
Subject to 2x + y 4
4x + 4y12
x0, y0munotes.in
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EXAMPLE 4:
A machine is used for producing two product sA and B.
Product A is produced by using 4 units of chemical salt and 2 unit s of
chemical mixture. Product B is produced by using 2 unit chemical salt
and 3 unit sof chemical mixture. Only 100 units of chemical salt and
1500 unit sof chemical mixture are available. The profit on product A
is Rs. 30 and on B is Rs. 20 per unit. Formulate this L.P.P. to
maximize the profit.
Solution: Let x = no. of unit of product A be produce.
y = no. of unit of product B be produce.
Profit on product A is Rs. 30 per unit of chemical mixture are
available.
Product A Product B Availability
Chemical salt 4 2 1000
Chemical mixture 2 3 1500
xand y are numbers of units x0, y0.
Mathematical form
Max z= 30x + 20y
Subject to 4x + 2y1000 (chemical salt)
2x + 3y1500 (chemical
mixture)
x0, y0.
3.4SKETCHING OF GRAPHS :
Before we begin to solve the problem of L.P.P. graphically, we
shall first see how to sketch the graph of
1.Linear equation
2.Linear inequality
3.4.1.Graph of a linear equation:
A linear equation ax + by = c in two variables x and y where a,
b,andc are constants. Here a and b are constant coefficient of x and y
respectively not all zero simultaneously, the graph of the linear
equation ax + by = c represent sa straight line in xy plane, intercepting
the x -axis and y -axis. There are four possibilities.
Case (i): If a = 0, thelinear equation reduces to by=c. The graph of
this equation is a straight line parallel to x -axis and intercepting y -axis
at c/b.munotes.in
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y-axis
bc,0 by = c
(0, 0) x-axis
Particularly ,
Sketching of Graph of 3y = 6
y= 2
y-axis
4-
3-
2- y = 2
1-
(0, 0) 1 2 3 4 5 x -axis
Case (ii): If a≠0, b = 0 and c ≠0. Linear equation reduces to ax = c .
The graph of this equation is a straight line parallel to y -axis
intercepting x -axis at c/a
Y
ax = c
(0, 0)
0'acXmunotes.in
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Partic ularly, Sketching of Graph of 9x = 27
x=3
Y
4-
3-
2- x = 3
1-
(0, 0) 1 2 3 4 5 X
Case (iii): If a = 0, b =0, c = 0, t he equation reduces to ax + by = 0.
The graph of this equation is a straight line passing through the origin
as shown in the figure.
y-axis
ax + by = 0
(0,0) x-axis
Particularly, Sketching of Graph of 5x + 2y = 0.
x-intercept and y –intercept are ‘0’munotes.in
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Graph of equation 5x + 2y = 0 is a straight line passing
through the origin.
5-y-axis 5x + 2y = 0
4-
3-
2-
1-
-3-2-1 0 -1 1 2 3 4 5 x -axis
-2
-3
-4
-5
Case (iv): If a≠ 0, b ≠ 0, c ≠ 0, the equation reduces to ax + by = c.
The graph of this equation is shown below.
y-axis
bc,0
0
0'acx–axisx -2 2
y 5 -5
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Particularly, Sketching of Graph of 2x + y = 6
Y
6-B (0, 6)
5-
4- 2x + y = 6
3-
2-
1- A (3, 0)
(0, 0) 1 2 3 45 X
Exercise
Draw the graph of the following linear equation s:
1.x + 3y = 6;
2.2x–y = 3;
3.3x–5y = 8;
4.x= 5;
5.x=-3;
6.y=-2.
7.
Answers :
1 y
4-
3-
2-B (0, 2) x + 3y = 6
1- A (6,0)
(0, 0) 1 2 3 4 5 6 7 8 9 xx y points
0 6 (0,6)
3 0 (3,0)
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2
y
1- A (1.5, 0)
0 1 2 3 4 5 x
-1- 2x–y = 3
-2-
-3-B (0, -3)
-4-
3.
y 3x–5y = 8
1- A (232, 0) = (2.7, 0)
0 1 2 3 4 5 x
-1-
-2-B (0, -1.6)
-3-
-4-
4.
y
x = 5
2-
1-
(0, 0) 1 2 3 4 5 6 x
-1-munotes.in
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5
y
x =-3
2-
1-
-3-2-1 0 1 2 3 4 x
-1
6. y-axis
0 1 2 3 x-axis
-1-
-2- y =-2
3.4.2.Graphs of Linear Inequalities:
ax +by +cis called a linear expression. We have seen that if
ax +by+c= 0, then it is a linear equation and it represents a straight
line, if a,b,care real numbers and a,bare both not zero.
Linear inequalities are of four types:
1.ax +by +c0,
2.ax +by +c0,
3.ax +by +c0,
4.ax +by +c0
These inequalities re present region of the plane, ifa,b,care
real numbers and a,bare not both zero. We are interested in knowing
what these regions are.
First draw the line ax +by +c= 0 on the graph paper. We
know that the plane is now divided into three mutually disjoint set viz.
the line itself and the two half plane sone on each side of the line. Let
us denote by P1, the plane which contain the origin and by P2the half
plane which does not contain the origin.munotes.in
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Now consider the origin (0, 0). Putting x = 0 an d y = 0 in the
linear expression ax +by +c, we get c.We consider two cases.
i.c< 0 and
ii.c> 0.
iii.c = 0
iv.
Case (I ): if c< 0, then the half plane P1which contain the origin
represents the inequality ax +by +cb< 0. Naturally, the half plane P2
which does not contain the origin represents the inequality ax +by +c
> 0
Further, the inequality ax + by + c0is represented
graphically by the union of the half plane P1and the line ax +by +c=
0. Whe reas, the inequality ax +by +c0 is repres ented graphically
by the union of the half plane P2and the line ax +by +c= 0.
Case (II): Ifc> 0, then the half plane P1which contain the origin
represent the in inequality ax +by +c> 0. Naturally, the half plane P2
which does not contain the or igin represents the inequality
ax +by +c< 0
Further, the inequality ax +by +c0 is represented by the
union of the half plane P1and the line ax +by +c= 0. Whereas, the
inequality ax +by +c0 is represented graphically by the union of
the half plane P2and the line ax +by +c= 0.
EXAMPLE 1 :Represent the inequality 2x + 3y 6 graphically.
Solution: Draw the line 2x + 3y = 6 on the graph paper. Consider
the origin (0, 0). Putting x = 0 and y = 0 in the inequality, left hand
side we get 0 < 6. Thus (0, 0) satisfies the inequality 2x + 3y 6.munotes.in
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Hence shaded area including the line will be 2x + 3y ≤6
5–
4–
3 –
P2
P1 2–
1–
-4-3-2-1 0 1 2 3 4 5
-1-
EXAMPLE 2:
Represent the inequality 2x + y5 graphically.
Solution:
Draw the line 2x + y = 5 on the graph paper. Consider
the origin O (0, 0). Putting x = 0 and y = 0 in the left hand side we get
0 > 5. But we want 2x + y 5.
Hence the shaded area will be away from origin above the line.
EXAMPLE 3. Represent the inequality y 4graph ically.
Solution:
(y = 4)
3– y≤4
2–
1–
x’ 0 1 2 3 4 x
y’ (y = 0)x y points
0 2 (0,2)
3 0 (3,0)
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(y = 0) satisfies the inequality.
Region represented by y4 is towards origin. Below the line y = 4.
EXAMPLE 4:Represent the inequality y ≥4 graphically.
y y≥4
4 ( y = 4)
3
2
1
x’ 0 1 2 3 4
x
(y
= 0) y’
y= 0 does not satisfy the inequality .Region represented by
y4 is above the line y = 4
EXAMPLE 5: Repres ent the inequality x 3
y
5– x = 3
4–
3–
2–
1–
x’-4-3-2-1 0 1 2 3 4 5 x
-1-
-2-
-3-
(x = 0) -4-x≤3
y’munotes.in
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O (0, 0), x = 0 satisfies the inequality
Region is to be left side of the line x = 3
EXAMPLE 5:Represent the inequality x ≥2 graphically
Solution: y
5–
4–
3– x≥2
2–
1–
x’-4-3-2-1 0 1 2 3 4 5 x
-1-
-2-
-3-
-4-
y’
x= 0 does not satisfy the inequality x ≥2
Region is to be right side of the line x = 2
EXAMPLE 6: Represent the in equality 3x + 4y ≥12 graphically.
Solution:
y
2- 3x + 4y≥12
3-
2-
1-
x’ (0, 0) 1 2 3 4 5 6 7 8 9
xx y points
0 3 (0,3)
4 0 (4,0)munotes.in
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First draw graph of equation 3x + 4y = 12.
Origin O(0, 0) does not satisfy the inequality 3x + 4y 2.
Region represented by the inequality 3x + 4y does not have
origin as a point.
Exercise
Sketch the graph of the following linear inequalities .
1)x + 3y6
2)x + 3y6
3)2x + 5y10
4)2x + 5y10
5)x2
6)y1
7)x0
8)y0
3.5SOLUTION OF L.P.P. BY GRAPHICAL METHOD
In the earlier section we studied how to formulate L.P.P.
However, we need to find solution for the L.P.P. i.e. we have to find
the values of the variable which will optimize (Maximize or minimize)
the objective function and satisfy all the inequality constraints as well
as non –negativity r estrictions.
Let us revise some concept we studied in the earlier section in
linear inequalities. These concepts will be useful to find solution of
L.P.P. by graphical method.
We studied how to find the points in two dimensional co –
ordinate geometry , which will satisfy the gi ven linear inequality. If
this feasible region is bounded, it is in the form of polyhedron. The set
of point sof feasible region, which is polyhedron is called polyhedral
set. A polyhedral set is called a convex set if a line joining any two
points from the set lies entirely in the set. Hence, the feasible region is
the convex set if it is a polyhedral set and the line joining any two
points from polyhedral set entirely lie in the polyhedral set (feasible
region). In the term inology of L.P.P. we can say that feasible region is
the set of points, which satifies the inequality constraints as well as non
–negative restriction. Such as set of points is called feasible solution
of L.P.P. To find optimal solution of L.P.P. the fol lowing result is
extremely useful.
Result:
If the convex set of feasible solution is a convex polyhedron set, then
at least one of the extreme points giving an optimal solution.
To find solution to L.P.P. graphically by using these results these are
the steps in the method.
Following steps are involved in solving L.P.P. graphically.munotes.in
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The steps involved in this method are:
1)Draw the graph for the inequality restriction.
2)Indicate the area of feasible solution (feasible region)
3)Determine the co -ordinates of a ll points at the corners (points)
of all the feasible region.
4)Find the value of objective function corresponding to all the
solution points determine din (3).
5)Determine the feasible solution which optimizes the value of
the objective function.
Example1: Solve graphically the L.P. Problem.
Maximize z = 9x + 12y,
subject to 2x + 3y 18,
2x + y10,
x0, y0
Solution: For the line 2x + 3y 18 For the line 2x + y 10
10-C(10,0)
9-
8-
7- Feasible region
6-A(0,6)
5-
4- P( 3,4)
3-
2-A
1- D(5,0) B(9,0)
O(0,0) 1 2 3 4 5 6 7 8 9
x1
-1- 2x+ 3y = 18 2x + y = 10
We first find the region, each point of which satisfies all the constraints
by drawing graphs of all the constraints. This is called the region ofx y Points
0 10 (0,10)
5 0 (5,0)x y Points
0 6 (0,6)
9 0 (9,0)
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feasible solution. The region of feasible solution is the quadrilateral
OAPD. The corner point of this region are O, A, PandD.
We find the co -ordinates of P by solving the following equations
simultaneously .
2x + 3y = 18,
2x + y = 10,
2y = 8
y = 4
Substituting value of y in 2x + y = 10, w e get 2x + 4 = 10. x = 3
So the point of intersecting = P (3, 4)
Objective function is Z = 9x + 12y,
To check the value of Z at end points of shaded feasible region O, A, P
and D
At O (0, 0), Z = 9(0) + 12(0) = 0
AtA(0,6), Z = 9(0) +12( 6)=72
At P(3, 4), Z = 9 (3) +12(4) = 75
At D ( 5, 0), Z = 9(5)+12(0) = 45
So, at P ( 3,4), Z is Maximized and the maximum value is 75forx= 3
and y=4.
Example 2:Solve graphically the L.P. Problem.
Maximize Z = 5x 1+ 3x 2
subje ct to the constraints 3x 1+ 5x 215,
5x1+ 2x 210,
x10, x 20.
Solution: For the line 3x 1+ 5x 215 For the line 5x 1+ 2x 2
10
x2
(0, 5) Feasible Region
(0, 3) B C
(0, 0) A(2,0) (5, 0) x1
3x1+ 5x 2= 15
5x1+ 2x 2=10x1 x2 Points
0 5 (0,5)
2 0 (2,0)x1 x2 Points
0 3 (3,0)
5 0 (5,0)munotes.in
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We first find the region, each point of which satisfies all the constraints
by drawing graphs of all the constraints. This is called the region of
feasible solution. The region of feasible solution is the quadrilateral
OACB. The corner point of this region are O, A, B and C.
We find the co -ordinates of C by so lving the following equations
simultaneously.
3x1+ 5x 2= 15 ……….(1) 5
and 5x 1+ 2x 2=10………..(2) 3
15x 1+ 25x 2= 75
15x 1+ 6x 2= 30
-- -
19x 2=45
x2 =1945
substituting x2value in any one of the equation, we get x 1=1920
The objective function is Z = 5x 1+ 3x 2.
We find the value of Z at O, A, B, C.
At O (0,0), Z = 5 × 0 + 3 × 0 = 0
At A (2,0), Z = 5 ×2 + 3 × 0 = 10
AtC
1945,1920, Z =5 ×1920+ 3 ×1945=19235
Z has maximum value19235at C
1945,1920.
i.e. x 1=1920,x2=1945
Example 3:
Maximize Z = 4x 1-2x2.
Subject to x1+ 3x 26
x1+ x 22
x10, x 20.
Solve the above L.P. problem graphically.
Solution:
Consider the given condition as e quality
x1+ 3x 2= 6 ( I )
x1+ x 2= 2 ( II)
For point of intersection, solve the equations simultaneously
x1+ 3x 2= 6
x1+ x 2 = 2
- + -
4x2 = 4
x2 =1
Substituting x 2= 1 in (ii ), x 1= 3
So the point of intersecting = P (3 ,1)
To draw the straight line representing (i), (ii) considermunotes.in
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(i) (ii)
By plotting the point A (0,2), B(6, 0), C( 0,-2) and D(2, 0) and
joining we get straight line AB and CD representing equation (i) and
(ii) and the feasible region is shaded as
3x + 4y≥12
y2
5-
5-Feasible Region
3-
2-A
1- P B
(0, 0) 1 2 3 4 5 6 7 8 9
x1
-1- D
-2-C
Note : To get P, the line CD is extended.
Too check the value of Zat end points of shaded feasible reg ion A, P
and D
At o (0, 0), Z = 4(0) –2(0) = 0
At (0, 2), Z = 4x 1–2x2= 4 (0) -2 (2) = -4
At P(3, 1), Z = 4 (3) –2 (1) = 10
At D (2, 0), Z = 4(2) –2(0) = 8
So, at P (3, 1), Z is Maximized and the maximum value is 10 for x 1= 3
and x2= 1
Exam ple4:
Maximize z = 800x + 100y,
subject to 4x + 6y120,
10x + 3y 0, y0
Solution :Now we solve this problem by the graphical method as
follows: First we draw the graph of the solution set of the linear
inequalities.
4x + 6y120, 10x + 180, x0, y0x1 x2 Points
0 2 (0,2)
6 0 (2,0)x1 x2 Points
0 -2 (0,-2)
2 0 (2,0)
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At each point of this region, we can find the values of the
objective function z = 800x + 100y.
y
60-B (0, 6)
50-
40-
30-
20-B(0,20)
10- A (30,0)
(0, 0) 10 C20 30 40 x
(18, 0)
10x + 3y = 180 4x + 6y = 120
Hence this region is called a feasible region. We want to find
that point at which the objective function z = 800x + 100y has a
maximum value. For this purpose , we us e the following result.
If the feasible region is a polygonal one, then the maximum and
minimum value of the objective function lie at some vertices of the
region.
Now we draw the line AB and CD whose equations are 4x + 6y
= 120 and 10x + 3y = 180 resp ectively. For this purpose, we find their
point of intersection with the co -ordinate axes. It is convenient to from
a table for their co -ordinates as follows:
For the line For the line
4x + 6y = 120 10x + 3y = 180
Shade the feasible region OCPBO by horizontal lines. Its
vertices are O (0, 0), C (18, 0), B (0, 20) and P which is the point of
intersection of the line AB and CD. Hence to get the co -ordina tes of
P, we solve their equation:
4x + 6y = 120 …..(1)
10x + 3y = 180 …..(2)
Multiplying equation (2) by 2 and
subtracting equation (1) from it,
20x + 6y = 360
4x + 6y = 120
- - -
16x = 240x y points
18 0 C(18,0)
0 60 D(0,60)x y points
30 0 A(30,0)
0 20 B(0,20)munotes.in
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x = 15
from (2)
150 + 3y = 1803y = 30y = 10P is (15, 10).
We find the values of the objective function at these vertices:
z (O) = 800 × 0 + 100 × 0 = 0
z(P) = 800 × 15 + 10 0 × 10 = 13000
z (C) = 800 (18) + 100 × 0 = 14400
z (B) = 800 × 0 + 100 × 20 = 2000
z has the maximum value 14400 at the point C (18, 0).
the manufactu rer should produce 18 scooters and 0 bicycles,
in order to have maximum profit of Rs. 14400.
Example 5:
Minimize z = 40x + 37y
Subject to constraints: 10x + 3y 180
2x + 3y60
x0, y0
Solution:
We first draw the line AB and CD who se equation are
10x + 3y = 180 and 2x + 3y = 60.
For the line 10x + 3y = 180 For the line 2x + 3y = 60
The feasible region is shaded in the figure. Its vertices are C
(30, 0), B (0, 60) and E, where E is the point of intersection of the
lines AB and CD.
y
60-B (0, 60)
50-
40-
30-
20-D E(15,10)
10-
(0, 0) 10 20 30 A(30,0)
2x + 3y≥60
10x + 3y ≥180
For the point E, we solve the two equations simultaneously.x y Points
18 0 (18,0)
0 60 (0,60)x y Points
30 0 (30,0)
0 20 (0,20)
munotes.in
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10x + 3y = 180
2x + 3y = 60
-- -
8x = 120
x =15
2x + 3y = 60 gives 30 + 3y = 60,
i.e., 3y = 30, i.e. y = 10
E is (15, 10)
The values of the objective function z = 40x + 37y at these vertices are:
Vertex (x, y) z = 40x + 37y
C (30, 0) z (C) = 40 × 30 + 37 × 0 = 1200
B (0, 60) z (B) = 40 × 0 + 37 × 60 = 220
E (15, 10) z (E) = 40 × 15 + 37 × 10 = 600 + 370 = 970.
z has minimum value 970 at the point E (15, 10), where x = 15
and y = 10.
Example 6:Minimize z = 4x + 2y
Subject to constraints: x + 3y3,2x + y2,x0, y0
Solution:
We first draw the line AB and CD .
For the line x + 3y = 3 For the line 2x + y = 2
The feasible region is shaded in the figure. Its vertices are A
(3, 0), D (0, 2) and E, where E is the point of intersection of the
lines AB and CD.
y
3-B (0, 60)
2-D (0,2)
1-B E(0.6,0.8)
(0, 0) 1 2 3 A(3,0) x
2x + y = 2 x + 3y = 3
To find the point E, we solve the two eq uations(1) and (2)
simultaneously.x y Point
3 0 A(3,0)
0 1 B(0,1)x y Point
1 0 C(1,0)
0 2 D(0,2)
munotes.in
51
2x + 6y = 6
2x + y = 2
-- -
5y = 4
x = 8.054
Substituting y = 0.8 in (1), we get, x + 3 × 0.8 = 3 x + 2.4 = 3 x = 3–2.4 = 0.6
Thus E is (0.6, 0.8)
The values of the objective function z = 4x + 2y at the vertices are
calculated below:
Vertex (x, y) z = 4x+ 2y
C (3, 0) z (A) = 4× 3 + 2× 0 = 12
E (0.6, 0.8) z (E) = 4 × 0.6 + 2 × 0.8 = 2.4 + 1.6 = 4
D(0, 2) z (D) = 4 × 0+ 2 × 2= 4
We can see the minimum value of z is 4, at two vertices E(0.6, 0.8) and
D(0, 2). Thus z is minimum at any point on the line segment ED.
Exercise
Maximize:
1)z = 5x + 10y, subject to
5x + 8y40, 3x + y 12, x0, y0
2)z = 7x + 6y, subject to
2x + 3y13, x + y 5, x0, y0
3)z = 5x + 3y, subject to
2x + y≤27, 3x + 2y ≤48, x≥0, y≥0
4)z = 20x + 30y, subject to
x + y6, 3x + y 12, x0, y0
5)z = 6x + 7y, subject to
2x + 3y12, 2x + y 8, x0, y0
6)z = 30x + 20y, subject to
2x + y20, x + 3y 15, x0, y0
7)z = 20x + 25y, subject to
5x + 2y50, x + y 12, x0, y0
8)z = 90x + 130y, subject to
2x + 3y18, 2x + y 12, x0, y0
Minimize:
1)z = x + 4y, su bject to
x + 3y3, 2x + y 2, x0, y0
2)z = 5x + 2y, subject to
10x + 2y ≥20, 5x +5 y 30, x0, y0
3)z = 9x + 10y, subject to
x + 2y30, 3x + y 30, x0, y0
4)z = 50x + 55y, subject tomunotes.in
52
x + 3y30, 2x + y 20, x0, y0
5)z = 80x + 90y, subject to
6x + 5y300, 2x +3y 120, x0, y0
6)z = 13x + 15y, su bject to
3x + 4y360, 2x +y 100, x0, y0
7)z = 12x + 20y, subject to
x + y7, 5x +2y 20, x0, y0
Answer:
Maximize:
1) Max. value is 50 at the point (0,5)
y
12-D
10-
8-
6-B
4- P1960,1956
2- C A
(0, 0) 1 2 3 4 5 6 7 8 x
5x + 8y =
40
3x + y = 12
2) Max. value is 35 at the point (5,0)
y
6-D
5-B
4- P(2,3)
3-
2-
6- C A
(0, 0) 1 2 3 4 5 6 x
x + y = 5 2x + 3y = 13munotes.in
53
3) Max. value is 75 at the point. (6,15)
y
30-B(0,27)
25-D (0,24)
20-
15- E(6,15)
10-
5- C
(0, 0) 5 10 15 20 x
A(13.5,0) 3x + 2y =48
2x + y = 27
4) Max. v alue is 180 at the point (0,6)
y
12-D(0,12)
10-
8-
6-B
4- E(3,3)
2- C(4,0) A(6,0)
(0, 0) 2 4 6 8 x
x + y = 6
3x + y = 12munotes.in
54
5) Max. value is 32 at the point (3,2)
y8-D(0,8)
7-
6- 2x + y = 8
5-
4-B(0,4)
3-
2- E(3,2) 2x + 3y = 12
1- C (4,0) A(6,0)
(0, 0) 1 2 3 4 5 6 x
6) Max. value is 310 at the point (9,2)
y
30-
25-
20-B(0,20)
15-
10-
5-D(0,5) E(9,2) A(10,0) C(15,0)
(0, 0) 5 10 15 20 x x + 3y =15
2x + y =20munotes.in
55
7) Max. value is 300 at the point (0,12)
y
30-
25-B(0,25)
20- 5x + 2y = 50
15-D(0,12)
10- E
310,326
5- C(12,0)
(0, 0) 5 10 15 20 x
A(10,0) x + y = 12
8) Max. value is 795 the point (4.5, 3)
y
14-
12-D
10- 2x + y = 12
8-
6-B(0,6)
4- E(4.5,3)
2- 2x + 3y = 18
(0, 0) 2 4 6 8 A 10
x
C(6,0)munotes.in
56
Minimize:
1)Min. value is 3 at the point (3,0)
y
3-
2-D
1-B P
54
53
C A
(0, 0) 1 2 3 x + 3y≥3 x
2x + y≥2
2)Min. value is 15 at the point (1,5)
y
10-B
-
8-
-
6-D
- P(1,5)
4-
- 5x + 5y = 30
2-10x + 2y = 20
-\ A C
(0, 0) 2 4 6 8 A 10
xmunotes.in
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3)Min. value is 174 at the point (6,12)
y
30-D(0,30)
25-3x + y = 30
20-
15-B(0,15)
10- E(6,12) x + 2y = 30
5- C(10,0) A(30,0)
(0, 0) 5 10 15 20 25 30 x
4)Min. value is 740 at the point (6,8)
y
30-
25-
20-D(0,20)
15-2x + y = 20
B(0,10) 10 - E(6,8)
5- C(10,0) A(30,0)
(0, 0) 5 10 15 20 25 30 x x + 3y = 30
5)Min. val ue is 4350 at the point (37.5,15)
y
60-B(0,60)
50-
40-D(0,40)
30-
20- 2x + 3y = 120 E(37.5,15)
10- A(50,0) C(60,0)
(0, 0) 10 20 30 40 50 60 x
6x + 5y = 300munotes.in
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6)Min. value is 1364 at the point (8,84)
y
120-
100-D(0,100)
80-E(8,84)
60-
40-
20- C A
(0, 0) 20 40 60 80 100 120 x
3x + 4y = 360 2x + y = 100
7)Min. value is 84 at the point (7,0)
y
10-D
9-
8-
7-B
6-
5- P(2,5)
4-
3-
2-
1- C A
(0, 0) 1 2 3 4 5 6 7
x
5x + 2y = 20 x + y = 7
munotes.in
59
UNIT I II
Unit -4
MEASURES OF CENTRAL TENDENCY
Unit Structures :
4.0 Objectives
4.1 Introduction
4.2 Requisites foran Ideal Average
4.3 Types of Averages
4.4 Arithmetic Mean (A.M)
4.5 Weighted Arithmetic Mean
4.6 Combined Mean
4.7 Median
4.8 Quartiles
4.9 Mode
4.10 Graphical Location of Mode
4.0OBJECTIVES
After reading this chapter you will be able to :
1)Define anaverage.
2)Understand different t ypes of average s.
3)Calculate Arithmetic mean andweighted Arithmetic mean.
4)Calculate Median andQuartiles.
5)Calculate Mode by formula.
6)Graphical location of Medi anandMode.
4.1 INTRODUCTION
In the earlier chapter we have discussed the Diagrams and Graphs
which are used for presentation of data. Diagrams furnish only
approximate information. Graphs are more p recise and accurat e than
diagrams. But they cannot beeffectively used for further statistical
analysis. To study the characteristics in detail, the data must be further
analysed.
There is a tendency in almost every s tatistical data that most of the
values concentrate at the centre which is referr ed as ‘central tendency ’.
The typical values which measure the central tendency are called
measures of central tendency. Measures of central tendency are commonlymunotes.in
60
known as ‘Averages.’ They are also known as first order measures.
Averages always lie between the lowest and the lightest observation.
The purpose for computing an average value for a set of
observations is to obtain asingle value which is representative of all the
items an d which the mind can grasp simply and quickly. The single value
is the point or location around which the individual items cluster.
The word average is very commonly used in day to day
conversation. For example, we often talk of average boy in a class,
average height of students, average income, etc. In statistics, the term
average has a different meaning. Averages are very much useful for
describing the distribution in concise manner. The averages are extremely
useful for comparative study of different distribution s.
4.2 REQUISIT ES FOR AN IDEAL AVERAGE
The following are the characteristics which must be satisfied by an
ideal average:
(i) It should be rigidly defined.
(ii) It should be easy to calculate and easy to understand.
(iii)It should be based on all observations.
(iv)It should be suitable for further mathematical treatment.
(v) It should be least affected by the fluctuations of sampling.
(vi) It should not be affected by extreme values.
4.3 TYPES OF AVERAGES
The following are the five types of averages which are commonly
used in practice.
1.Arithmetic Mean or Mean (A.M)
2.Median
3.Mode
4.Geometric Mean (G.M)
5.Harmonic Mean (H.M)
Of these, arithmetic mean, geometric mean and harmonic mean are
called mathematical averages; median and mode are called
positional averages. Her e we shall discuss only the first three of
them in detail, one by one.
4.4 ARITHMETIC MEAN (A.M)
(i)For Simple or U ngrouped data :(where frequencies are not given)
Arithmetic Mean is def ined as the sum of all the observations divided by
the total number of observations in the data and is denoted by x, which is
read as ‘ x-bar’munotes.in
61
i.e. x=sum of all observations
number of observations
In general, if x1x2……., xnare the nobservations of variable x, then the
arithmetic m ean is given by
x=1 2 ........n x x x
n
Ifwe denote the sum x1+x2+………………+ xnas∑x,then
x=x
n
Note : The symbol ∑ is the Greek letter capital sigma and is used in
Mathematics to denote the sum of the values.
Steps :(i) Add together all the values of the variable xand o btain the
total, i.e, ∑x.
(ii) Divide this total by the number of observations.
Example 1:
Find the arithmetic mean for the following data representing marks in six
subjects at the H.S.C examination of a student.
The marks are 74, 89,93,68,85 and 76.
Solution : n=6
x=x
n x=74 89 93 68 85 76
6=485
6
x = 80.83
Example 2:Find arithmetic mean for the following data .
425 , 408 , 441 , 435 , 418
Solution : n= 5and x=x
n
x =425 408 441 435 418
5=2127
5
x= 425.4
(ii) For Grouped data (or) discrete da ta:
(values and frequencies are given )
Ifx1,x2…… xnare the values of the variable xwith corresponding
frequencies ƒ1,ƒ2,………ƒnthen the arithmetic mean of xis given by
x=1 1 2 2
1 2........
.......n n
nf x f x f x
f f f
munotes.in
62
Or
Whe re∑ƒ=ƒ1+ƒ2+…+ƒn= sum of the frequencies
If we denote N =∑ƒ,then x=fx
N
Steps : (1) Multiply the frequency of each row with the variable xand
obtain the total ∑fx.
(2) Divide the total ∑fxby the total frequency.
Example 3:Calculate arithmetic mean for the following data .
Age in years: 11 12 13 14 15 16 17
No.ofStudents: 7 10 16 12 8 11 5
Solution:
Age in years
(x)No.of Students
(ƒ)fx
11 7 77
12 10 120
13 16 208
14 12 168
15 8 120
16 11 176
17 5 85
Total N=69 ∑fx=954
∑ƒ=N=69,∑fx=954
x=fx
N=13.83 years.
Example 4:Calculate the average bonus paid per member from the
following data :
Bonus (in Rs.) 40 50 60 70 80 90 100
No. of persons 2 5 7 6 4 8 3
Solution :
Bonus (in Rs) x No. of persons ( ƒ) fx
40 2 80
50 5 250
60 7 420
70 6 420x=fx
f
munotes.in
63
80 4 320
90 8 720
100 3 300
Total N= 35 ∑fx= 2510
Now, N= 35 and∑fx= 2510
x=fx
N=2510
35=Rs.71.71
(iii)Grouped data (class intervals and frequencies are given )
Formula : x=fx
Nwhere N = ƒandx= Mid -point of t he class
intervals
Steps : (1) Obtain the mid -point of each class interval
( Mid -point = lower limit + upper limit )
2
(2) Multiply these mid -points by the respective f requency of
each class interval and obtain the total ∑fx.
(3) Divide the total obtained by step (2) by the total frequency N.
Example 5:
Find the arithmetic mean for the following data representing marks of 60
students .
Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
No.ofStudents 8 15 13 10 7 4 3
Solution :
Marks No. of students
(ƒ)Mid-point ( x) fx
10-20 8 15 120
20-30 15 25 375
30-40 13 35 455
40-50 10 45 450
50-60 7 55 385
60-70 4 65 260
70-80 3 75 225
Total N=60 ∑fx= 2270
x=fx
N=2270
60=37.83
The average marks are 37.83munotes.in
64
Example 6:
Calculate the arithmetic mean of heights of 80 Students for the following
data.
Heights
in cms130-
134135-
139140-
144145-
149150-
154155-
159160-
164
No. of
Students7 11 15 21 16 6 4
Solution:
The class intervals are of inclusive type. We first make them continuous
by finding the class boundaries. 0.5 is added to the upper class limits and
0.5 is subtracted from the lower class limits to obtain the class boundaries
as
130-0.5= 129.5 and 134+0.5=134.5
135-0.5= 134.5 and 139+0.5= 139.5 and so on.
Heights ( in
cms)Class
boundariesNo. of
Students( ƒ)Mid-
point( x)fx
130-134 129.5 -134.5 7 132 524
135-139 134.5 -139.5 11 137 1507
140-144 139.5 -144.5 15 142 2130
145-149 144.5 -149.5 21 147 3087
150-154 149.5 -154.5 16 152 2432
155-159 154.5 -159.5 6 157 942
160-164 159.5 -164.5 4 162 648
Total N=80 ∑fx= 1270
N=∑ƒ=80and∑fx= 11270
x=fx
N=11270
80=140.88
The average height of students = 140.88 cm.
4.5 WEIGHTED ARITHMETIC MEAN
One of the limitations of the ar ithmetic mean discussed above is
that it gives equal importance to all the items. But these are cases where
the relative importance of the different items is not the s ame. In these
cases, weights are a ssigned to different items according to their
importanc e. The term ‘weight’ stands for the relative importance of the
different items.
Ifx1,x2,…., xnare the nvalues of the variable xwith the
corresponding weights w 1, w 2,…… …., w n, then the weighted mean is
given bymunotes.in
65
wx=1 1 2 2
1 2......
......n n
nw x w x w x
w w w
=wx
w
Wherewx= Weighted Arithmetic Mean and∑w= Sum of the weights.
Steps :
(1)Multiply the weights wby the variable xand obtain the total ∑xw.
(2) Divide the total ∑xwby the sum of the weights ∑w.
Example 7:
Find the weighted mean for the following data.
x 28 25 20 32 40
W 3 6 4 5 8
Solution :
x W xw
28 3 84
25 6 150
20 4 80
32 5 160
40 8 320
Total 26 794
wx=wx
w
=794
26=30.54 units
Example 8:
A candidate obtained the following marks in percentages in an
examination. English 64 , Mathematics 93, Economics 72, Accountancy
85 and Statistics 79. The weights of these subjects are 2, 3, 3, 4, 1
respectively. Find the candidate’s weighted mean.
Solution :
Subject Percentage of
Marks ( x)Weights (w) wx
English 64 2 128
Mathematics 93 3 279
Economics 72 3 216
Accountancy 85 4 340
Statistics 79 1 79
Total 13 1042
∑w = 13 , ∑wx= 1042munotes.in
66
wx=wx
w
=1042
13=80.15Weighted Mean Marks= 80.15
4.6 COMBINED MEAN
If there are two groups containing n 1and n 2observations with means
x1andx2respectively , then the combined arithmetic mean of two groups
is given by
x=1 1 2 2
1 2n x n x
n n
The above formula can be generalized for more than two groups. If n1,n2
,……, nkare sizes of kgroups with means x1,x2……., xkrespectively
then the mean xof the combined group is given by
x=1 1 2 2
1 2......
......k k
kn x n x n x
n n n
Example 9:
If average salaries of two groups of employees are Rs . 1500 and Rs .
2200 and there are 80 and 70 employees in the two groups. Find the mean
of the combined group.
Solution :
Given : Group I Group II
n1=80 n 2=70
x1= 1500 x2= 2200
x=1 1 2 2
1 2n x n x
n n
=80 (1500 ) + 70 (2200)
80 +70
=120000+ 154000
150
=274000
150
= 1826.67
The average monthly salary of the combined group of 150 employees
is Rs . 1826 .67
Example 10:
The mean weight of a group of 50 workers is 58 kgs. The second group
consists of 60 workers with av erage weight 62 kgs. and there are 90
workers in the third group with average weight 56 kgs. Find the average
weight of the combined group.munotes.in
67
Solution :
Given Group I Group II Group III
No. of workers n 1=50 n2=60 n3=90
Mean we ight x1=58 x2=62 x3=56
x=1 1 2 2 3 3
1 2 3n x n x n x
n n n
=50 (58) + 60 (62) + 90 (56)
50 + 60 + 90
=11660 =58.3
200
The average weight of th e combined group of 200 workers is 58.3 kgs
Merits of Arithmetic Mean .
(1)It is rigidly defined .
(2)It is easy to understand and easy to calculate .
(3)It is based on each and every observation of the series .
(4)It is capable for further mathematical treatment .
(5)It is least affected by sampling fluct uations.
Demerits of Arithmetic Mean .
(1)It is very much affected by extreme observations.
(2)It can not be used in case of open end classes.
(3)It can not be determined by inspection nor it can be located
graphically.
(4)It can not be obtained if a single observation is missing .
(5)It is a value which may not be present in the data .
Exercise 5.1
1.Find the arithmetic mean for the following sets of observations.
(i) 48, 55,83,65,38,74,58
(ii) 154,138,165,172,160,145,157,185
(iii)2254,2357,22 41,2012,2125
Ans: (i)x= 60.14 (ii) x=159.5 (iii) x=2197.8
2.Calculate the mean for the following distribution:
x: 15 17 19 21 23 25
ƒ: 6 11 8 15 5 4
Ans:x=19.57
3.Find the mean for the following data.
Size of Shoe: 5 6 7 8 9 10
No.ofpairs 22 35 28 42 15 12
Ans:x=7.19munotes.in
68
4.Calculate the arithmetic mean for the following data
Daily Wages
inRs:40-80 80-20 120-60 160-00 200-40 240-80 280-20
No. of
workers8 15 13 19 12 14 10
Ans: x=181.32
5.Calculate the mean for the following data r epresenting monthly
salary of a group of employees.
Salary in
Rs:500-
10001000 -
15001500 -
20002000 -
25002500 -
30003000 -
3500
No. of
persons:7 11 13 8 5 4
Ans: x=1802.08
6.Find the mean for the following data.
Weekly
wages in Rs:50-99 100-149 150-199 200-249 250-299 300-349
No. of
worke rs12 17 15 20 8 7
Ans: x=184.63
7.Find the arithmetic mean for the following data
Age in
yrs:10-19 20-29 30-39 40-49 50-59 60-69 70-79
No. of
persons:6 12 19 23 15 8 6
Ans: x=43.15
8.Find the weig hted mean for the following data.
x: 25 22 15 30 18
w: 6 4 3 8 2
Ans: xw= 24.30
9.AStudent Scores 44 in Test I, 32 in Test II, 27 in Test III and 38 in
Test IV, These are to be weighted with weights 4,3,3 and 4
respectively. Find the average score.
Ans:xw=36.07
10.The average daily wages for 120 workers in a factory are Rs. 78. The
average wage for 80 male workers out of them is Rs.92 Find the
average wage for the remaining female workers.
Ans: Rs.50
11.Thereare three groups in a class of 200 Students. T hefirst groups
contains 80 Students with average marks 65, the second groupmunotes.in
69
consists of 70 Students with average marks 74. Find the average
marks of the Students from the third group if the av erage for the
entire class is 68 .
Ans: 64.4
12. Find the com bined arithmetic mean for the following data:
Group I Group II Group III
Mean 125 141 131
No.of
observations.150 100 50
Ans: 131.67
4.7 MEDIAN
The median by definition refers to the middle value in a
distribution. Median is the value of the variable which divides the
distribution into two equal parts. The 50% observations lie below the
value of the median and 50% observations lie above it. Median is called a
positional average. Median is denoted by M.
(i) For Simple data or ungrouped data:
Median is defined as the value of the middle item of a series when
the observations have been arranged in ascending or descending order of
magnitude.
Steps: 1) Arrange the data in ascending or descending order of magnitude.
(Both arrangements would give the same answer).
2) When n is odd:
Median= value of n+1thterm.
2
3) When n is even:
Median= Arithmetic Mean of value of n&,n+ 1thterms.
2 2
i.e, adding the two middle values and divided by two.
where n = number of observations.
Example 11
Find the median for the following set of observations
65, 38, 79, 85, 54, 47, 72
Solution :Arrange the values in ascending order:
38, 47, 54, 65, 72, 79, 85
n= 7 (odd number)munotes.in
70
The middle observation is 65
Median= 65 units .
using formula:
Median = value of n+1thterm.
2
= value of 7+1 th term.
2
= 4thterm
Median = 65
Example 12:Find the median for the following data.
25, 98, 67, 18, 45, 83, 76, 35
Solution : Arrange the values in ascending order
18, 25, 35, 45, 67, 76, 83, 98
n = 8 (even number)
The pair 45, 67 can be considered as the middle pair.
Median = A.M of the pair =45+6 7
2
Median = 56
Using formula :
Median = A. M. of values of nth,n+1thterms
2 2
= A. M. of 8,8+1thterms
2 2
= A. M. of ( 4 , 4 +1 )thterms
= A. M. of ( 4 , 5 )thterms
= A. M. of 45 and 67
=45 + 67
2
Median = 56
(ii) For ungrouped frequency distribution :
Steps :
1) Arrange the data in ascending or descending order of magnitude with
respective frequencies .
2) Find the cumulative frequency (c. ƒ) less than type .
3) Find N/2 , N = total frequency.
4) See t he c.ƒcolumn either equal or greater than N/2 and determine the
value of the variable corresponding to it . That gives the value of Median .munotes.in
71
Example 13:From the following data find the value of Median.
Income in Rs: 1500 2500 3000 3500 4500
No. of persons 12 8 15 6 5
Solution :
Income (in Rs.) ( ҳ) No. of persons ( ƒ) c.ƒless than type
1500 12 12
2500 8 20
3000 15 35
3500 6 41
4500 5 46
N= 46
N=46 = 23
2 2
In c.ƒcolumn, we get 35 as the first cumulative frequency greater than 23.
The value of ҳ corresponding to the c. ƒ. 35 is 3000.
Median = Rs.3000
Example : 14
Find the median for the following data .
ҳ : 25 20 30 40 35 15
ƒ: 19 14 23 12 20 8
Solution : Arrange the values in ascending order .
ҳ ƒ c.ƒless t han type
15 8 8
20 14 22
25 19 41
30 23 64
35 20 84
40 12 96
N= 96
N=96= 48
2 2
In c.f column, greater than 48 is 64.
The corresponding value of xto the c.f is 30.
Median =30units .
(iii) For grouped data.munotes.in
72
Steps:
1) Find the c.f less than type
2) Find N/2, N= total frequency.
3) See the c.f column just greater than N/2.
4) The Corresponding class interval is called the Median class.
5) To find Median, use the following formula.
M=ll+2N cf
f
(l2–l1)
Where
ll= lower class -bounding of the median class.
l2= upper class -bounding of the median class.
ƒ= frequency of the median class.
c.ƒ= cumulative frequency of the class interval preceding the median
class.
Example : 15
The following data relate to the number of patients visiting a government
hospital daily:
No. of
patients :1000 -
12001200 -
14001400 -
16001600 -
18001800 -
20002000 -
2200
No. of
days :15 21 24 18 12 10
Solution :
No. of patients No. of days c.ƒ. less than type
1000 -1200 15 15
1200 -1400 21 36
1400 -1600 24 60
1600 -1800 18 78
1800 -2000 12 90
2000 -2200 10 100
N=100
N/2 =100/2=50
See the c. ƒcolumn greater than 50 is 60. The corresponding class interval
1400 -1600 is the median class.
l1=1400; l 2=1600;ƒ=24; c.ƒ= 36
M=ll+2N cf
f
(l2–l1)= 1400+ [50 36
24](1600 -1400)
= 1400+[ 14](200) =1400+116.67
M= 1516.67 patientsmunotes.in
73
Example 16:Find the median for the following distribution:
Weig hts
(in kgs)30-34 35-39 40-44 45-49 50-54 55-59 60-64
No. of
students4 7 15 21 18 10 5
Solution : Convert the given class intervals into exclusive type
Weights No. of students c.ƒ. less than type
29.5-34.5 4 4
34.5-39.5 7 11
39.5-44.5 15 26
44.5-49.5 21 47
49.5-54.5 18 65
54.5-59.5 10 75
59.5-64.5 5 80
N=80
N/2 =80/2 = 40
Median class is 44.5 -49.5
l1= 44.5 ; l 2= 49.5 ,ƒ=21, c.ƒ. =26
M=ll+2N cf
f
(l2–l1)
= 44.5 + [ 40-26] (49.5 -44.5)
21
= 44.5 + [14/21] (5)
= 44.5 + 3.33
M = 47.83 units.
Merits of Median :
1.It is rigidly defined .
2.It is easy to understand and easy to calculate .
3.It is not affected by extreme observations as it is a positional
average.
4.It can be calculated , even if the extreme values are not known .
5.It can be located by mere inspection and can also be located
graphically.
6.It is the only average to be used while dealing with qualitative
characteristics which can not be measured numerically .
Demerits of Med ian :
1.It is not a good representative in many cases.
2.It is not based on all observations.
3.It is not capable of further mathematical treatment .
4.It is affected by sampling fluctuations.munotes.in
74
5.For continuous data case , the formula is obtained on the
assumption of uniform distribution of frequencies over the class
intervals. This assumption may not be true.
4.8QUARTILES
When a distribution is divided into fo urequal parts, each point of
division is called as quartile and each part is of 25% (one -fourth) of the
total observations. There are three partition values such as Q 1, Q2, and Q 3.
Q1is called first quartile or lower quartile.
Q3is called third quartile or upper quartile.
Q2is called second quartile which coincides with median. Therefore Q 2is
nothing b ut Median.
Q1. Q2. Q3
N/4 N/4 N/4 N/4
The Steps involved for computing the quartiles is basically the same as
that of comput ing median.
To find Q 1:(First Quartile)
Steps: 1) Find the c. ƒless than type.
2) Find N/4,N=total frequency.
3) See the c. ƒ. column just greater than N/4.
4) The corresponding class interval is called th e quartile class.
5) To find Q 1, use the following formula:
Q1=l1+4N cf
f
(l2–l1)
Where l 1= lower limit of the quartile class
l2= upper limit of the quartile class.
ƒ= frequency of the q uartile class.
c.ƒ= cumulative frequency of the class interval prec eding the
quartile class.
To find Q 3: (Third Quartile)
Steps : 1) Find the c. ƒless than type.
2) Find 3N/4, N=total frequency.
3) See the c. ƒcolumn just greater than 3N/4
4) The corresponding class interval is called the quartile class.
5) To find Q 3, use the following formula.
Q3=l1+3
4N cf
f
(l2–l1)
Where l 1= lower limit of the quartile class.
l2=upper limit of the quartile class .munotes.in
75
ƒ= frequency of the quartile class.
c.ƒ.= cumulative frequency of the class interval preceding the
quartile class.
To find Q 2(Second Quartile )
Q2= Median (Discussed a bove )
Example 17:Find the three quartiles for the following data.
Commiss ion
(in Rs):100-
120120-
140140-
160160-
180180-
200200-
220220-
240
No. of
Salesmen:13 38 45 56 27 17 12
Solution:
Commission (in Rs.) No. of Salesmen c.f.less than type
100-120 13 13
120-140 38 51
140-160 45 96
160-180 56 152
180-200 27 179
200-220 17 196
220-240 12 208
n= 208
To find Q 1:
N=208= 52
4 4
See the c.f. column just greater than 52.
Quartile class: 140 -160
l1= 140, l 2= 160,ƒ=45, c.ƒ=51
Q1= l1+[N/4-c.ƒ](l2-l1)
ƒ
= 140+[ 52-51(160-140)
45
= 140+[1/45](20)
= 140+0.44
Q1=140.44
To find Q 2= Median :
N/2 = 208/2 = 104
See the c. ƒcolumn just greater than 104.
Median class =160 -180
l1= 160; l 2=180;ƒ=56; c.ƒ=96munotes.in
76
M=l 1+ [N/2-c.ƒ](l2-l1)
ƒ
= 160+[ 104-96](180 -160)
56
= 160+[8/56] (20)
= 160+2.86
Q2= 162.86
To find Q 3:
3N/4=3(208/4) =156
See the c. ƒcolumn just greater than 156.
Quartile class=180 -200
l1=180; l 2=200;ƒ=27; c.ƒ=152
Q3= l1+[3N/4-c.f](l2-l1)
ƒ
= 180+[ 156-152] (200 -180)
27
= 180+[4/27] (20)
= 180+2.96
Q3= 182.96
Q1= 140.44
Q2=M= 162.86
Q3= 182.96
Graphical location of Quartiles :
The med ian and the quartiles obtained graphically from the c. ƒ. curve
less than type curve as follows :
First draw a cumulative frequency curve less than type.
To locate Q 2= Median :
Locate n/2 on the Y –axis and from it draw a perpendicular on the c.
ƒ. curve. From the point where it meets the c. ƒ. curve draw another
perpendicular on the X -axis and the point where it meets the X -axis is
called the median .
To locate Q 1and Q 3:
Locate n/4 and 3N/4 on the Y -axis and proceed as above to obtain
Q1and Q 3respectively on the X -axis.munotes.in
77
Note that the values obtained from the graph are approximate figures.
They do not represent exact figures.
Example 18:
Draw a ‘less than “ cumulative frequency curve for the following data and
hence locate the three quartiles graphically.
Age in
yrs.0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
No. of
persons15 13 25 22 25 10 5 5
Solution :
Age (in yrs.) No. of persons. c.ƒ. less than type
0-10 15 15
10-20 13 28
20-30 25 53
30-40 22 75
40-50 25 100
50-60 10 110
60-70 5 115
70-80 5 120
N=120
X-axis (upper limit ) Y-axis c.ƒ. < type
10 15
20 28
30 53
40 75
50 100
60 110
70 115
80 120
N/4= 120/4 = 30
N/2 = 120/2 =60
3N/4 = 3 (120/4)
3N/4 =90munotes.in
78
Q1M Q3 Age in years
From the graph :Q1 = 21 approximately
M = 33 approximately
Q3 = 46 approximately
Exercise 5.2
1. Find the median for the following data
(i)50 , 28 , 35 , 98 , 75 , 44 , 58
(ii) 16 , 22 , 10 , 12 , 30 , 37 , 28 , 40 , 15 , 20
Ans: (i) M=50 (ii) M= 21
2 . Calculate the median for the following distribution :
x: 25 30 35 40 45 50 55
ƒ: 8 14 23 28 10 6 4
Ans: M= 40
3. Find the median for the following data.
Weekly
Wages (in
Rs)50-100 100-150 150-200 200-250 250-300 300-350
No. of
Workers:20 18 32 18 12 15
Ans: M= 180.47munotes.in
79
4. Find the median height for the following distribution:
Height in
cms:150-154 154-158 158-162 162-166 166-170 170-174
No. of
Students:6 15 23 20 12 10
Ans M=161.83
5. Find the three quartiles for the following distribution:
Income
in Rs:0-500 500-
10001000 -
15001500 -
20002000 -
25002500 -
30003000 -
3500
No.of
families3 5 10 18 15 9 6
Ans:Q 1=1425, M=1916.67, Q 3= 2450
6. Find the median and the two quartiles for the following data. Also
locate them graphically.
Rainfall
(in cms):15-20 20-25 25-30 30-35 35-40 40-45 45-50
No.of
years:3 7 10 16 12 8 4
Ans:Q 1= 27.5, M=33.13, Q 3=38.75
6. The following is the data representing profits in thousands of Rs.
of some companies. Find the quartiles and hence locate them graphically.
Profit: 70-80 80-90 90-100 100-110 110-120 120-130
No.of
Companies11 15 19 23 22 10
Ans:Q 1=89.33, M=102.17,Q 3=113.18
4.9MODE
(i) For Raw data:
Mode is the value which occurs most frequently ,in a set of observations.
It is a value which is repeated maximum number of times and is denoted
by Z.
Example 19:Find mode for the following da ta.
64, 38,35,68,35,94,42,35,52,35
Solution :
As the number 35 is repeated maximum number of times that is 4 times.
Mode=35 units.
(ii)For ungrouped frequency distribution :
Mode is the value of the variable corresponding to the highest frequen cy.munotes.in
80
Example 20:Calculate the mode for the following data.
Size of Shoe: 5 6 7 8 9 10
No.of Pairs: 38 43 48 56 25 22
Solution :Here thehighest frequency is 56 against the size 8.
Modal size =7
(iii) For Grouped data:
In a Continuous dist ribution first the modal class is determined.
The class interval corresponding to the highest frequency is called modal
class.
Formula:
=l1+1 0
2 1
1 0 1 2( )( ) ( )f fl lf f f f
Where l 1,= lower boundary of the modal class
l2= upper boundary of the modal class
ƒ1= frequency of the modal class
ƒo= frequency of the class interval immediately preceding the
modal class.
ƒ2=frequency of the class interval immediately succeeding the
modal class.
Example 21
Find the mode f or the following data.
Daily
wages(in
Rs):20-
4040-
6060-
8080-
100100-
120120-
140140-160
No.of
employees:21 28 35 40 24 18 10
Solution:
Daily wages (in Rs.) No. of employee s
20-40 21
40-60 28
60-80 35ƒo
80-100 40ƒ1
100-120 24ƒ2
120-140 18
140-160 10
The maximum frequency is 40.
Modal class is 80 -100.
l1= 80; l2=100; ƒo=35;ƒ1=40,ƒ2 =24munotes.in
81
Z=l1+1 0
2 1
1 0 1 2( )( ) ( )f fl lf f f f
= 80 +40 35(100 80)(40 35) (40 24)
= 80 +5
5 16
x 20 = 80+4.76
Z= 84.76
Modal daily wages = Rs. 84.76
Example : 21
Find the mode for the following data.
Marks: 10-19 20-29 30-39 40-49 50-59 60-69 70-79
No of
Students:8 22 31 44 15 13 10
Solution:
Marks Class boundaries No. of Students
10-19 9.5-19.5 8
20-29 19.5-29.5 22
30-39 29.5-39.5 31ƒo
40-49 39.5-49.5 44ƒ1
50-59 49.5-59.5 15ƒ2
60-69 59.5-69.5 13
70-79 69.5-79.5 10
The maximum frequency is 44.
The modal class is 39.5 -49.5
l1=39.5ƒ0=31
l2=49.5ƒ1=44,ƒ2=15
Z=l1+1 0
2 1
1 0 1 2( )( ) ( )f fl lf f f f
= 39.5 +44 31(49.5 39.5)(44 31) (44 15) = 39.5 +13
13 29
x 10
= 39.5 + 3.095
= 42.595
Z = 42.60
Modal Marks = 42.60
4.10 GRAPHICAL LOCATION OF MODE
Mode can be obtained from a histogram as follows . The method
can be applied for class intervals of equal length having a unique
maximum frequency .munotes.in
82
In the histogram , the rectangle with the maximum height
represents the modal class . The right upper corner of this rectangle is
connected with right upper corner of the preceding rectangle by a straight
line . Similarly , the left uppe r corner of the maximum height of the
rectangle is connected with left upper corner of the succeeding rectangle
by a straight line . These two straight lines are intersecting at a point .
From the point of intersection, a perpendicular is drawn on x-axis, the
foot of which gives the value of Mode.
Example 23:
Draw a histogram for the following data and hence locate mode
graphically.
Profit (in
lakhs):10-20 20-30 30-40 40-50 50-60 60-70
No. of :
Companies12 25 31 45 20 15
Mode Profit (in lakhs)
Modal profits = Rs. 44 lakhs.
Merits and Demerits of Mode .
Merits :
1.It is easy to understand and easy to calculate.
2.It is unaffected by the presence of extreme values.
3.It can be obtained graphically from a histogram.
4.It can be calculated f rom frequency dis tribution with open -end classes .
5.It is not necessary to know all the items. Only the point of ma ximum
concentration is required .munotes.in
83
Demerits :
(1)It is not rigidly defined .
(2)It is not based on all observations.
(3)It is affected by sampling flu ctuations.
(4)It is not suitable for further mathematical treatment .
Exercise 4.3
(1) Find the mode for the following data .
(i) 85, 40 , 55 , 35 , 42 , 67 , 75 , 63 , 35 , 10 , 35
(ii) 250 , 300 , 450 , 300 , 290 , 410 . 350 , 300
Ans: (i) Z = 35 (ii ) Z= 300
(2) Find the mode for the following data.
x: 15 18 20 22 24
ƒ: 8 6 13 18 10
Ans:= 22
(3) Calculate the modal wages for the following distribution.
Wages in
Rs:50-65 65-80 80-95 95-110 110-125 125-140
No. of
workers4 7 15 28 11 5
Ans:= 101.5
(4)Find mode for the following data.
Life in
hrs:200-
400400-
600600-
800800-
10001000 -
12001200 -
14001400 -
1600
No. of
bulbs:15 22 40 65 38 26 10
Ans: z=896.15
(5)Find the mode for the following data and hence locate mode
graphically.
Salary in
Rs:1000 -
15001500 -
20002000 -
25002500 -
30003000 -
35003500 -
4000
No. of
persons:23 35 50 30 18 12
Ans: z=2214.29
(6)Find the mode for the following data, also locate mode graphically.
Class
Interval :100-
150150-
200200-
250250-
300300-
350350-
400400-
450
Frequency: 15 29 31 42 56 35 17
Ans:= z= 320munotes.in
84
(7)Find Mean Median and mode for the following data.
Income (in
Rs.):1000 -
20002000 -
30003000 -
40004000 -
50005000 -
60006000 -
7000
No.of
employees:15 26 45 32 12 10
Ans:x= 3714.29; M=3644.44; Z= 3593.75
(8)Prove that median lies between mean and mode from the following
data:
Weekly
(wages
in Rs.)100-
200200-
300300-
400400-
500500-
600600-
700700-
800
No. of
workers.18 23 32 41 35 20 15
Ans:x= 443.48; M=446.34; Z=460
(9)Calculate mean median and mode for the following data.
Height
(in.cms)125-130 130-135 135-140 140-145 145-150 150-155
No. of
children6 13 19 26 15 7
Ans:x=140.52; M=140.96; Z=141.94
(10) Find arithmetic mean median and mode for the following data.
Age in
yrs:10-20 20-30 30-40 40-50 50-60 60-70 70-80
No. of
persons:12 18 23 28 20 16 7
Ans:x=43.23; M=43.21; Z= 43.85
munotes.in
85
UNIT I II
Unit -5
MEASURES OF DISPERSION
Unit Structures :
5.0 Objectives
5.1 Introduction
5.2 Range
5.3 Quartile Deviation
5.4 Mean Deviation
5.5 Standard Deviation
5.0OBJECTIVES
After reading this chapter you will be able to:
Compute different t ypes of deviations like; Range, Quartile
Deviation, Standard Deviation and Mean Deviation .
Compute relative measures of deviation like; coefficient of
Range/Quartile Deviation/Mean Deviation/Variation .
Compute combined standard deviation .
5.1INTRODUCTION
Why Do We Need to Study the Measure of Dispersion?
The common averages or measures of central tendency indicate the
general magnitude of the data and give us a single value which
represents the data ,but they do not tell us the degree of spread out or
the extent of variability in individual items in a distribution. This can
beknown by certain other measures called measures of dispersion .
We will discuss the most commonly used statistical measures showing
the degree and the characters of variations in data.
Dispersion in particular helps in finding out the variability of the data
i.e., the extent of dispersal or scatter of individual items in a given
distribution. Such dispersal may be known with reference to the central
values or the common averages such as mean, median and mode or a
standard value; or with reference t oother values in the distribution.
The need for such a measure arises because mean or even median and
mode may be the same in two or more distribution but the composition
of individual items in the series may vary widely. We give an example
to illustrate this.
Region Rainfall (cms)
East 106 106 106 106 106
West 90 98 106 114 122
North 120 130 84 126 70munotes.in
86
In the above example note that the average rainfall in all the
three regions is t he same i.e. 106 cms however it can be easily seen
that it would be wrong to conclude that all the three regions have the
same climatic pattern. This is so because in East all the values are
equal to the average; whereas in the western region they are cent ered
around their mean, and in north they are widely scattered. It may thus
be misleading to describe a situation simply with the aid of an average
value.
In measuring dispersion, it is necessary to know the amount of
variation and the degree of variatio n. The former is designated as
absolute measures o f dispersion and expressed in the denomination of
original variates such as inches, cms, tons, kilograms etc while the
latter is designated as related measures of dispersion. We use the
absolute measures of dispersion when we compare two sets of data
with the same units and the same average type. If the two sets of data
do not have the same units then we cannot use the absolute measures
and we use the relative measures of dispersion.
Absolute measures can b e divided into positional measures
based on some items of the series such as (I) Range, (ii) Quartile
deviation or semi –inter quartile range and those which are based on all
items in series such as (I) Mean deviation, (ii)Standard deviation. The
relative m easure in each of the above cases are called the coefficients
of the respective measures. For purposes of comparison between two
or more series with varying size or number of items, varying central
values or units of calculation, only relatives measures ca n be used .
5.2 RANGE
Range is the simplest measure of dispersion.
When the data are arranged in an array the difference between the
largest and the smallest values in the group is called the Range .
Symbolically: Absolute Range = L -S, [where L is the largest value
and S is the smallest value]
Relative Range =Absolute Range
sum of the two extremes
Amongst all the methods of studying dispersion range is the
simples t to calculate and to understand but it isnotused generally
because of the following reasons:
1)Since it is based on the smallest and the largest values of the
distribution, it is unduly influenced by two unusual values at
either end. On this account, range is usually not used to
describe a sample having one or a few unusu al values at one or
the other end.
2)It is not affected by the values of various items comprised in
the dis tribution. Thus, it cannot give anyinformation about the
general characters of the distribution within the two extreme
observations.munotes.in
87
For example, let us consider the following three series:
Series: A
6 46 46 46 46 46 46 46
Series: B
6 6 6 6 46 46 46 46
Series: C
6 10 15 25 30 32 40 46
It can be noted that in all three series the range is the same, i.e. 40,
however the distributions are not alike: the averages in each case is
also quite different. It is because range is notsensitive to the values of
individual items included in the distribution. It thus cannot be
depended upon to give any guidance for determining the dispersion of
the values within a distribution.
5.3 QUARTILE DEVIATION
The dependence of the range on extreme items can be avoided
by adopting this measure. Quartiles together with the median are the
points that divide the whole series of observations into approximately
four equal parts so that quartile measures give a rough idea of the
distribution on either side of the average.
Since under most circumstances, the central half of the
distribution tends to be fairly typical, quartile range Q 3–Q1affords a
convenient measure of absolut e variation. The lowest quarter of the
data (upto Q1) and the highest quarter (beyond Q3) are ignored. The
remai nder, the middle half of the data above Q1and below Q3or (Q3–
Q1) are considered. This, when divided by 2, gives the semi -
interquartile range or quartile deviation .
Q.D. =3 1
2Q Q
In a symmetrical distribution, when Q 3(75%) plus Q 1(25%) is
halved, the value reached would give Median, i.e., the mid -point of
75% and 25%. Thus, with semi -interquartile range 50% of data is
distributed around the median. It gives the expected range between
which 50% of all data should lie. The Quartile Deviation is an absolute
measure of dispersion . The corresponding relative measure of
dispersion is
Coefficient of Q.D. =3 1
3 1Q Q
Q Q
For the following Fr equency Distribution we show how to calculate the
Quartile Deviation and the coefficient of Quartile Deviation.
xƒ c.ƒ.
10 6 6
15 17 23
20 29 52
25 38 90munotes.in
88
30 25 115
35 14 129
40 9 138
90 1 139
It may be noted that:
Corresponding to Q1, the c.f. =1
4n
Corresponding to Q 2, the c.f. =2( 1)
4n
Corresponding to Q3, the c.f. =3( 1)
4n
Thus from the given data :
Q3= Value of the variates correspond ing to c.f.3 140
4x= 105 which
corresponds to 30
Q2= Value of the variates corresponding to c.f.2 140
4x= 70 which
corresponds to 25
Q1= Value of the variates corresponding to c.f.140
4= 35, which
corresponds to 20
Quartile deviation or Q.D. =3 1
2Q Q=30 20
2= 5
And Co -efficient of Quartile Deviation
=3 1
3 1Q Q
Q Q
=30 20
30 20
=10
50= 0.2
Example
Find the Quartile Deviation of the daily wages (in Rs.) of 11 workers
given as follows: 125, 75, 80, 50, 60, 40, 50, 100, 85, 90, 45.
Solution :Arranging the data in ascending order we have the wages of
the 11 workers as follows:
40, 45, 50, 50, 60, 75, 8 0, 85, 90, 100, 125
Since the number of observations is odd (11), the 1stQuartile is given
by:
Q1= (11 + 1)/4 = 3rdobservation = 50.
… (1)
Q3= 3(11 + 1)/4 = 9thobservation = 90.
… (2)
Q.D. =3 1
2Q Q= (90 –50)/2 = 20 …
from (1) and (2)munotes.in
89
Example
The following data gives the weight of 60 students in a class. Find the
range of the weights of central 50% students.
Solution :
Tofind r ange of the weight’s of central 50 % students means to find
the inter quartile range. For that we require Q1andQ3. The column of
less than c fis introduced as follows:
Q1class Q3class
To find Q1:N= 56. Thus m=N/4 = 14.
Thecfjust greater than 14 is 20, so 35 –40 is the 1stquartile class and
l1= 35, l2= 40, i= 40 –35 = 5, f= 16 and pcf= 4.
1 114 4x 35 x 5 35 3.125 38.12516 m pcfQ l if
To find Q3:N= 56. Thus m= 3N/4 = 42.
Thecfjust greater than 42 is 50, so 50 –55 is the 3rdquartile class and
l1= 50, l2= 55, i= 55 –50 = 5, f= 10 and pcf= 40.
1 142 40x 50 x 5 50 1 5110 m pcfQ l if
inter quartile range = Q 3–Q1= 51 –38.125 = 12.875 kg
Thus, the range of weight for the central 50% students = 12.875kg
Example
Find the semi –inter quartile range and its coefficient for the following
data:Weight in kg30–
3535–
4040–
4545–
5050–
5555–
60
No. of
students4 16 12 8 10 5
Weight in kg30–
3535–
4040–
4545–
5050–
5555–
60
No. of
students4 16 12 8 10 6
cf 4 20 32 40 50 56
Size of
shoe01 2 3 4 5 6 7 8 9 10
No. of
boys710 15 11 18 10 16 5 12 6 2munotes.in
90
Solution : The less than c f are computed and the table is
completed as follows:
Here N= 112.
To find Q1:m=N/4 = 28.
The first c fjust greater than 28 is 32, so the 1stquartile is Q1=
2
To find Q3:m= 3N/4 = 84.
The first c fjust greater than 84 is 87, so the 3rdquartile is Q3=
6
the semi inter quartile range i.e.Q.D. =3 1 6 222 2Q Q
Coefficient of Q.D. =3 1
3 16 20.56 2Q Q
Q Q
5.4 MEAN DEVIATION
Both the range and quartile deviation do not show the scatterness
around an average and as such do not give a clear idea of the
dispersion of the distribution , these measures also exclude some data
and consequently do not give a complete picture based on the entire
data. . To study the deviations around the average we now introduce
two more measures of dispersion the mean deviation and the standard
deviation. The mean deviation is also called as the average deviation.
The essence of average deviation lies in the concept of dispersion,
which is the average amount of scatter of individual items from either
the mean or the median ignoring the algebraic signs.
This measure takes in to account the whole data. When it is calculated
by averaging the deviations of the individual items from their
arithmetic mean ,taking all deviation to be positive, the measures is
called mean deviation. It may be pointed that we are concerned with
the di stance of the individual items from their averages and not with
their position, which may be either above or below the average.
Suppose that we have sample of six observations 0,2,3,4,4, and 5. The
mean of these observations is:0 2 3 4 4 5
6= 3
Now we obtain the deviation of each observation from the mean of 3.
For the first observation, for example, this gives a deviation of 0 –3 =
-3. Similarly, the other observations are : -1, 0, + 1,+ 1 and + 2 .It
may seem that a good way to measure dispersion would be to take theSize of
shoe01 2 3 4 5 6 7 8 9 10
No. of
boys710 15 11 18 10 16 5 12 6 2
c f 717 32 43 61 71 87 92 104 110 112
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mean of six deviations. But this gives 0, since, -3-1+0+1+1+2. For
practice in using symbols, we give symbolic definition of the mean
deviations. To avoid the cancelling off of the positive deviations with
the negative dev iations we take only the magnitude ( absolute value )
of each of these deviations .
So in the above case the M.D. =ix x
n=3 1 0 1 1 2
6=8
6
For a frequency distribution the calculation of M.D. is done as follows
M.D.=f d
f
where f is the frequency of the observation and d=
ix xwhere xis the mean.
The coefficient of M.D. =. .M D
Mean
We now take an exampl e to calculate the Mean deviation and the
coefficient of mean deviation for a frequency distribution.
Example
Find mean deviation from mean for the following distribution:
X 10 15 20 25 30 25 40 90
ƒ 6 17 29 38 25 14 9 1
Solution
Calcu lation of mean deviation from mean :
x ƒ f . x dx
= 25.43xƒ.dx
10 6 60 15.43 92.58
15 17 255 10.43 177.31
20 29 580 5.43 157.47
25 38 950 0.43 16.34
30 25 750 4.57 114.25
35 14 490 9.57 133.98
40 9 360 14.57 131.13
90 1 90 64.57 64.57
139 3535 ∑ƒdx =
887.63
In the above example the mean is =.f xxn=3535
139= 25.43 n =f
And mean Deviation from Mean =( . )f dx
n=887.63
139= 6.386 n =fmunotes.in
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And Coefficient of mean deviation is =6.3860.2525.43
The mean deviation tells us that some values were above the mean and
some below. On the average, the deviation of all values combined was
6.386. Co-efficient of mean deviation o r 0.25 .
In case of continuous frequency distribution we take the mid value of
each class as the value of x and proceed to do the calculations in the
same manner as above.
Example
Find the mean deviation from mean and its coefficient for the
following dat a giving the rainfall in cm in different areas in
Maharshtra: 105, 90, 102, 67, 71, 52, 80, 30, 70 and 48.
Solution : Since we have to compute M.D. from mean, we first prepare
the table for finding mean and then introduce columns of absolute
deviations fro m the mean.
105 +90 102 67 71 52 80 30 70 48. 71571.510 10x
Example
The marks obtained by 10 students in a test are given below. Find the
M.D. from median and its relative measure.
Solution: The marks of 10 students are arranged in ascending order
and its median is found. The column of absolute deviations from
median is introduced and its sum is computed. Using the formula
mentioned above, M.D. from median and its coefficient is calculated.x d=x x
105 33.5
90 18.5
102 30.5
67 4.5
71 0.5
52 19.5
80 8.5
30 41.5
70 1.5
48 23.5
∑x= 715 d= 182
Marks: 15 10 10 03 06 04 11 17 13 05Now,
M.D. from mean =d
n
x=182
10
x= 18.2
Coefficient of M.D. from
mean
=x
x=18.2
71.5munotes.in
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x d=x M
03 5
04 4
05 3
06 2
10 2
10 2
11 3
13 5
15 7
17 9
Total d= 42
Example
On the Mumbai –Nashik highway the number of accidents per day in
6 months are given below. Find the mean deviation and coefficient of
M.D.
No. of
accident s0 1 2 3 4 5 6 7 8 910
No. of
days26 32 41 12 22 10 05 01 06 15 10
Solution:
Now, N= 180m=N/2 = 180/2 = 90.
Thecfjust greater than 90 is 99. The corresponding observation
is 2.M= 2
M.D. from median = M=fd
N=4162.31180
Coefficient of M.D. =M
M=2.311.152
Example
The following data gives the wages of 200 w orkers in a factory with
minimum wages Rs. 60 and maximum wages as Rs. 200. Find the
mean deviation and compute its relative measure.No. of
accidentsNo. of days
(f)cf d=x M f.d
0 26 26 2 52
1 32 58 1 32
2 41 99 0 0
3 12 111 1 12
4 22 133 2 44
5 10 143 3 30
6 05 148 4 20
7 01 149 5 05
8 06 155 6 36
9 15 170 7 105
10 10 180 8 80
Total N= 180 - - fd= 416Since N= 10,
Median = A.M. of 5thand 6thobservation
M=6 1082
M.D. from median =d
n=42
10
M= 4.2
Coefficient of M.D. =4.20.5258M
M
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Wages
less than80 100 120 140 160 180 200
No of
workers30 45 77 98 128 172 200
Solution :The data is given with less than cf, we first convert them to
frequencies then find the median and follow the steps to compute M.D.
as mentioned above.
Now, N= 200m=N/2 = 200/2 = 100.
The median class is 140 –160.l1= 140, l2= 160, i= 160 -140
= 20, f= 30 and pcf= 98
M =
1100 98x 140 x 20 140 1.33 141.3330 m pcfl if
M.D. from med ian = M=fd
N=7034.6835.1734200
Coefficient of M.D. =M
M=35.17340.25141.33
5.5 STANDARD DEVIATION
As we have seen range is unstable, quartile deviation excludes ha lf the
data arbitrarily and mean deviation neglects algebraic signs of the
deviations, a measure of dispersion that does not suffer from any of
these defects and is at the same time useful in statistic work is
standard deviation .In 1893 Karl Pearson first introduced the concept.
It is considered as one of the best measures of dispersion as it satisfies
the requisites of a good measure of dispersion. The standard deviation
measures the absolute dispersion or variability of a distribution. The
greater the a mount of variability or dispersion greater is the value of
standard deviation. In common language a small value of standard
deviation means greater uniformity of the data and homogeneity of the
distribution. It is due to this reason that standard deviation is
considered as a good indicator of the representativeness of the mean.
It is represented by σ(read as ‘sigma’); σ2i.e., the square of the
standard deviation is called variance . Here, each deviation is squared.Wages in
RsNo. of
workers ( f)cf xd=
141.33xfd
60–80 30 30 70 71.33 2139.9
80–100 15 45 90 51.33 769.95
100–120 32 77 110 31.33 1002.56
120–140 21 98 130 11.33 237.93
140–160 30 128 150 8.67 260.1
160–180 44 172 170 28.67 1261.48
180–200 28 200 190 48.67 1362.76
Total N= 200 - - - fd= 7034 .68
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The measure is calculated as the average of deviations from arithmetic
mean. To avoid positive and negative signs, the deviations are s quared.
Further, squaring gives added weight to extreme measures, which is a
desirable feature for some types of data. It is a square root of arithmetic
mean of the squared deviations of individual items from their
arithmetic mean.
The mean of squared dev iation, i.e., the square of standard deviation is
known as variance. Standard deviation is one of the most important
measures of variation used in Statistics. Let us see how to compute the
measure in different situation.
s.d =σ=2( )x x
n
For a frequency distribution standard deviation is
2( )f x x
f
=22fx fx
f f
We will now take an example of a frequency distribution and calculate
the standard deviation.
Example
From the following frequency distribution, find the standard deviation
using the formula for grouped data:
Class interval Frequency
10–20 9
20–30 18
30–40 31
40–50 17
50–60 16
60-70 9
Total 100
Interval Mid-
point
XFrequency
ƒƒx dx dx2ƒdx2
10–20 15 9 135 -24 576 5184
20–30 25 18 450 -14 196 3528
30–40 35 31 1085 -4 16 496
40–50 45 17 765 6 36 612
50–60 55 16 880 16 256 4096
60-70 65 9 585 26 676 6084
Total 100 3900 20000munotes.in
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Standard Deviation σ=2fdx
f
=20000
100=200=14.1
The same problem can also be solved with the step deviation method
which is useful when the numbers are large.
Taking assumed mean, A = 35,35xdxc ,where c = 10
Interval Mid-
point
XFrequency
ƒ35xdxcdx2fdxƒdx2
10–20 15 9 -2 4 -18 36
20–30 25 18 -1 1 -18 18
30–40 35 31 0 0 0 0
40–50 45 17 1 1 17 17
50–60 55 16 2 4 32 64
60-70 65 9 3 9 27 81
Total 100 Total 40 216
S.D. =σ=22. .f dx f dxx cf f
=2216 4010100 100x
=2 10x = 14.1421
Relative measure of standard deviation or coefficient of variation
CV=.S D
meanx 100 =14.14
39x 100 = 36.26
Exam ple
The marks of internal assessment obtained by FYBMS students in a
college are given below. Find the mean marks and standard deviation.
22 30 36 12 15 25 18 10 33 29
Solution :We first sum all the observations and find the mean. Then
the differences of the observations from the mean are computed and
squared. The positive square root average of sum of square of the
differences is the required standard deviation.
x d x x d2
22 -1 1
30 7 49
36 13 169
12 -11 121
15 -8 64
25 2 4
18 -5 25
10 -13 169
33 10 100
29 6 36
x= 230 -2d=
738I.xxn=2302310.
II.2738
10d
n
=73.8 8.59
8.59munotes.in
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Example
Find the standard deviation for the following data:
03 12 17 29 10 05 18 14 12 20
Solution :We find the sum of the obser vations and the sum of its
squares. Using formula 2.2, S.D. is computed as follows:
x x2
3 9
12 144
17 289
29 841
10 100
05 25
18 324
14 196
12 144
20 400
x= 1402x=
2472
Example
Compute the st andard deviation for the following data
x 100 102 104 106 108 110 112
f 5 11 7 9 13 10 12
Solution : Short -cut method :
In problems were the value of xis large (consequently its square also
will be very large to compute), we use the short -cut method. In this
method, a fixed number x0(which is usually the central value among x)
is subtracted from each observation. This difference is denoted as
u=x–x0. Now the columns of fuandfu2are computed and the S.D. is
calculated by the formula:2 2.f u fu
N N . One can observe that
this formula is similar to that mentioned in 2.3. This formula is called
as change of Origin formula.
In this problem we assume x0= 106. The table of calculations is as
follows:
x 106 u x f fu fu2
100 -6 5 -30 180
102 -4 11 -44 176
104 -2 7 -14 28
106 0 9 0 0
108 2 13 26 52
110 4 10 40 160
112 6 12 72 432
Total -N=
67fu=
502.f u =
1028The formula 2.2gives the S.D. as below:
2 2x x
n n =22472 140
10 10
=247.2 196 51.2
In this table, the
column of fu2is
computed by
multipl ying the
entries of the
columns fuand
u.munotes.in
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From the table we have:2.f u= 1028,fu= 50 and N= 67.
2 2 2. 1028 50
67 67 f u fu
N N
15.34 0.56 14.78
3.84
Some important points to be noted regarding Standard deviation :
The standard deviation being an algebraic quantity, it possesses the
following important characteristics :
1. It is rigidly defined
2. It is based on all the observations, i.e. the value of the standard
deviation will change if any one of the observations is changed.
3. In the case of the value which lies close to the mean, the
deviations are small and therefore variance and standard deviation
are also small. Variance and standard deviation would thus be zero
when all the values are equal.
4. If the same amount is added to or subtracted from all the values,
the mean sh all increase or decrease by the same amount; also
deviations from the mean in the case of each value would remain
unchanged and hence variance and standard deviation shall remain
unchanged
5. In case a number of samples are drawn from the same population,
it may be observed that standard deviation is least affected from
sample to sample as compared to other three measures of
dispersion.
Limit ations : Standard deviation lays down the limits of variability by
which the individual observation in a distributio n will vary from the
mean. In other words, Mean ± 1 standard deviation will indicate the
range within which a given percentage of values of the total are likely
to fall i.e., nearly 68.27% will lie within mean ± 1 standard deviation,
95.45% within mean ± 2 standard deviation and 99.73% within mean ±
3 standard devi ation.
The point may be illustrated by taking an example of distribution of
weight of 1000 school students with a mean height of 40 Kgs and
standard deviation of 6 Kgs. If the groups of students is a normal one,
about two thirds (68.26) of the students wo uld fall within ± 1 standard
deviation from the mean. Thus 683 students would weigh between 34
and 46 i.e., 40 ± 6 Kg. Further, when 2 standard deviations are added
and subtracted from the mean, the total population covered would be
95.44% and in the case of 3 standard deviation it would cover 99.73%.
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UNIT IV
Unit -6
ELEMENTARY PROBABILITY THEORY
Unit Structures :
6.0 Objectives
6.1 Introduction
6.2 Types of Events
6.3 Algebra of Events
6.4 Solved Examples
6.5 Counting Principle and Combination
6.6 Random Variable and Expected Value
6.7 Normal Distri bution
6.0OBJECTIVE S
To understand the uncertainty (chance) involved in the
unpredictable events.
To find the probability (numeric value of the uncertainty) and
various rules of probability to measure the uncertainty.
To find expected value (Mean) and variance in random
experiments.
Use of normal distribution to find proportions and percentage of
the observations referred to certain continuous variables.
6.1 INTRODUCTION
In our day -to-day life conversation we here the statements like Most
probably it will rain today. Or a sales manager makes the claim the
sales will cross Rs.500 cores.
Both these statements show that the claims are subject to uncertainty
and cannot be predicted in advance with 100% guaranty.
Probability measures the certainty in suc h type unpredicted
events. The origin of probability lies in Gambling or the games of
choices such as tossing a fair coin, throwing a cubic die or removing a
card from a pack of playing cards. Today probability plays an
important role in the field of Econo mics, Finance, and Medicine etc.
for making inferences and predictions.
To understand the concept of probability and learn the methods
of calculating the probabilities, we should first define understand some
basic terms and concepts related to probabilit y.munotes.in
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Random Experiment: Any act or trial in which we are not sure
about the result is called as the random experiment.
e.g.. Tossing a fair coin. Throwing a cubic die. Removing a number in
the game of Housie.
Outcome: The possible result of the random exp eriment is called an
outcome.
e.g. When we toss a coin, there are two possible outcomes Head(H)
and Tail(T).
or when we throw a cubic die the possible outcomes of the no of dots
oen the uppermost face are 1,2,3,4,5,or6.
Sample Space: The collection of al l the possible outcomes in the
Random experiment is called the sample space. It is denoted by S. The
outcomes listed in the sample space are called the sample points. The
sample space may be finite, countable infinite or infinite in nature. The
no of sampl e points in the sample space is denoted by n(S).
e.g. When we toss a pair of unbiased coins, the sample space is
S =HH,HT, TH, TT n(S) =4
Or when a cubic die is thrown the sample space is
S =1,2,3,4,5,6 n(S) =6
Event: An event is a well -defined subset of the sample space. It is
denoted by the letters like A,B,C etc. The no of sample points in the
event is denoted by n(A).
e.g. In the experiment of throwing a cubic die when the sample space
is
S =1,2,3,4,5,6 . We can define the events as follow s
Event A: The no of dots appeared is multiple of 3.
A =3,6 n(A)=2
B: The no of dots appeared is a prime number.
B =2,3,5n(B) =3munotes.in
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6.2TYPE OF EVENTS
Simple event -The event containing only one sample point is called a
simple event.
e.g Tossing a pair of coins S = HH,HT, TH, TT
Now the event defined as
A: Both coins show Head.
Is a simple event.
Null event: It is the event containing no sample point in it is called a
null or impossible event. It is the impossible happening and is denoted
by ‘’. e.g in the experiment of throwing a cubic die when the sample
space is
S =1,2,3,4,5,6 .We define th e event
Event A: The no of dots appeared is a two digit number
A =i.e. A=and n(A)=0.
6.3ALGEBRA OF EVENTS
Union of two events A and B (AUB): union AUB is the event
containing all the sample points in A and B together.
AUB = {Elements in eit her A or B or in both A and B together}
= {x such that x A or xB}orAlB both
e.g. When A={ 1,2,3} and B={2,3,5}
AUB= {1,2,3,5}
Intersection of two events A and B (A B): For the events A and B
defined on the sample space S associated with the random experimentRandom Experiment Event
Tossing a pair of coins
S =HH,HT, TH, TT ,n(S) =4A:Both coins show Head. A = {HH}
B: At lest one coin show Head.
B = {HT,TH,HH} n(B)=3
Tossing three coines at a
time
S =HHH, TTT, HHT, HTH,
THH,HTT,THT,TTH , n(S) =8
________ __________________A: exactly two Head turns up.
A: {HHT, HTH, T HH}
Throwing a pair of cubic
dice.
S=(1,1), (1,2), (1,3),( 1,4),
(1,5), (1,6)… (5,6), (6,6)
n(S)= 36A: The sum of the dots on the uppermost
faces is 6 or 10
A:{(1,5) (5,1), (2,4) (4,2) (3,3) (4,6)
(6,4) (5,5)} n(A)= 8
B: The sum o f the dots on the uppermost
faces is divisible by 4.
B:{(1,3),(3,1),(2,2),(2,6),(6,2),(3,5),(5,3)
,(4,4),(6,6)} n(B)=9
Selecting a two digit number
S= {10,11,12,…….99}
n(S)=100}A: The number is a perfect square.
A:{ 16,25,36,49,64,81} n(A)=6
B:The number is >99. B:{ } n(B) =0
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E, intersection of A and B is the event containing all the sample points
common to A and B both.
AB = {Elements in A and B both}
= {x such that x A and xB}
e.g. When A={ 1,2,3} and B={2,3,5}
AB= {2,3}
Mutually exclusive events: The two e vents A and B are said to be
mutually exclusive or disjoint if they have no common element in
them. i.e. their intersection is an empty set.
Disjoint events cannot occur simultaneously.
e.g. When A={ 1,2,3} and B={4,5}
Mutually exclusive and exhaustive e vents: The two events A and B
are said to be mutually exclusive and exhaustive if they are disjoint and
their union is S.
e.g.when a cubic die is thrown the sample space is
S =1,2,3,4,5,6
Now if the events A and B defined on S have the sample points as
follows
A={1,2,3} and B={4,5,6} then A B=and AUB=S
Hence A and B are mutually exclusive and exhaustive.
Complement of an event A : Let A be any event defined on the sample
space S, then it’s complement A’ is the event containing all; the
sample points i n S which are not in A.
A’={elements in S which are not in A}
A= {x s.t. x S but xA}
e.g. When S = 1,2,3,4,5,6 and A={2,3}
A’={1,4,5,6}.
Probability of an event A: (Classical definition) Suppose S is the
sample space associated with the random experim ent E, and A is any
event defined on the sample space S, then it’s probability P(A), is
defined as
P(A) =( )
( )n A
n S
In other words probability of A is the proportion of A in S.
Example: A cubic die is thrown up find the probability that ,the no of
dots appeared is a prime number.
Solution: When a cubic die is thrown the sample space is
S =1,2,3,4,5,6 n(S) =6
Now we define the event A as,
A: the no of dots appeared is a prime number.
A=2,3,5 n(A) =3
Using the above formula, we get
P(A) =( )
( )n A
n S=3
6=1
2munotes.in
103
From the above we can note that,
Probability of event A always lies between 0 and 1 i.e. 0 P(A)1.
P(Sample spa ce)=P(S)=1 and P( ) = 0.
Also, when events A and B are such that A B then, P(A) P(B).
6.4 SOLVED EXAMPLES
Example 1:
An unbiased die is thrown, find the probability that, i) the no of dots is
less than 3 ii) the no of dots in divisible by 3.
Solution: when a cubic die is thrown the sample space is
S =1,2,3,4,5,6 n(S)=6
(i)Let A denote the event that no of dots <3. A={1,2} i.e. n(A)=2
...P(A) =( )
( )n A
n S=2
6= 0.33
.
(ii)Now let B denote the event of no of dots is divisible by 3.
B: {3,6} and n(B) = 2
...P(B)=( )
( )n B
n S=2
6= 0.33
Example 2:
Three unbiased coins are tossed at a time. Find the probability that, (a)
exactly two Head t urns up and (b ) at most two Head turns up.
Solution : When three coins are tossed up at a time the sample is
S =HHH, HHT, HTH, THH,HTT,THT, TTH, TTT n(S) =8.
Now to find the required probability, we define the events as follows,
Event A: Exactly two Head turns up.
A= { HHT, THH, HTH} n(A)= 3
...P(A) =( )
( )n A
n S=3
8
Event B: At most two Head turns up.
B:HHT, HTH, THH,HTT,THT, TTH, TTT n(B) =7.
...P(B) =( )
( )n B
n S=7
8.
Exampl e 3:
A pair of fair dice is rolled. Write down the sample space and find the
probability that, a) the sum of dots o n the uppermost face is 6 or 10, b)
the sum of dots is multiple of 4 and c) the sum of the dots is < 6.
Solution: When a pair of dice is rol led, the sample space is
S=(1,1), (1,2), (1,3),( 1,4), (1,5), (1,6)
(2,1)…………………………..(2,6)
……………………….. (5,6), (6,6) n(S)= 36.
To find the probability we define the events,munotes.in
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a) event A: the sum of dots on the uppermost face is 6 or 10.
A= {(1, 5) (5,1), (2,4) (4,2) (3,3) (4,6) (6,4) (5,5)} n(A)= 8.
...P(A) =( )
( )n A
n S=8
36
b) event B: The sum of the dots on the uppermost faces is divisible
by 4.
B: {(1,3),(3,1),(2,2),(2,6),(6,2),(3,5),(5,3),(4,4) ,(6,6)} n(B)=9.
...P(B) =( )
( )n B
n S=9
36=0.25.
c) event C: the sum of the dots is < 6.
C: { (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1)} n(C) = 8.
...P(C)=( )
( )n C
n S=8
36.
6.5COUN TING PRINCIPLE AND COMBINATION
In some experiments like selecting cards, balls or players we
cannot list out the complete sample space but count the no of sample
points in the given sample space. To count the no of points in the
sample space in these experiments, we define the concepts of
combination. First, we state the Counting principle (Fundamental
Principle of Mathematics), as follows
Counting principle: If A and B are two different things can be
independ ently performed in m and n different possible ways then, by
counting principle
(a)both A and B together can be performed in m.n possible ways.
(b)any one of them i.e. A or B can be performed in m+n possible
ways.
Combination : Combination of r different things from n things e.g.
selecting 3 balls from 5 balls, or 4 students from the group of 10
students of a class. It is calculation by the formula,
n
rC=!
! !n
r n rwhere n!= n (n -1) (n -2)-----3x2x1.
(e.g. 5!= 5x4x3 x2x1=1 20)
Illustrations -
(i) We can select a group of 3 students from 5 students in5C3ways and
5C3=5!
2!(5 2)!=5!
2!3!=5 4 3 2 1
2 13 2 1x x x x
x x x=10.
(ii) A student can select 4 different questions from 6 indepen dently in
6
4Cways.
6C4=6!
2!(6 4)!=6!
2!4!6 5 4 3 2 1
2 1 4 3 2 1x x x x x
x x x x x=15.munotes.in
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(iii) A box contains 4 red and 6 green balls then by counting principle,
a) 2 red and3 green together can be drawn in4C2x6C3ways.
b) 3 red or3 green can be drawn in4C3+6C3ways.
Example 4:
Three cards are drawn from the pack of 52 playing cards.Find the
probability that (a) all three are spade cards, (b) all three are of same
suit, and (c)there are two Kings and one Quee n.
Solution: When three cards are drawn from the pack of 52 cards, no of
points in the sample is
n(S) =52C3=52!
3! 52 3 !=51x51x50
3x2x1=22100.
Now, we define the event,
a) A: all three are spade cards.
There are 1 3 spade cards, so 3 of them can be drawn in13C3
n(A)=13C3=13! 13x12x11
3!10! 3x2x1 =286
...P(A)=( )
( )n A
n S=286
22100=0.0129.
a)event B: all three are of same suit
There are four suits ( Club, Diamond, Heart & Spade ) of 13 cards each.
Three from them can be drawn in13C3ways each. Now by the
counting principle,
n(B) =13C3+13C3+13C3+13C3= 4x 286 = 1144. (refer to
illustrations above)
Hence P(B) =1144
22100=0.051.
b)event C: there are two K ings and one Queen.
There are 4 kings and 4 Queens in the pack. So 2 kings and 1 Queen
can be drawn in4C2and4C1respectively.
Therefore by counting principle,
n(C) =4C2x4C1= 6x4 = 24.
Where,4C2=4! 4x3x2x1
2!2! 2x1x2x1 =6 &4C14! 4x3x2
1!3! 3x2x1 =4.
...P(C) =( )
( )n C
n S=24
22100
Example 5:
A box contains 5 Red and 4 Green balls. Two balls are drawn at
random from the box, find the probability that I) Both are of same
color ii) onl y red balls are drawn.
Solution : In the box there 9(4+5) balls in total, so 2 of them can be
drawn in9C2ways.
n(S)=9C2=9! 9x8
2!7! 2 1x=36.munotes.in
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(i) We define the event A: Both are of same color
n(A)=4C2+5C2=4! 5!
2!2! 2!3! = 6+10 =16.
...P(A) =( )
( )n A
n S=16
36=0.44
Hence probability of same colour is 0.44
(ii) Let us define event B: only red balls are drawn
Now both the balls will be drawn from 4 red balls only
n(B)=4C2=6
...P(B)=( )
( )n B
n S=6
36=0.16
Hence probability of both red balls is 0.16.
Exercise
1.Define the terms, i) Sample space. ii) An event. iii)
Mutually exclusive ev ents
2.Define the probability of an event. Also state the properties of the
probability of event.
3.An unbiased coin is tossed three times, write down sample points
w.r.t. following events: (a) Head occur only two times ,(b)Head
occur at least 2 times and (c )There are more Heads than Tail.
4.A pair of fair dice is rolled, write down the sample points w.r .t.
following events: The sum of the no of dots appearing on the
uppermost faces is (i) 7 or 11 (ii) multiple of 3 (iii) a
perfect square.
5.A pair of coins is t ossed at a time find the probability that,
6.Both the coins show Head. ii) No coin show sHead. iii) Only one
Head turns up.
7.A cubic die is thrown, find the probability that the no of dots
appeared is (a)A prime number, (b) A number multiple of 2.
8.A box co ntains 20 tickets numbered 1 -20. A ticket is drawn at
random from the box, find the probability that, i) the ticket bares a
number < 5. ii) the number on the ticket is divisible by 4. iii) it is a
cube of a natural number.
9.A card is drawn from the pack of 52 playing cards find the
probability that, (a)The card is a king card (b) It is a face
card.
10.If two fair dice are thrown , what is the probability the sum of the
no of dots on the dice is, a) greater than 8. b) between 5 and 8.
11.Three unbiased coins are tossed at a time find the probability that,
(a)exactly one Head turns up. (b) At most 2 Heads turn up. (c)All
3 coins show Heads.
12.Two cards are drawn from the pack of 52 playing cards find the
probability that, (a)Both the cards are of same suits. (b)Both are
Ace cards. (c) Both are of different colour.munotes.in
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13.A pair of dice is rolled, write down the sample space and find the
probability that, the sum of the no of dots appearing on the
uppermost faces is, (a) 6 or 10. (b) multiple of 5. (c) a
perfect square.
14.Four unbiased coins are tossed at a time find the probability that,
(i)exactly 2 Heads turn up. (ii) at most 3 Heads turn up. (iii) at least
3 coins show Heads.
15.Three cards are drawn from the pack of 52 playing cards find the
probability that, (i)all 3 c ards are of same colour. (ii) Two cards
are face cards. 9iii) Only face cards are drawn.
16.Two boxes identical in size and shape respectively contain 3 red, 4
blue and 5 red, 2 blue balls. One ball is drawn at random from each
box. What is the probability th at both the balls are of same colour.
17.A committee of 4 is to be formed from 3 Professors and 7 students
in a college. Find the probability that it includes, a) only 2
Professors. b) there are at least 3 students.
18.A box contains 6 red, 4 green and 3 white balls. Two balls are
drawn at random, find the probability that, a) both are of same
colour. b) no white ball is drawn. c)the balls are of different
colour.
19.Given P(A)=0.6, P(B)=0.5, and P(A B)= 0.4. Find P(AUB);
P(A/B); P(B/A) .
20.For two events A and B; P(A)=2
5, P(B’) =1
3, P(AUB)=5
6.Find
P(AB); P(A/B); P(only A); P(Only one).
21.For two mutually exclusive events A and B, P(A)=0.7 and
P(B)=0.5, find P(A B)andP(A/B)
22.For the independent events A and B, P(A)=1
2, P(B) =2
5. Find
P(AB); P(AUB);P(only B).
23.One of the two purses contains 4 Gold coins and 5 Silver coins,
another purse contains 3 Gold and 6 Silver coins. A coin is drawn
at random from one of the purses, find the probability that it is a
silver coin.
24.Two students A and B are solving a problem on Mathematics
independently. Th eir chances of solving the problem are1
2and
1
3respectively. Find the probability that, i) the problem is solved.
ii) it is solved by only oneof them.
25.A government contractor applies for 2 tenders of suppl ying
breakfast supply and lunch box supply. His chances of getting the
contracts are 0.6 and 0.5 respectively. Find the probability that, (i)
hewill get either of the contract (ii) he will get only one contract.munotes.in
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6.6RANDOM VARIABLE AND EXPECTED VALUE
Random variable is a real valued function defined on the sample space.
Suppose S is the sample space associated with the random experiment
E, then to every sample point in S we can assign a real number denoted
by a variable X called as a random variable on S. e.g. When we toss a
coin three times the sample is
S={HHH, HHT, HTH, THH,HTT,THT, TTH, TTT , now if we define
a variable
X: No of tosses showing Heads.
Then X takes values 0,1,2,3 where
{X=0}{TTT}; {X=1}{HTT, THT, TTH};
{X=3}{HHH} {X=2} {HHT, HTH,
THH}.
Hence to each sample points in S we have assigned a real number,
which uniquely determine the sample point.
The variable X is called as the random variable defined on the sample
space S.
We can also find the probabilities of value s 0,1,2,3 of the r.v. X as
follows
P ({X=0}) = P ({TTT}) = 1/8, P ({X=1}) = P ({HTT, THT, TTH) =
3/8,
P ({X=3}) = P ({HHH}) = 1/8,P ({X=2}) = P ({HHT, HTH, THH}) =
3/8.
Now we can express these probabilities in the form of a table,
X: 0 1 2 3
P (x) 1/8 3/8 3/8 1/8
This is called as th e probability distribution of
random variable X.
In general probability distribution of X satisf iesthe following
conditions;
(i)allp(x) are positive. i.e. p(x) 0
(ii)p(x) =1for all x .
Arandom variable Xdefined on the sample space S may be finite or
infinite, at the same time it may take only countable values (without
decimal) such variables are called as discrete random variables. On the
other hand some variables like height, weight, income do take the
fractional values also and called as the continuous random variables.
Expected value of X ,E(X) :
Suppose a random variable X defined on sample space S takes values
x1,x2,x3,-----xn.with respective pro babilities p 1,p2,p3,-----pn; P(x= x 1)
= p 1,it’s expected value is defined as,
E(X) =x.p(x)
Expected value is also called as the mean of X.X P (x)
0 1/8
1 3/8
2 3/8
3 1/8
Total 1munotes.in
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Variance of X ; V(X): For the random variable X, variance is defined
as,
V(X) = E( X -E(X))2
=E(X2)–[E(X)]2.
=x2.p(x) –(x.p(x))2
Root of variance is called as standard deviation S.D.
Solved examples :
Example 6:
A discrete random variable X has the following probability
distribution.
x: -2 -1 0 1 2
p(x) k 0.2 2k 2k 0.1
Find k. Also find the expected value of random variable .X
Solution : Since X is a random variable with given p(x), it must satisfy
the conditions of a probability distribution.
p(x) = 1 5k+0.3=1 k=
0.7/5=0.14
p(x) values are 0.14;0.2;0.28;0.28;0.1
Now we calculate the expected value
by
the formula,
E(X) =x.p(x) =0------ from the table.
Example 7:
A random v ariable follows the probability distribution given below,
X 0 1 2 3 4
p(x) 0.12 0.23 0.35 0.20 0.10.
Obtain the expected value and variance of X.
Solution : The expected value and variance are given by the formula,
E(X) =x.p(x) and V(X) =x2.p(x) –(x.p(x))2
Now from the table,
E(X) =x.p(x) = 1.93.
V(X) =x2.p(x) –(x.p(x))2
= 4.03 -(1.93)2
V(X) =0.35.
Hence, Mean E(X) =1.93 units and V(X) = 0.35 units.x p(x) xp(x)
-2 0.14 -2x0.14
=-0.28
-1 0.2 -0.2
0 0.28 0
1 0.28 0.28
2 0.1 0.2
Total 1 0=x.p(x)
x p(x) xp(x) x2.p(x)
0 0.12 0x0.12=0 0x0=0
1 0.23 0.23 0.23
2 0.35 0.70 1.40
3 0.20 0.60 1.80
4 0.10 0.40 1.60
Total 1.00 1.93=x.p(x) 4.03 =
x2.p(x)munotes.in
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Example 8:
Find mean and variance of the random variable X whose probability
distribution is given by
X: -2 -1 0 1 2
P(x) 1/16 1/8 5/8 1/8 1/16.
Solution : For the random variable X, we have
From the table we get,
E(X) =x.p(x) = 0.
And V(X) = x2.p(x) –(x.p(x))2
=12
16-0 =12
16
Example 9:
A uniform die is thrown find the expcted value of the random variable
X denoting the n o on the uppermost face.
Solution : When a uniform die is thrown the random variable
X: the no on the uppermost face, takes the possible values 1,2,3,4,5 or
6.
With the same probability of occurrence.
Therefore we can find the mean or expected value of X b y using the
formula,
E(X) =x.p(x)
=21
6=3.5 ------ from the
table.
Hence the mean of X is 3.5.
EXERCISE II
1. A random variable X has the following probability distribution:
X: -2 -1 0 1 2 3
P(x) 0.1 k 0.2 2k 0.3 k
Find the value of k. Find the expected value and variance of x.x p(x) xp(x) x2.p(x)
-2 1/16 -2/16 -2x(-2/16) = 4/16
-1 1/8 -1/8 1/8
0 5/8 0 0
1 1/8 1/8 1/8
2 1/16 2/16 4/16
Total 1.00 0=x.p(x) 12/16 =x2.p(x)
x p(x) xp(x)
1 1/6 1/6
2 1/6 2/6
3 1/6 3/6
4 1/6 4/6
5 1/6 5/6
6 1/6 6/6
Total 6/6 =1 21/6
=x.p(x)munotes.in
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2. A random variable X has the following probability distribution:
X: 0 1 2 3 4 5
P(x) 0.1 0.1 0.2 0.3 0.2 0.1
Find the expected value and variance of x.
3.An unbiased coin is tossed four times. Find the expected value and
variance of the random variable defined as number of Heads.
6.7NORMAL DISTRIBUTION
Normal distribution deals with the calculation of probabilities for a
continuous random variable like Height of players, Marks of students,
or Wages of workers. We define the normal distribution as follows.
A continuous random variable X is said follow a normal distribution
with param etersand 6, written as X N(, 62) if it’s probability
function is given by
p(x) =1
221
2x
e
where x,R,σ>0
= 0 otherwise.
Here the constants are = Mean(X); σ= S.D.(X)
= 3.142 and e= 2.718 (appro x).
Before we learn to calculate the probabilities on normal distribution,
we state the charac teristics of the normal distribution stated below.
Characteristics of the normal distribution
i)The graph of normal distribution is a bell shaped curve.
ii)The area under the curve reads the probabilities of normal
distribution hence total area
is 1 ( one).
iii)The curve is symmetric about it’s mean . Hence,
Area on l.h.s. of = Area on r.h.s. of = 0.5. Since area reads
probability,
P(X<) = P(X> ) = 0.5= 50%.
iv)Hence mean divides the curve into two equal parts so it is also
the median.
The curve has it’s maximum height at x = , therefore it the mo de of
the distribution.
v)Hence for normal distribution Mean = Median = Mode =
For the probability calculations, we define the variable
Z =x
. .Mean
S D=x
Mean (Z) = 0 and S.D.(Z)= 1. Z is called a standard normal variable
(s.n.v.)munotes.in
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Also P (X) = P (Z) = Area (Z). The area (probability) values of z are
tabulated.
vi)The lower (Q 1) and upper quartiles (Q 3) are equidistant from the
mean
. i.e.-Q1= Q 3-=1 3
2Q Q
vii)The mean deviation (M.D.) of normal distribution is4
5
viii) The quartile deviation (Q.D.) of normal distribution is 0.67 .
Area under the normal curve between
(i)+is 68.27% (ii)+2is 95.45%
(iii)+3is 99.73%. (iv)+1.645is 90%
(v)+1.96is 95% (vi)+0.67is 50%.
Solved examples:
Example 10:
A continuous random variable X follows a normal distribution with
mean 50 and S.D. of 10. Find the following probabilities for X,
a) P (X55) b) P(45X60) c) P(X45).
Solution : For the normal variable X , we have
Mean (X)= = 50 and standard deviation = 6= 5.
...X (N (, 62) = N (50, 102).
We define the variable Z =x
=x 50
5
Mean (Z) = 0 and S.D.(Z)= 1. Z is called a stan dard normal variable
(s.n.v.)
Also P (X) = P (Z) = Area (Z).
a) P (X55)= P(x 50
555 50
5)
= P(Z1)
= Area r.h.s. of +1
= 0.5 -Area from 0 to 1
= 0.5 -0.3413 = 0.1587.
b) P(45X60) = P(45 50
5x 50
560 50
5)
= P(-1Z2)
= Area between -1 & +2
= Area from -1 to 0+ Area from 0 to 2.
= 0.3413 + 0.4772 = 0.8185.
-1 0 2
c) P(X45)=P(x 50
545 50
5)
= P(Z-1)
= Area on l.h.s. of -1.
= 0.5 -Area from -1 to 0.
= 0.5 -0.3413 = 0.1587.
munotes.in
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Example 11:
The marks of 150 students in the class is said to follow a normal with
mean 60 and S.D. of 10. Find, the expected no of students scoring
marks below 45.Percentage of students scoring marks between 55 and
70.
Solution : Let X: Marks of students;
Mean(X) = 60 and S.D.(X) = 6 = 10.
X has normal distribution with =60 and 6 = 10 .
i.e. X( N ( , 62) = N (60, 102)
We define Z=x
=x 60
10.
...Mean (Z) = 0 and S.D.(Z)= 1. Z is called a standard normal variable
(s.n.v.)
Also P (X) = P (Z) = Area (Z).
Now, to find the expected no of studen ts scoring marks below 45, we
find
P(marks less than 45)
= P(X45) =.P(x 60
1045 60
10)
= P(Z-1.5)
= Area on l.h.s. of -1.5.
= 0.5 -Area from -1.5 to 0.
= 0.5 -0.4332 = 0.0668 = 6.68%.
Expected no of stude nts = 6.68%(150) =10.
Similarly, to find percentage of students scoring marks between 55 and
70.
Consider, P(marks between 55 and 70)
= P(55X70)
=P(55 60
10x 60
1070 60
10)
= P(-0.5Z1)
= Area between -0.5 & +1
= Area from -0.5 to 0+ Area from 0 to 1.
= 0.1915 + 0.3413= 0.5328 = 53.28%.
...53.28% students have scored marks between 55 and 70.
Example 12:
The height of 250 soldiers in a military camp confirm s a normal
distribution with mean height of 155cms.and S.D. of 20cms. Find the
proportion of soldiers with height above 170 cms. Also find the height
of the shortest soldier in the group of tallest 20% soldiers.
Solution : Let r.v. X denotes the height of a soldier.
Mean(X) = 155 and S.D.(X) = = 20.
...X has normal distribution with =155 and = 20.
i.e. X N ( ,2) = N (155, 202)
munotes.in
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We define Z=x
=x 155
20.
...Mean (Z) = 0 and S.D.(Z)= 1. Z is called a standard normal variable
(s.n.v.)
Also P (X) = P (Z) = Area (Z).
Now, to find the proportion of soldiers with height above 170 cms, we
find,
P(soldier's height is above 170 cms)
P (X170) = P155 170 155
20 20x
= P(Z0.75)
` = Area r.h.s. of 0.75
= 0.5 -Area from 0 to 0.75.
= 0.5 -0.2734 = 0.2766.
...Proportion of soldiers with height above 170 cms is
(0.2766x250 ):250
i.e.69:250.
Now, let the height of the shortest soldier in the group of tallest 20%
soldiers be h.
P(height less than h) =20%=0.2 i.e. P(Xh) =.0.2
Consider, P(X h) = P(x 155
20h 155
20= t say) i.e. P(Z t) = 0.2.
Area on l.h.s. of t =0.2( t is less than 0 i.e. negative.( since area on
l.h.s.< 0.5)
area from t to 0 = 0.3.
Now from the normal area table, area from 0 to 0.84 is 0.3.
Hence, t = -0.84 ( t=h 155
20=-0.84 i.e. h = 155 + 20( -0.84) =155 -
16.8= 138.2.
Therefore, the height of the shortest soldier in the g roup of tallest
20% soldiers is 138 cms.
Example 13:
The daily wages of 300 workers in a factory are normally distributed
with the average wages of Rs.2500 and S.D. of wages equals to
Rs.500. Find the percentage of workers earning wages between
Rs.3000 an d Rs.4000. Also find the wages of the lowest paid worker in
the group of highest paid 30% workers.
Solution : Let r.v. X denotes the wages of a worker.
Mean(X) = 2500 and S.D.(X) = = 500.
...X has normal distribution with =2500 and = 500.
i.e. X( N (,2) = N (2500, 5002)
We define Z=x
=x 2500
500
...Mean (Z) = 0 and S.D.(Z)= 1. Z is called a standard normal variable
(s.n.v.)
Also P (X) = P (Z) = Area (Z).
munotes.in
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Now, to find the percentage of workers earning wages between Rs
3000 and Rs 4000.
Consider, P(wages between Rs.3000 and Rs 4000)
= P(3000 X4000)
=P(3000 2500
500x 2500
5004000 2500
500)
= P(1Z1.5 )
= Area between +1 & +1.5
= Area from 0 to 1. 5-Area from 0 to 1.
= 0.4332 -0.3413= 0.0919 = 9.19%.
...9.91% workers are earning wages between Rs 3000 and Rs 4000.
Now, let the wages of the lowest paid worker in the group of highest
paid 30% workers be h.
P(wages greater than h) =30%=0.3 i.e. P(Xh) =
0.3.
Consider, P(X h) = P(x 2500
500h 2500
500= t say)
i.e. P(Zt) = 0.3.
0
t=0.2
...Area on r.h.s. of t =0.3 ( t is greater than 0 i.e. positive.( since area
onr.h.s.< 0.5)
area from 0 to t = 0.2.
Now from the normal area table, area from 0 to 0.52 is 0.2.
Hence, t = 0.52 h 2500
500= 0.52h = 2500 + 500(0.52) = 2760.
Therefore, the wages of the lowest paid worker in the group of highest
paid 30% workers are Rs. 2760.
EXERCISE III
1.Define a normal variable. State the properties of normal
distribution.
2.What is mean by a standard normal variable. What are the mean and
standard of a a standard n ormal variable.
3.A continuous random variable X follows a normal distribution with
mean 50 and S.D. of 10. Find the following probabilities for X,
a) P (X55) b) P(45X60) c) P( X45).
Given, Area under the normal curve,
From 0 to 1 is 0.3413.
From 0 to 2 is 0.4772.
4.The marks of 150 students in the class is said to follow a normal
with mean 60 and S.D. of 10. Find, the expected no of students scoring
marks below 45.Percentage of students scoring marks between 55
and 70.
munotes.in
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Given, Area under the normal curve,
From 0 to 0.5 is 0.1915.
From 0 to 1 is 0.3413.
From 0 to 1.5 is 0.4332.
5.The height of 250 soldiers in a military camp confirms a normal
distribu tion with mean height of 155cms.and S.D. of 20cms. Find the
proportion of soldiers with height above 170 cms. Also find the height
of the shortest soldier in the group of tallest 20% soldiers.
Given, Area under the normal curve,
From 0 to 1.5 is 0.4332.
From 0 to 0.84 is 0.3.
6.The daily wages of 300 workers in a factory are normally
distributed with the average wages of Rs.2500 and S.D. of wages
equals to Rs.500. Find the percentage of workers earning wages
between Rs.3000 and Rs.4000. Also find the wag es of the highest paid
worker in the group of lowest paid 30% workers.
Given, Area under the normal curve,
From 0 to 1 is 0.3413.
From 0 to 1.5 is 0.4332.
From 0 to 0.52 is 0.2.
7.A normal distribution has mean = 15 and 6= 5. Find the following
probabilities.
P(X20) P(10X17.5) P(X12).
Given, Area under the normal curve,
From 0 to 0.4 is 0.1554.
From 0 to 0.5 is 0.19 15.
From 0 to 1 is 0.3413.
8.The weights of 450 students in a school are normally distributed
with the average weight of 50 kg. and S.D.5 kg. Find the percentage of
students with weight:
i)less than 45 kg. ii) Between 40 and 47 kg.
Given, Area under the normal curve,
From 0 to 0.4 is 0.1554. From 0 to 0.5 is 0.1915.From 0 to 1 is
0.3413.
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UNIT V
Unit -7
STATISTICAL DECISION THEORY
Unit Structure:
7.0 Objectives
7.1 Introduction
7.0OBJECTIVES
After going through this chapter you will be able to understand:
The decision making situation.
The decision making criteria to arrive at opti mum decision.
The decision tree technique for multi -stage decision making.
7.1INTRODUCTION
The decision theory may be applied to problems whether it involves
financial management or a plant assembly line, whether the time span is
ten years, five years o r one day and whether it is in public or private
sector. In each of such decision -making problems, there are certain
common elements which are called ingredients of decision problems.
These ingredients are:
1. Alternative Courses of Action: The process of decision -making
involves the selection of a single act from among some set of alternative
acts. Decision is needed in a problem situation where two or more
alternative courses of action are available, and where only one of these
actions can be taken. Ob viously, if there is only one course of action
available, no decision is required since that action must be taken in order
to solve the problem. For the sake of simplicity t he possible actions are
symbolis ed by a1a2, a3, ... etc. The totality of all possib le actions is called
action space denoted by A.If there are only three possible actions, we
write A = action s pace = (a 1a2, a3,). The decision procedure involves
selecting among the alternatives a single course of action that can be
actually carried out. If such a course of action is selected that cannot be
carried out in the existing situation and circumstance, it will then amount
to waste of time and resources. Q uite often the objective of precision is to
select an act which will accomplish some predesi gnated purpose. The
decision taken may be regarded as satisfactory or not depending upon
whether it has helped in the attainment of that objective.munotes.in
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2. Uncertainty: In all decision problems "uncertainty" is found to be a
common element. When the outcome of some action is not known in
advance. the outcome is said to be uncertain. When there are many
possible outcomes of an event (also called states ofnature ) one cannot
predict with certainty what will happen —it is only in terms of
probabilities we may be ab le to talk. The term "state of nature" does not
mean nature in the ordinary sense of the word. it is a general term which is
used to encompass all those factors beyond the control of the decision -
maker that affect the outcome of his decision. The various s tates of nature
(outcomes) are symbolised by θ,1,θ2θ3...etc. Totality of all outcomes is
called nature spac e or state space symbolised by Ω. If an action leads to
three outcomes θ1,θ2andθ3then we write : Ω= (θ1,θ2,θ3).
For example, if a product is marketed it may be highly appreciated
(outcome θ1), it may not app eal to the customers (outcome θ2) or it may be
liked by a certain fraction of th e customers say, 25% (outcome θ3).
It must be pointed out that sometimes a distinction is made between
decision-making under "risk" and decision -making under "uncertainty".
When the state of nature is unknown, but objective or empirical data is
available so that the decision -maker can use these data to assign
probabilities to the various states of nature the pr ocedure is generally
referred to as decision -making under 'risk.' When the state of nature is
unknown and there is no objective information on which probabilities can
be based, the pro cedure is referred to as decision -making under
'uncertainty'. It may , however, be noted that even w hen no objective
information is available, the decision -maker may, in Bayesian decision
theory, assign subjective probability to the sta tes of nature to help in
taking a decision. Once probabilities are assigned, regardless of th e
manner in which they were obtained, the decision procedure that follows
isexactly the same. Hence for practica l purposes risk and uncertainty
areessentially the same and decision -making under both circumstances
will be referred to as decision -making under uncertainty.
3. Pay Offs: In order to evaluate each possible course of action, the result
ofeach event with each course of action ha savalue (or pay off) placed
upon it . A number of consequences result from each action under
different conditi ons , the conditions being various states of na ture, the
consequences will be more in number. In practical situations, p articularly
in business and economic problems, consequences can be expressed .in
terms of money and utility .
The consequences may be ev aluated in several ways such as:
(i) in terms of profit,
(ii) in terms of cost.
(iii)in terms of opportunity loss :(The opportunity loss is defined as the
difference between the highest possible profit for an event and the actual
profit obtained for the actual action taken .)
(iv) units of satisfaction or utility.munotes.in
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4. Decision criteria: The decision -maker must -determine how to select
thebest course of action. In most decision p roblems the expected
monetary value is used as a decision criterion. When consequences are
evaluated in terms of profit, they are called payoffs. A pay off table is
prepared and it shows the relation between all possible states of nat ure, all
possible actions and the values associated with the consequences. A
specimen of payoff ta ble is given below;
GENERAL FORMAT OF A PAYOFF TABLE
Acts
States of
naturea1 a2 a3 ...... ak ...... am
θ1 p11 p12 p13 p1k p1m
θ2 p21 p22 p23 p2k p2m
θ3 p31 p32 p33 p3k p3m
θi p/1 p/2 p/3
θn pn1 pn2 pn3 pnk pnm
In the above table, the column heading d esignate the various actions out of
which the decision -maker may choose while the row heading show sthe
admissible states of nature under which the decis ion-maker has to take
decision .
A payoff may be thought of as a conditional value or conditional profit
(loss). It is condi tional value in the sense that isassociated with each
course of action there is a certain profit (or loss), given that a specific state
of nature has occurred. A payoff table thus contains all conditional values
of all possible combinations and states of nature. The payoff table
indicates that there is no single act which is best for all the states of natue.
Therefore, in order to ma ke an optimal decision some criterion and
additional information are necessary.
The calculation of payoff depends on the problem . Very often it is a
relative ly easy matter and sometimes a bit algebraic reasoning is required .
With the payoff table, the decision -maker may be able to reach the optimal
solution of a problem, if he has a knowledge of what event is going to
occur. Since there is uncertainty about occurrence of events, a decision -
maker must make some prediction or forecast usually in terms of
probability of occurrence of events. With the probabilities of occurrence
assigned, the last st ep of statis tical decision theo ry is to analyse these
probabilities by calculating expected payoff (EP) of expected monetarymunotes.in
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value for each course of action. Th e decision criterion here is to choose
the O ptimal Act(OA) ,the act that yields the highest EP.
An alternative decision criterion of stati stical decision theory is what i s
called expected opportunity loss, EOL. This criterion also leads to the
same result as obtained from expected profits (EP).
Calculations of EOL are the same as those for EP except for the fact that
we have to use conditional opportunity loss (COL) instead of payoffs. It
may be pointed out that COL of the optimal acts is zero, C OL of any act
other than OAis positive and is the difference between the payoff of OA
and the act taken.
If we replace the payoffs by their corresponding opportunity losses we
get a new table called loss table.
Example 1:
A baker produces a certain type of special pastry at a total average cost of
Rs. 3 and sells it at a price of Rs. 5. This pastry is produced over the
weekend and is sold during the following week; such pastry being
produced but not sold during a week's time istotall y spoiled and h asto be
thrown away. According to past experience the weekly demand for these
pastries is never less than 78 or greater than 80. You are required to
formulate action space, payoff table and opportunity loss table.
Solution :It is clear from the problem g iven that the manufacturer will not
produce less than 78 or more than 80 pastries. Thus, there are three
courses of action open to him:
a1= produce 78 pastries
a2= produce 79 pastries
a3= produce 80 pastries
thus the action space or A= (a1,a2,a3)
The state of nature is the weekly demand for pastries. There are three
possible states of nature, i.e.,
θ1= demand is 78 pastries
θ2= demand is 79 pastries
θ3= demand is 80 pastries
Hence the state space Ω= (1,2,3)
The uncertainty element in the problem i s the weekly deman d. The bakery
profit conditioned by the weekly demand. Cell values of payoff table are
computed as follows:'
Demand x Price -Production x Cost
P11 = payoff when action a1is taken but the state of nature is θ1
= Rs. [5 x 78 -3 x 78] = Rs. 156.
P12 = payoff when action a2is taken but the state of nature is θ1
= Rs. [5 x 78 -3 x 79] = Rs. 153.munotes.in
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P13 =payoff when action a 3is taken but the state of nature is θ1
= Rs. [5 x 78 -3 x 80] = Rs.150.
Similarly P21= payoff when action a1is taken and the state of nature is θ2
= Rs. [5 x 78 -3 x 78] = Rs.156.
P22=Rs. [5x79 -3x79]= Rs. 158.
P23=Rs. [5x79 -3x80] = Rs. 155.
Also, P31= payoff when action a1is taken and the state of nature is θ3
P31=Rs. [5x78 -3x78]= Rs. 156.
P32=Rs. [5x79 -3x79] = Rs. 158.
P33=Rs. [5 x 80 -3x80] = Rs. 160
AnEMV indicates the average profit that would be gained If a particular
alternative were selected in many similar decision -making situations.
Since decisions arc often made on a one time basis, the decision criterion
is to choose the alternative course of action that maximizes the expected
monetary value.
These values are tabulated below:
PAYOFF TABLE
Action
State of nature a1=78 a2=79 a3=80
θ1= 78 156 153 150
θ2= 79 156 158 155
θ3=80 156 158 160
To calculate oppo rtunity loss, we first determine maximum payoff in each
state of natur e.
In first state of nature1,maximum payoff = Rs. 156
In second state of nature2, maximum payoff = Rs. 158
In third state of nature3, maximum payoff = Rs. 160
L11156-156 = 0,L12= 156 -153 = 3 ,L13= 156 -150 = 6
L21= 158 -156 = 2 ,L22= 158 -158 = 0, L23= 158 -155 = 3
L31= 160 -156 = 4, L32= 160 -158 = 2, L33= 158 -158 = 0.
The loss table corresponding to payoff table is given below.
Actio n
State of
naturea1 a2 a3
1 0 3 6
2 2 0 3
3 4 2 0
Example 2:Suppose in a process of producing bulbs, the number of
producing defe ctive bulbs is constant but unknown. Cons ider a lot of 100
bulbs which is either to be sold at Rs. 5 each giving a double money backmunotes.in
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guarantee for each defective item or to be junked at a cost of Rs. 100 for
the lot. Construct action space, state space, and payoff table.
Solution: Two possible a ctions are open to the manufacturer, namely,
a1= Junk the lot at a loss of Rs. 100 for the lot,
a2= Sell the lot at Rs. 5 each giving double money back guarantee for
each defective ite m.
Thus A =(a1a2)=action space
The number of defective bulbs in the lot designates the status of nature.
The number of defective bulbs can be 0, 1, 2, 3. .... 100. i.e.there are 101
possible states of nature. Let 0 idenote that there are " i"defective bulbs in
the lot.
Thus state space or Ω= {0 0, 01,02, . . . , 0 100}
Since there are 101 states of nature, it is very difficult to work out all
possible payoff. Suppose themanufacturer takes action a 1, then he will
lose Rs. 100.
Pi1=-100 for all i=0,1,2.....100
Suppose the manufacturer takes action a 2and the state of nature is that,
there are defective bulbs in the lo t then the payoff is given by
Pi2=100x5 -ix10= 500 -10i For all i= 0, 1, 2. .... 100.
Thus his payoffs are
Pi1=-100 } for all i=0,1,2.....100
Pi2=500-10i} for all I = 0, 1, 2 …. 100
I (i) Maximax or Hurwicz Decision criterion: The maximax or Hurwicz
decision criterion is a criterion of super optimism. It is based upon the idea
that we do get some favourable or lucky breaks. Since, nature can be good
to us, the decision maker should select that state of nature which will yield
him the highest payoff for the selected strategy. Thus, the maximax
criterion attempt s to maxi mise the maximum gain i. e. maximax chooses
the act that is the "best of the best". The procedure involved in the
criterion is to look at the various payoffs for each strategy and select the
highest amount . Therefore , the maximum of th ese maximum payoffs is
selected . This payoff is referred to as a maximax (maximum of
maximums).
i)Example 3:Consider the following payoff table :
Strategies
States of NatureS1 S2 S3
N1 24 17 10
N2 24 27 20
N3 24 27 30
Maximum gains
Minimum gains24
(24)27
17(30)
10munotes.in
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For maximax criterion, first of all maximum payoff of each strategy is
identified. Maximum payoff for strategy S 1is 24, for S2is 27 and for S 3, it
is 30. Next, we must determine the largest payoff of these maximum
payoffs which is 30. Hence, base d on maximax criterion the best strategy
would be S 3.
(ii) Maximin or Wald decision criterion: The maximin criterion is a
criterion of pessimism. Wald suggested that the decision maker should
always be pessimistic or conservative, resulting in a maximin c riterion.
This maximin payoff appro ach means that the decision maker, under
continually adverse circumstances, should select the strategy that will give
him as large a minimum payoff as possible i.e. the criterion attempts to
maximise the minimum gains or it tries to "pick the best of the worst".
Under maximin criterion, minimum payoff of each strategy is selected. In
the example given in (i) part, minimum payoffs for S 1is 24, for S 2it is 17,
and for S 3, the minimum payoff is 10. Once the minimum payoff
corresponding to each strategy has been selected, the next step is to select
the maximum of these minimum payoffs (maximum of minimums) which
is 24 in .our above example. Hence, strategy S 1is the maximin gain
strategy; In essence, the worst state of natur e that could happen would
give a payoff of 24.
(iii)Minimax Regret Criterion: Suppose that the manager is concerned
about how the decision he makes might be viewed in the future after the
state of nature is known with certainty. In such situation, the mi nimax
regret criterion can be used. T o employ this criterion, one must transform
the payoff matrix into a regret matrix by replacing every payo ff in a row
of the payoff matrix with the difference obtained by subtracting the payoff
from the row's maximum pa yoff. This regret or opportunity loss represents
the amount of profit that a person lost because he did hot select the most
profitable act .When the minimax cr iterion is used, the decision m aker
expects theworst event to materialise and so he selects the act that will
give the minimu m of the maximum opportunity losses.
Example 4:
Given the following pay -off function for each act a 1and a 2Qa1=-30 + 50 x
Qa2=-90 + 20 x
(i) What is the break -even value of x?
(ii) If x= 10, which is the bet ter act?
(iii) If x= 10, what is the regret of the poor strategy?
(iv) Ifx=-5 which is the better act?
(v) If x=-5 what is the regret of the poorer strategy?
Solution: Equating Qa 1and Qa 2
-30 + 50 x=-90 + 20 x
or x=-2
-2 is the break even point.
(ii) Substituting x= 10 in Qa 1and Qa 2
Qa1=470 Qa2=110
Hence a 1is a better strategy.munotes.in
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(iii) Regret with x= 10 is equal to 470 -110 = 360
This happens when strategy a 2is adopted.
(iv) for x=-5 Qa1=-280
Qa2=-190
a2is a better strategy.
(v) Regret in (iv) = -190–(-280)
=-190 + 280
= 90
Example 5:
A food products company is contemplating the introduction of a
revolutionary new product with new packaging to replace the exis ting
product at much higher price (S 1)ora moderate change in the composition
of the existing product with a new packaging at a small increase in price
(S2) or a small change in the composition of the existing except the word
'New' with a negligible incre ase in price (S 2). The three possible s tates of
nature of events are ( i) high increase in sales ( N), (ii) no change in sales
(N2),and ( iii) decrease in sales( N3).The marketing department of the
company worked out the payoffs in terms of yearly net profi ts for each of
the strategies for these events (expected sales) This is represented in the
following table:
State of nature Payoffs (in Rs.)
Strategies N1 N2 N3
S1 7,00.000 3,00,000 1 ,50,000
S2 5,00,000 4,50,000 0
S3 3,00,000 , 3,00.000 3,00,000
Which strategy should the executive concerned choose on the basis of
(a)Maximin Criterion,
(b)Maximax Criterion,
(c) Minimax Regret Criterion,
(d) Laplace Criterion ?
Solution. a)Maximin Criterion. When this criterion is adopted that course
of action is selected which maximises the minimum payoffs.
Strategy Minimum Payoffs Rs
S1 1,50,000
S2 0
S3 3.00.000
The executive would choose strategy S2
(b) Maximax C riterion. In this criterion we select that strategy which
gives the maximum payoffs.
Strategy Maximum Payoffs
S1 7,00,000
S2 5.00,000
S3 3,00,000munotes.in
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The executive should choose S 1
(c)Minimax Regret Criterion. When this crite rion is adopted three st eps
are necessary :
(i) to determine the opportunity loss for each strategy by subtracting from
maximum pay -off of a state of nature, the payoffs of all the strategies.
(ii) to determine the maximum opportunity loss for each strategy.
(iii) to select t he strategy which minimises the maximum of the loss.
OPPORTUNITY LOSS TABLE
N1 N2 N3
S17,00,000 -7,00.000 =
04,50,000 -3,00,000 =
1,50,0003.00,000 -1,50,000
= 1.50,000
S2 7,00,000 -5,00,000 =
2,00,0004,50,000 -4,50,000
= 03,00,000 -0
= 3,00,000
S3 7,00,000 -3.00,000 =
4,00,0004,50,000 -3,00.000 =
1,50,0003,00.000 -3.00.0 00
= 0
Maximum opportunity loss
S1 = 1,50,000
S2 = 3,00,000
S3 = 4,00,000
The exec utive should choose strategy S 1for it minimises the maximum
opportunity loss.
Laplace criterion. This involves three things:
(i) assigning equal opportunity to each state of nature.
(ii) calculating the expected monetary value E1MV.
(iii) se lecting that strategy whose E1MVis maximum.
E1MV TABLE
S1:1(7,00,000 + 3,00,000 + 1 ,50,000) = 11,50,000 =3,83,333.33
3 3
S2:1(5,00.000 + 4,50,000 + 0) =9,50,000 =3,16, 666.67
3 3
S3:1(3,00.000 + 3,00,000 + 3,00,000) = 9,00,000 = 3,00,000
3 3
Since the E 2MVis highest for strategy 1, hence the executive should select
strategy S 1.
Example 6:
A management is faced with the problem of choosing one of three
products for manufacturing. The potential demand for each product maymunotes.in
126
turn out to be go od,moderate or poor. The probabilities for each of
thestate of nature were estimated as follows :
Nature of Demand
Product Good Moderate poor
X 0.70 0.20 0.10
Y 0.50 0.30 0.20
Z 0.40 0.50 0.10
The estimated profit or loss under the three states may be taken as
X 30,000 20,000 10,000
Y 60,000 30,000 20,000
Z 40,000 10,000 -15,000 (loss)
Prepare the expected value table and advise the management about the
choice of produc t.
Computation of EV for various Acts
Expected payoff (in Rs. ‘0000) for various acts
X Y Z States of Nature
x1j p1jx1ip1jx2j p2jx2ip2jx3j p3jx3ip3j
Good 30 0.7 21 60 0.5 30 40 0.4 16
Moderate 20 0.2 04 30 0.3 09 10 0.5 05
Poor 10 0.1 01 20 0.2 04 -15 0.1 -1.5
Expected
Monetary value26 43 19.5
Since the expected value is highest for second course of action, the
management is advised to produce Y.
Exam ple 7:
A TV dealer finds that the procurement c ost of a TV is Rs. 20 and th e cos t
ofa unit shortage is Rs. 50. For one particular model of TV the probability
distribution of weekly sales is as follows:
Weekly Sales 0 1 2 3 4 5 6
Probability 0.10 0.10 0.20 0.20 020 0.15 0.05
How many units per w eek should the dealer buy? Also find E. V.P.I.
Solution : The cost matrix is constructed below. Also derived from it are
the expected costs for the various strategies.munotes.in
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Demand Prob. Strategies (buy so many)
0 1 2 3 4 5 6
0 0.10 0 20 40 60 80 100 120
1 0.10 50 20 40 60 80 100 120
2 0.20 100 70 40 60 80 100 120
3 0.20 150 120 90 60 80 100 120
4 0.20 200 170 140 110 80 100 120
5 0.15 250 220 190 160 130 100 120
6 0.05 300 270 240 210 180 150 120
Expected 147.5 122.5 102.5 92.5 92.5 102.5 120
Thus either buy 3 or 4.
Example 8:Following are t he records o f demand of an item for the past
300 days.
θ= Demand in units No. of days Prob (θ)
10,000 18 0.06
11,000 90 0.30
12,000 120 0.40
13,000 . 60 0.20
14,000 12 0.04
300 1.00
(i) What is the expected demand?
(ii) It costs Rs. 15 to make an item which sells for Rs. 20 normally but at
the end of the day su rplus has to be disposed at Rs. 10 per item. What is
the, optimum output?
Solution : Expected demand = 10x0.06 = 0.60 .
= 11 x0.30 = 3.3 0
= 12 x 0.40 = 4.80
=13x 0.20 = 2.60
=14 x 0.04 =0.56
11.86
Expected demand = 11,860 (Answer)
The daily pay-off matrix is constructed be low:
Strate gies
P(θ) states of
nature10 1112 13 14
0.06 10 50 45 40 35 30
0.30 11 50 55 50 45 40
0.40 12 50 55 60 55 50
0.20 13 50 55 60 65 60
0.04 14 50 55 60 65 70
Expected
Payoff50 54.4 55.8 53.2 48.6
Demand of 12,000 is the optimum wth an expected payoff of55.8x 1000 =
55,800 rupees. ;munotes.in
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Example 9:The sales manager of Beta Co . is highly experienced in the
fad market. He is sure that the sales of JUMBO (during the period it has
special appeal) will not be less than 25,000 units. Plant capacity limits
total pr oduction to a maximum of 80000 units during JUMBO 's, brief life.
According to the sales manager, there are 2 chances i n 5 for a sales
volume of 50,000 units. The probability that it will be more than 50,000 is
four times the probability that it will be less than 50.000. If sales exceed
50,000 units volumes of 60,000 and 80,000 are equally likely. A 70,000
unitvolume is 4 times as likely as e ither. It costs Rs. 30 to produce a unit
of JUMBO whereas its selling price is estimated at Rs. 50 per unit. Initial
investment is estimated at Rs. 8,00,000. Should the venture of production
be undertaken?
Solution:
Prob. (50,000) = 2= 0.40
5
Prob. (less than or more than 50,000) = 0.60
Prob. (less than 50,000): Prob. (More than 50,000): 1:4
Thus Prob. (less than 50,000) = 0.60 x 1= 0.12
1+4
Prob. (More than 50,000) = 0.60 x 4= 0.48
1+4
Prob. (60,000): Prob. (70, 000): Prob. (80,000):: 1:4:1
Thus Prob (60,000) = 0.48 x1 = 0.08
1+4+1
Prob (80,000) = 0.08
and Prob 70,000 =4x 0.48 = 0.32
1+4+1
Now we have complete probability distribution of sales except that for
sales less than 50,000 we have a summarised probability of 0.12. The
payoff matrix is compiled below. We used 25,000 in place of less than
50,000 as the worst contingency.
Pay-off Table
(Rs.in'000 )
Demand Prob < 50,000
(take it
25,000)50,000 60,000 70,000 80,000
< 50,000
(worst
25,000)0.12 500 -250 -550 -850 -1150
50,000 0.40 500 1,000 700 400 100
60,000 0.08 500 1,000 1200 900 600
70,000 0.32 500 1,000 1200 1400 1100
80,000 0.08 500 1,000 1200 1400 1600
Expected Payoff 500 850 790munotes.in
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Expected Payoff is, thus, Rs. 850,000. This exceeds 800,000 the initial
investment; therefore, the venture oug ht to be undertaken.
2 Decision Trees
A Decision Tree diagram is a graphical representation of various
alternatives and the sequence of events in a decision problem. In
constructing a decision tree, there are certain conventions to be followed.
The tree is constructed starting from left and moving towards right. The
square box denotes a decision point at which the available strategie s are
considered . The circle O represents the chance node or event, the various
states of nature or outcomes emanate from this chance event. At each
decision or chance node, there can be one or more branches represented by
a straight line. In figure b elow, there are two b ranches at the decision
node, each branch represents a strategy. Similarly, at each of the two
chance nodes, there are three branches. Any branch that is not followed by
another decision or chance node is called a terminal branch.
Let us illustrate the de cision tree with an example.
Example 1 0: A manufacture r of toys is interested to know whether he
should launch a deluxe model or a popular model of a toy. If the deluxe
model is launched, the probabilities that the market will be good, fair or
poor are giv en by 0 .3,0.4 and 0 .3 respectively with payoffs Rs. 1,40,000,
Rs. 70,000 and Rs. ( -10,000). If the popular model is introduced, the
corresponding probabilities are given by 0.4, 0.3 and 0.3 with respective
payoffs Rs 1,50,000, Rs. 80,000 and Rs. ( -15,000). The problem is to
decide which model should be launched. The decision tree for the given
problem is drawn below:
Market Good P = 0.3 Rs. 1,40,000
Introduce
Delux Model Marke t Fair P= 0.4 Rs.70,000
Market Poor P=0.3 (Rs.-10,000)
Introduce Market Good P= 0.4 Rs. 1,50.000
Popular Model
Market Fair P=0.3 Rs.80,000
Market Poor P =0.3 Rs.(-15,000 )
Fig. 1
Thus the decision tree shows that structure of the decision problem. To
carry out the decision tree analysis, conditional payoffs are e stimated for
every combination of actions and events (i.e., for every path through the
tree). These payoffs can be either positive or negative. Also the
probabilities for each event must be assessed by the decision maker. To
analyse the decision tree begin at the end of the tree and work backward.
For each set of events branches, the Expected Monetary Value (EMV) ismunotes.in
130
calculated and for each set of decision branches, the one with the highest
EMV is selected. This EMV now represents the EMV for the decision
point from which this set of event branches emanates. Following the same
procedure we move back on the decision tree to the next decision point.
This technique of analysing tree is called the roll -back technique. Thus,
there are two rules concerning roll -back technique:
(i) If branches emanate from a circle, the total expected pay off
may be calculated by summing the expected value of all the
branches.
(ii) If branches emanate from a square, we calculate the total
expected benefit for each branch emanating from that square
and let the total expected pay -off be equal to the value of the
branch with the highest expected benefit.
Let us analyse the tree given above by the roll back technique. Here point
1 is decision point and C & D are the chance nodes.
EMV (at C) = .3 x Rs. 1,40,000 + .4 xRs. 70,000 +.3 x ( -Rs.10,000)
= Rs. 42,000 + Rs. 28,000 -Rs. 3,000 = Rs. 67,000
EMV (at D) = .4 x Rs. 1,50,000 +. 3 x Rs. 80,000 + .3 x ( -Rs. 15,000)
= Rs. 60,000 + Rs.24,000 -Rs. 4 ,500 = Rs. 79,500
The decis ion at point 1 is to introduce the popular model since it results in
the highest EMV
Example 1 1: Mr. X is trying to decide whether to travel to Sri Lanka from
Delhi to negotiate the sale of a shipment of china novelties. He hol ds the
novelties s tock and is fairly confident, but by no means sur e that if he
makes the trip, he will sell the novelties at price that will give him profit of
Rs. 30,000. He puts the probability of obtaining the order at 0.6. If he does
not make th etrip,he will certainly not get the order.
If the novelties are not sold in Sri Lanka there is an Ind ian customer who
will certainly buy them at a price that leaves him a profit of Rs. 15,000
and his offer will be open at least till Mr. Xreturns from Sri Lanka. Mr. X
estimates the expenses of trip to Sri Lanka at Rs. 2,500. He is however,
concerned that his absence, even for only three days, m ay lead to
production inefficiencie s in the factory. These could cause him to miss the
deadline o n another contract, with t heconsequence that a late penalty of
Rs. 10,000 will be invoked. Mr. X assesses the probability missing the
deadline under these circumstances at 0.4. Further, he believes that in his
absen cethere will be a lower standard of hou se-keeping in the factor y, and
the raw material and labour costs on the other contract will rise by about
Rs. 2,000 above the budgeted figure.
Draw an appropriate decision tree for Mr. X's problem and using EMV as
the appropri atecriterion for decision, f ind the appropriate initial decision.munotes.in
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Solution:
Fig. 2
The decision tree is dra wnintheabovefigure . Calculations of EMV
proceed from ri ght to left.
Go Not Go
30,000 x 0.6 = 18,000 15,000 X 1 = 15,000
15,000 x 0.4 = 6000 (where 1 is the probability)
24000
24,000 x 0.6 = 14,400
plus
24,000 x 0.4 = 9600
minus
10,000 x 0.4 = 4000 Hence Mr. X should proceed to Sri Lanka
20,000
minus 4,500
15,500
Steps in Decision Tree Analysis
In a decision an alysis, the decision maker has usually to proceed through
the following six steps:
1. Define the problem in structured terms. First of all. the factors
relevant to the solution should be determined. Then probability
distribi utions that are appropr iate to describe future behaviour of
those factor sestimated. The financial data concerning conditional
outcomes is collect ed.
2. Model the decision process. A decision tree that illustrates all
alternatives in the problem is constructed. The enti re decision proce ss is
presented schematically and in an organised step -by-step fashion.munotes.in
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3. Apply the appropriate probability values and financial data. Toeach
of the branches and sub -branches of the decision tree the appropriate
proba bility values and financial data are applied. This would enabl e to
distinguish between the probability value and conditional monetary value
associate d with each outcome.
4. "Solve" the decision tree. Using the methodology mentioned above
proceed to locate that particu lar branch of the tree that has the largest
expected value or that maximises the decision criterion.
5. Perform sensitivity analysis. Determine how the solution reac ts in
changes in inputs. Changing probability values and conditional fin ancial
values all ows the decision maker to test b oth the magnitude and the
direction of the reaction.
6. List the underlying assumptions. The accounting cost finding and
other assumptions used to arrive at a solution should be explained. This
would alsoenable others to k now what risks they are taking when they use
the resul ts of your decision tree analysis. The limits under which the
results obtained will be valid and not valid should be clearly specified.
Advantages of Decision Tree Approach
The decision tree analysis a s a tool of de cision -making is important
because of the following:
1. Decision trees are of great help in complicated kinds of decis ion
problems. However, in a simple problem th ere is no advantage of
constructing a decision tree.
2. It structures the d ecision proces s making managers, approach
decision making in an orderly sequential fashion.
3. It requires the decision -maker to examine all possible outcomes
desirable and undesirable.
4. It communicates clearly the decision -making process to others.
5. It allows a group to discuss alternat ives by focusing on each finan cial
figure, probability value and underlying assum ption one at a time.
Thus, group can move in orderly steps towards a consensus decision
instead ofdebating a decision in its entirety.
6. It can be used with a computer so that m any different sets of
consumptions can be simulated and their effects on the final
outcomes observed.
Example 1 2:M/s J. Bloggs & Co. is currently working with a process,
which, after paying for materials, labo ur etc. brings a profit of Rs. 10,000.
The following alternatives are made available to the company.
(i) The company can conduct research(R 1) which is expected to cost Rs.
10,000 and having 90% probability of success, the company gets a
gross income of Rs. 25,000. (ii) The c ompany can conduct researchmunotes.in
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(R2) expected to cost Rs. 5,000 and having a probability of60%
success. If successful, the gross income will be Rs. 25,000.
(iii) The company can pay Rs. 6,000 as royalty of a new process which
will brin g a gross inco meof 20,000.
(iv) The company continues the current process.
Because of limited resources, it is assumed, that only one of th e two types
of research can be carried out at a lime. Which alternative should be
accepted by the company?
Solutio n: The decision tree which represents the possible courses of action
is depicted in figure 4. Point 1 is a'decision box' located 'now' on t he time
scale. The fou r possibilities arising here are shown. Upon failure of a
particular research, say, R 1, there are again 3 original alte rnativ es to be
sorted out, that of R 1being excluded. lf R 2fails after failure of R 1, the
company is left with only two choices, i.e., either to pay royalty or
continue the existing process .
Branch Curre nt
NetReturn = Rs. 10,000
Branch Licence
Net Return = 20, 000-6,000 =Rs. 14,000.
Branch R1First
Value at Point 3
Current Rs. 10,000
Licence Rs. 14,000
Value at point C: Expected gross profit
= 25,000 x 0.6 + 14,000 x 0.4
= 20,600.
Value at Point 2
Licence Rs. 14,000
Current Rs. 10,000
R2 Rs. 20,600 -Rs. 5,000 = Rs.. 1 5,600
Value at Point A Rs. 25,000 x 0.9 + 0.1 x 15,600 = Rs. 24,060
Value at Branch R 1 Rs. 24,060 -Rs. 10,000 = Rs. 14,060
Branch R2First
Value at point 5
Current Rs. 10,000
Licence Rs. 1 4,000
Value at Point at D: Expected gross profit
= 0.9 x 25,000 + 0.1 X 14,000 = Rs. 23.900munotes.in
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R1SUCCEEDS 25,000
A R1FAILS CURR ENT PROCESS 10,000
R2 0.4 3LICENSE 20,000
R2
FAILS CURRENT 10,000
R1 PROCESS
Licence
CURRENT PROCESS 10,00 0
1
LICENS E 20,000
R2 CURRENT PROCESS 10,000
CURRENT
R1FAILS PROCESS 10,000
0.1 LICENSE 20,000
B R2FAILS R1 5
0.4 R1SUCEEDS 25,000
0.9
LICENSE 20,000
R2SUCEEDS 25,000
0.6
Fig 3
Value at point 4
Current Rs. 10,000
Licence Rs. 14,000
R1 = Rs. 23,900 -Rs. 10,00= Rs. 13,900
Value at Point B = 0.4 x 14,000 +0.6x25,000=Rs.20,600
Value of Branch R2 = Rs. 20,600 -Rs. 5,000 -Rs. 15,600
Thus R 2. followed by licence upon former's failure is the best course of
action.
EXERCISES
1.The research department of consumer products division has
recommended the marketing department to launch a soap with 3 different
perfumes. The marketing manager has to decide the type of perfume to
launch under the following estimated payoff for various levels of sales.
Type of perfume Estimated sales (units)
25,000 15.000 10,000
I 200 25 20
II 30 20 10
III 50 30 5
Find the best decision using (i) Maximax , (ii) Maximin ,
(iii) Minimax Regret and(iv) Laplace criteria.
Ans: (i) I, (ii) I, (iii) I, (iv) Imunotes.in
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2.A consumer product company is examining the introduction of a new
product with new packaging or replace the existing product at much
higher price (S 1) or moderate change in the composition of the e xisting
product with new packaging at a small increase in price (S 2) or a small
change in the composition of the existing product except the word 'new'
with the negligible increase in price (S 3). The three possible states of
nature are E 1: increase in sal es, E 2:no change in sales, E 3: decrease in
sales. The marketing department has worked out the following payoffs in
terms of profits. Which strategy should be considered under (i) Minimax
Regret (ii) optimistic (iii) equi -probability conditions?
Strategies States o f Nature
S1 S2 S3
E1
E2
E330.000
50,000
40.00040,000
45,000
40,00025,000
10,000
40,000
Ans: (i) S 2, (ii) S 1, (iii) S 2
3.Construct a payoff matrix for the following situati ons and find the best
decision using (i)Maximin, (ii)Maxi max, (iii)Laplace criteria .
Product
Fixed Cost Rs. Variable cost unit in Rs.
X
Y
Z250
350
50012
10
5
The likely demand (units) of products
Poor demand 300
Moderate demand 700
High demand 1000
Selling p rice of each product is Rs. 25.
Ans: (i) Z, (ii) Z, (iii) Z
4.The management of a company is faced with the problem of choosing
one of the three products A, B, C for manufacturing. The demand for these
products can be good, moderate or poor. The probabil ities of each state of
nature are estimated as follows :
Nature of demand
Product
Good Moderate Poor
A 0.7 0.2 0.1
B 0.4 0.5 0.1
C 0.5 0.3 0.2munotes.in
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The estimated profit/loss under the three states of nature are given as
follows:
Profit i n 10,000 Rs. Product
Good Moderate Poor
A 30 20 10
B 40 10 -15
C 60 30 20
Using EMV criterion, advice the management the optimum decision.
Ans: Select product C
5. You are given, the following payoff table for three acts A 1, A 2,A3and
the states of nature S1, S2, S3.
States of nature A1 A2 A3
E1 25 -10 -125
E2 400 440 400
E3 650 740 750
The probabilities of the states of nature are 0.1, 0.7, 0.2 respecti vely. Find
the optimum decision using EMV criterion.
Ans: Optimum decision at A 2
6.An investo r is given the following investment alternatives and
percentage rate of return.
State of Nature Course of Action
Regular Share Risky Share Property
Good 7% -10% -12%
Better 10% 12% 18%
Best 15% 18% 30%
Over past 100 days, 50 days have market in better condition and for 20
days market conditions are b est. Using this information stat e optimum
investment strategy for the investment. Use EMV criterio n.
Ans: Invest in property.
7.A retailer purchases grapes every morning for Rs. 50/ -a case nd sell s
for Rs. 80/ -a case. Unsold c ases at the end of the day are d onated to old
age homes. Past sales have ranged from 15 to 18 cases per d ay. The
following are the details of demand and probability.
Cases demanded : 15 16 17 18
Probability : 0.1 0.2 0.4 0.3munotes.in
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The retailer wants to know how many cases should he buy to m aximize
his profit. Use EMV criterion.
Ans: 17 cases
8.Following is the pay -off matrix correspo nding to four states of nature
S1, S2, S3, S4and four courses of action A 1,A2,A3, A4.
Course of Action
State of Nature
A1 A2 A3 A4Probability of State
s1 50 400 -50 0 0.15
S2 300 0 200 300 0.45
S3 -150 100 0 300 0.25
s4 50 0 100 0 0.15
(i) Calculate expected pay off and find best course of action using
EMV.
(ii) Calculate EOL for each course of action hence find best action using
EOL.
Ans: (i) A 4, (ii) A 4
9.Following is pay -off tab le corresponding to four acts A 1,A2,A3,A4and
four states of nature E1, E2, E3, E4with the probabili ty of the events of this
table P(E 1) = 0.20, P(E 2) = 0.15, P(E 3)=0.40, P(E 4) =0.25.Calculate the
expected pay off and expected opportunity loss andsuggest best course of
action.
Acts Events
E1 E2 E3 E4
A1 40 50 200 0
A2 300 200 0 100
A3 50 100 40 200
A4 300 0 100 50
Ans: A2
10.The flo rist shop promises its customers delivery within three hours on
all orders. All flowers are purchased the previo us day and delivered tomunotes.in
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florist by 8.00 a.m. the next morning. Demand distri bution of roses is as
follows.
Dozens of roses 7 8 9 10
Probability 0.1 0.2 0.4 0.3
Florist purchases roses for Rs. 10.00 per dozen and sells them at Rs. 30
per dozen. All unsold roses are distributed in a local hospital fre e of cost.
How many dozens of roses should the florist order each evening?
Ans: 9 dozens of roses
11.The probability of the demand for tourist cars for hiring on any day at
ABC travels and tours are as under:
No. of cars demanded 0 1 2 3 4
Probability 0.1 0.2 0.3 0.2 0.2
The cars have a fixed cost of Rs. 90 0 each day to keep and the daily hire
charges are Rs. 2000. (i)If the tour owner company owns 4 cars what i s its
daily expectation? (ii)Use EMV and EOL criterion to suggest how many
cars the company should keep.
Ans:(i) Rs. 800/ -, (ii) 2 cars.
12.Unique home appliances finds that the cost of holding a co oking ware
in stock for a month is Rs. 200.Customer who cannot obtain a cooking
ware immediately tends to go to other dealers and he estimates that for
every cus tomer who cannot get immediate delivery he loses an average of
Rs. 500. The probabilities of a demand of 0, 1, 2, 3,4, 5 cooking ware in a
month are 0.05, 0.1, 0.2, 0.3, 0.2, 0.15 respectively. Determine the
optimum stock level of cooking wares. Using EMV criterion.
Ans: 4 Cooking wares
13.A person has the choice of running a hot snack stall or an ice cream
stall at a certain holiday hotel during coming season. If weather is cool
and rainy he can expect to make a profit of Rs, 1,50,000 and if it is warm
he can expect to make profit of Rs. 40 ,000 by running hot snack stall. If he
runs ice -cream and cold drink shop he can make a profit of Rs. 1,60,000 if
the weather is warm and only Rs. 30,000 if whether is cool and rainy. The
odds in favour of warm weather are 2 : 3 and that of having cool and rainy
are 3 : 2. Use EMV to find Best f Action:
Ans: Hot snack stall.
14.You are given the following estimates concerning a Research and
Development programme:munotes.in
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Decision
DiProbability of
decision Di
given research R
P(Di \R)Outcome
numberProb ability
of outcome
Xi given Di
P(Xi \Di)Payoff value
of outcome,
Xi (Rs ‘000)
Develop 0.5 1
2
30.6
0.3
0.1600
-100
0
Do not
develop0.5 1
2
30.0
0.0
1.0600
-100
0
Construct and evaluate the decision tree diagram for the above data. Show
your working for evaluation.
The decision tree of the given problem along with necessary calculation is
shown in Fig. below
Decision tree
15.A glass factory specializing in crystal is developing a substantial
backlog and the firm’ management is considering tree courses of sub -
contracting (S1), being overtime production (S2), and construct new
facilities (S3). The correct choice depen ds largely upon future demand
which may be low, medium, or high. By consensus, management ranks the
respective probability as 0.10, 0.50 and 0.40. A cost analysis reveals effect
upon the profits that is shown in the table below:
Show this decision situati on in the f orm of a decision tree and indicate the
most preferred decision and corresponding expected value.
A decision tree which represents possible courses of action and
nature are shown in the Fig. In order to analyses the tree we start working
backw ard from the end branches.munotes.in
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He most preferred decision at the decision node 0 is found by calculating
expected value of each decision branch and selecting the path (course of
action) with high value.
Decision tree
Decision Point Outcome Probability Cond itional
value (Rs.)Expected
value
D3 (i) Accept A
(ii) StopSuccess
Failure
-0.6
-20,000
-10,000
-12,000
-4000
Rs.8,000
0
D2 (i) Accept B
(ii) StopSuccess
Failure
-0.4
0.6
-24,000
-10,000
-9,600
-6,000
Rs. 3,600
0
D1 (i) Accept A
(ii) Accept B
(ii) Do nothingSuccess
Failure
Success
Failure
-0.6
0.4
0.4
0.6
-20,000+3,600
-10,000
24,000+8,000
-10000
-14,160
-4,000
Rs. 10,160
12,800
-6,000
Rs. 6,8000
Rs. 0
Since node 3 has the highest EMV, therefore, the deci sion at node
0 will be to choose the course of action S3, i.e., construct new facilities.
16.A person wants to invest in two independent investment schemes: A
and B, but he can undertake only at a time due to certain constraints. He
can choose A first an d then stop, or if A is not successful then B or vice -
versa. The probability of success of A is 0.6, while for B it is 0.4. Themunotes.in
141
investment in both the schemes requires an initial capital outlay of Rs.
10,000 and both return nothing if the venture is unsucc essful. Successful
completion of A will return Rs. 20,000 (over cost) and successful
completion of B will return Rs. 24,000 (over cost). Draw decision tree and
determine the best strategy.
The decision tree corresponding to the given information is
depic ted in the Fig.
Decision tree
Since EMV = Rs 10,160 at node 1 is the highest the best strategy at node
D1 is to accept course of Action A first and if A is successful then Accept
B.
munotes.in
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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