STATISTICS AND BUSINESS MATHEMATICS B.com-1 Regular Annual Examination 2015

Size: px

Start display at page:

Download "STATISTICS AND BUSINESS MATHEMATICS B.com-1 Regular Annual Examination 2015"

Jonas Rose
5 years ago
Views:

1 B.com-1 STATISTICS AND BUSINESS MATHEMATICS B.com-1 Regular Annual Examination 2015 Compiled & Solved By: JAHANGEER KHAN

2 (SECTION A) Q.1 (a): Find the equation of straight line when x-intercept = 3 and y-intercept = 5. Also find the slope of the equation. SOLUTION (1-a): As we know that = By substituting values = Multiply the equation by on both sides =15(1) Comparing with Where Hence Solution Set = {, } Q.1 (b): For the derivatives in each of the problem. (any Two) (i) (ii) SOLUTION (1-b-i): (iii) pg. 1

3 + ( (3) (2 ) SOLUTION (1-b-ii): pg. 2

4 ( ) * + * + * + * + * + * + * + * + * + X * + * + pg. 3

5 * SOLUTION (1-b-iii): + Q.2 (a): Find the quadratic equation y =. Determine: (i) Which way parabola opens (ii) The Vertex (iii) The roots SOLUTION (2-a-i): SOLUTION (2-a-ii): The coordinates of vertex are Substituting values: * + pg. 4

6 Hence Solution Set = {, } SOLUTION (2-a-iii): As we know that By substituting values Hence Solution Set = {, } Q.2 (b): Find the inverse of the following square matrix A than verify that A -1 x A=I. A=* + SOLUTION (2-b): * + * + pg. 5

7 * + [ ] [ ] x [ ] x * +=* + [ ] =* + [ ] =* + [ ] =* + [ ] =* + pg. 6

8 [ ] =* + * + * + Q.3 (a): Given, A=[ ] and B=* + Find A x B SOLUTION (3-a): A x B =[ ] x * + A x B =[ ) ] A x B =[ ] A x B =[ ] Q.3 (b): Examine maximum and minimum value of the function y =. SOLUTION (3-b): Now taking. 0 = 27= = pg. 7

9 Since > 0, so attains minima at = = = = = (SECTION B) Q.4 (a): Calculate A.M, G.M, HM and Mode for the given frequency distribution SOLUTION (4-a): Total Note: Since data contains value zero so G.M and H.M cannot be calculated for the given data. pg. 8

10 Mode = The most repeated value in the data So, Mode = 4 Hence solution set = {, G.M and H.M cannot be calculated, Mode=4} Q.4(b): Find chain index for 2001 as base for the production of wheat from the data given below: Year production SOLUTION (4-b): Year Production Link Relatives Chain Indices x 100=100% = 100% x 100=86.80% = 86.80% x 100=120.16% =104.30% x 100=111.53% =116.33% x 100=117.02% =136.13% x 100=99.28% =135.15% x 100=87.52% =118.28% x 100=107.23% =126.83% x 100=93.45% =118.52% Q.4(c): If an investor buys shares of Rs.9000/- at a price of Rs.45/- per share of Rs.9000/- at price of Rs.36/- per share. Calculate the average price per share. SOLUTION (4-c): Type 1 Shares Type 2 Shares pg. 9

11 Q.5(a): For the following frequency distribution: C B f Find Mean Deviation from Mean. SOLUTION (5-a): C B Total , pg. 10

12 Q.5(b): In a moderately skewed frequency distribution: Mean = 62.5 and Median = 59.2 find Mode. SOLUTION (5-b): Mean = 62.5 Median = 59.2 Mode =? Q.5(c): Given = 20, ơ x = 4 find and ơ y (mean and sd of y). SOLUTION (5-c): ơ ơ ơ ơ Hence Solution Set = {, ơ } Q.6(a): The following table shows the heights of father and heights son of sons: Heights of fathers Heights of sons (i) Find the Karl Pearson Coefficient of Correlation. (ii) Find the equation of the regression line of son on father. SOLUTION (6-a-i): * +* + * +* + * +* + pg. 11

13 * +* + SOLUTION (6-a-ii): Q.6(b): The average is 68 and S.d is 4 of marks of section A. the average is 52 and S.d is 12 marks of section B. which is more consistent? SOLUTION (6-b): Section A Section B pg. 12

14 x x x x Conclusion: Since <, it means that section A is more consistent than section B. (SECTION C) Q.7(a): How many three digit numbers can be formed from the digit 1, 2, 5, 6 and 9 if each digit can be used once? SOLUTION (7-a): n Pr= 5 P3= 5 P 3 = 5 P3 = x x 5 P3 = Q.7(b): What is the probability of getting a total of 7 or 11, when a pair of dice is tossed? SOLUTION (7-b): { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) } tota of or * + pg. 13

15 Q.7(c): Find 90% confidence interval for the mean of a normal distribution if standard deviation is known to be 2 & if a sample of size 8 give the value 9, 14, 10, 12, 7, 13, 11, 12. SOLUTION (7-c): ơ ơ ơ ( ) ( ) pg. 14

16 Q.8(a): A type of 200 watt light bulb has been found to have a mean life of 2000 hours & S.d of 250 hours. What is the probability that a sample 81 bulbs will have an average life of fewer than 1920 hours? SOLUTION (8-a): ơ ơ Q.8(b): Find the expected value of X, where X represents the outcome when a die is tossed. SOLUTION (8-b): X P(x) 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 Total 1 pg. 15

17 Q.8(c): In survey of 400 infants chosen at random, it was found that 190were girls. Fit test to examine the hypotheses that boy and girl are equa y ike y α SOLUTION (8-c): 1. H 0 : Boy and girl are equally likely. 2. H A : Boy and girl are not equally likely. 3. α = Use χ 2 test: χ 2 cal= Boy Girl Total N= χ 2 cal=1 5. Critical Value: d.f=n-1=2-1=1 χ 2 tab= χ ; 1= Decision/Conclusion: Since < so accept H 0 and reject H A. It means that boy and girl are equally likely. Q.9(a): An unbiased coin is tossed times f x is a random variab e showing the number of heads than construct the Binomia distribution of x if the probabi ity of head in a single toss is 2/3. SOLUTION (9-a): n n Total pg. 16

18 Q.9(b): A random variable of 50 observations produced the following sums =20 = 10.9 Test the hypothesis that population mean is 0.45 against the alternative less than 0.45, use α SOLUTION (9-b): =20 = 10.9 = H 0 : µ= H A : µ< α 4. Use test: pg. 17

19 5. Critical Value: 6. Decision/Conclusion: Since > so reject H 0 and accept H A. it means that population mean is less than Q.10(a): For a normal random variable x with mean equal to 30 and standard deviation 5 find the probabilities (i) P ( x ) (ii) P (x ) SOLUTION (10-a-i): ơ ơ pg. 18

20 SOLUTION (10-a-ii): ơ ( ) Compiled & Solved By: JAHANGEER KHAN ơ ( ) ( ) ( ) ( ) ( ) Q.10(b): Draw all possible sample of size 2 with replacement from the population 2, 4, 10 verify that sample mean is an unbiased estimate of population mean. E = µ SOLUTION (10-b): S.no Samples Sample Mean 1. 2, , , , , , , , ,10 10 Total E = µ Verified pg. 19

STATISTICS AND BUSINESS MATHEMATICS B.com-1 Private Annual Examination 2015

B.com-1 STATISTICS AND BUSINESS MATHEMATICS B.com-1 Private Annual Examination 2015 Compiled & Solved By: JAHANGEER KHAN (SECTION A) Q.1 (a): Find the distance between the points (1, 2), (4, 5). SOLUTION