BUSINESS MATHEMATICS AND STATISTICS

SOLVED QUESTION PAPER OF SESSION : 2010-2011

SUBMITTED TO: SUBMITTED BY : Ms. Bikramjeet Kaur Bavanpreet Singh Roll No. :171 Manveen Kaur Roll no. : 172

SECTION-A

QUES.1(a):Difference between Determinant and Matrix

DETERMINANT

MATRIX

A determinant is always a square

Determinant has a definite value

Two vertical lines are used to denote determinants

A matrix may be rectangular or a square

A matrix has no definite value

We use square brackets to denote matrix

(b):Define Regression coefficients and Regression lines

1.Regression coefficients Just as there are two regression

equations ,similarly there are two regression coefficients .regression coefficient measures the average change in the value of one variable for a unit change in value of another variable.

(1) Regression coefficient of X on Y :this coefficient shows that with a unit change in the value of X ,what will be the average change in the value of Y.

byx=r. deviation of x deviation of y

(2)Regression coefficient of X on Y :this coefficient shows that with a unit change in the value of X, what will be the average change in the value of X.

Bxy=r.deviation of x deviation of y

2.Regression lines The regression line shows the average

relationship between two variables . This is also known as the line of best fit on the basis of regression line , we can predict the value of a dependent variable .

(1):Regression line of X on Y:the regression line of X and Y gives the best estimate for the value of X for any given value of Y .

(2):Regression line of Y on X:the regression line of Y on X gives the best estimate for the value of Y for any given value of X.

(c).Distinguish between imple random sampling and stratified

random sampling. SIMPLE RANDOM SAMPLING STRATIFIED RANDOM SAMPLING

It is that method in which each item of universe has an equal chance of being selected in the sample. Which item will put in sample and which not , such decision is not made by an investigator on his will but selection of units is left on chance.

This method is used when units of universe are hetrogenous rather than homogenous . under this method , first of all units of the population are divided into different strata in accordance with their characterstics . then the samples are selected on random basis.

(d).Show that 1 a bc

1 b ca =(b-c)(c-a)(a-b) 1 c ab

Sol. Operating r₂-r₂-r₁-r₃-r₃-r₁ 1 a bc 1 a bc 1 b ca = 0 b-c ca-bc 1 c ab 0c-a ab-bc Expanding along c₁, we get=(a-b)(c-a)(b-c)=(a-b)(b-c)(c-a)

(e).Simple interest on a certain sum of money for 1.5 years is ₹150 and

compound interest is ₹153.33.find rate

Simple interest for 1.5 years =₹150Simple interest for 1 year = 150X²/₃ =100Now , C.I. For 1st year =S.I. Of the 1st year=₹100153.33-100=₹ 53.33=C.I. For remaining ⅟₂years-S.I. For ⅟₂years=₹53.33-₹50=₹3.33Rate =3.33 X 100 =6.66% 100X ⅟₂

(f).The arithmatic mean ,mode and the median of a group of 75 observations was calculated to be 27,34 and 29.ut was later

discovered that 1 observation was wrongly read as 43 instead

of 53.examine the error

ƩX=NXN=75,X=27ƩX=75X27=2025CORRECTED ƩX=2025-43+53=2035C0RRECTED X=2035/75=27.13

SECTION-B

QUES : Calculate mean,median and mode from the following data:

X F

0.5 1.0 3.0 5-7 7-10 10-15 15-20 20-25

2 3 2 4 4 12 6 2

Make equal intervals Class intervals

0-5 5-10 10-15

15-20 20-25

f 7 8 12 6 2

Class intervals

f M.V. fX c.f.

0-5 7 2.5 17.5

7

5-10 8 7.5 60.0

15

10-15 12 12.5 150.0

27

15-20 6 17.5 105.0

33

20-25 2 22.5 45.0

35

N=35 377.5

X=ƩfX =377.5 =10.79 N 35Median=17.5th itemMedian class=10-15M=10+17.5-15X5=11.041 12 Modal class:10-15Z=l₁ + f₁+f₀ X i 2f₁-f₀-f₂

X=10.79,M=11.04,Z=12

Ques : Calculate chain base index numbers chained to 2003 from the average price of following three commodities:

Commodities

2003 2004 2005

2006 2007

Wheat 4 6 8 10 12

Rice 16 20 24 30 36

Sugar 8 10 16 20 24

Chain base index numbers chained to 2003

Commodities

2003 p₀ LR

2004 p₁ LR

2005 p₂ LR

2006 p₃ LR

2007 p₄ LR

Wheat 4 100

6 150

8 133.3

10 125

12 120

Rice 16 100

20 125

24 120

30 125

36 120

Sugar 8 100

10 125

16 160

20 125

24 120

Total 300

400

413

375

360

AV of LR 100

133.3

137

125

120

CBI 100

133.3

183

228

274

Solution.

year 2001 2001 2003 2004 2005 2006 2007

production

77 88 94 85 91 98 90

year Y X XY X²

2001 77 -3 -231 9

2002 88 -2 -176 4

2003 94 -1 -94 1

2004 85 0 0 0

Ques : Below are given the figures of production of sugar factory:

(a)Fit a straight line by the least squares method and tabulate the trend values.

(b)Eliminate the trend . What components of the series are thus left over?

(c)What is monthly increase in the production of sugar?

The equation of straight line curve isY=a+bXa=ƩY=89 Nb=ƩXY=2 ƩX²

2005 91 +1 91 1

2006 98 +2 196 4

2007 90 +3 270 9

N=7 ƩY=623 Ʃx=0

ƩXY=56

ƩX²=28

Y = 89 = 2X origin:2004, X unit = 1 year Calculation of trend values For 2001, X = -3 2002, X = -2 2003, X = -1 2004, X = 0 2005, X = +1 2006, X = +2 2007, X = +3

(b). Elimination of trend

Year 2001 2002

2003

2004

2005 2006

2007

Y 77 88 94 85 91 98 90

Y 83 85 87 89 91 93 95

Y-Y -6 3 7 -4 0 5 -5

(c).Monthly increase = b = 2 = 0.166 12 12 = 166.66 tonnes.

SECTION - C

Ques : A computer while calculating the correlation coefficient between two variables X and Y from 25 pairs of observation obtained from following results :

N = 25 ƩX = 125 ƩX² = 650 Ʃy = 100 Ʃy² = 460 Ʃxy = 508 Incorrect correct X Y X Y 6 14 8 12 8 6 6 8

Correct ƩX = 125 - 6 – 8 + 6 = 125Correct ƩY = 100 - 14 - 6 + 12 + 8= 100Correct ƩX² = 436Correct ƩY² = 520Correct ƩXY = 520r = 0.667

Ques: the following data relates to the scores obtained by a salesman of a company in an intelligence test and their weekly sales in thousand rupees :

intelligence

A B C D E F G H I

Test scores

50 60 50 60 80 50 80 40 70

Weekly sales

30 60 40 50 60 30 70 50 60

(a). Obtain the regression equation of sales on intelligence test scores of the salesman.

(b). If he intelligence test scores of a salesman is 65, what would be his expected weekly sales?

Solution :

X

x x² Y y

y² xy

50 -10 100 30 -20 400 200

60 0 0 60 10 100 0

50 -10 100 40 -10 100 100

60 0 0 50 0 0 0

80 20 400 60 10 100 200

50 -10 100 30 -20 400 200

80 20 400 70 20 400 400

40 -20 400 50 0 0 0

70 10 100 60 10 100 200

ƩX=540 Ʃx=0 Ʃx²=1600

ƩY=450 Ʃy=0 Ʃy²=1600

Ʃxy=1200

X = 60 , Y = 50Byx = 0.75

Regression equation of Y and X Y= 50 + 0.75X(b). For X = 65 , Y₆₅ = 50 + 0.75(65)

=53.75

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Ques : 12 dice were thrown 4096 times . Each 4,5 or 6 spot appearing was considered to be a success . while 1,2 or 3 spot was a failure calculate the theoretical frequencies for 0,1,2,....,12 successes.

Solution : N = 12 , P = probability of success N = =4096The probability of 0,1,2,....,12 will be given by P(X)ᶰCᵪq = ᶰ-ᵡpᵡ for n = 1,2,3,...,12.

The expected frequencies are obtained by increasing the probability by N as follows :

X

fe9x0 = N .ᶰ Cᵪ q ᶰ-ᵡpᵡ

1 1

2 12

3 66

4 220

5 495

6 792

7 924

8 792

9 495

10 220

11 66

12 12

Total 4096

THANK YOU