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
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