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 (1-a): As we know that Substituting values: Q.1 (b): Find the roots of the function. SOLUTION (1-b): As we know that By substituting values pg. 1
Hence Solution Set = {, } Q.2 (a): Find the equation of straight line passing through the points (3, -2), (5, 1). SOLUTION (2-a): As we know that Substituting values: Q.2 (b): Find the derivative of the function (i); (ii) SOLUTION (2-b-i): pg. 2
SOLUTION (2-b-ii): Q.3 (a): Solve the equation by Cramer s rule:, SOLUTION (3-a): pg. 3
As we know that Substituting values: Hence Solution Set = {, } Q.3 (b): Given two matrices * +, * + Show that SOLUTION (3-b): * + * + [ ] * + * + * + * + pg. 4
* + * + * + * + [ ] * + * + * + * + (SECTION B) Q.4 (a): Fill in the blanks: (i) Statistics mainly concerned with data. (ii) Ogive is used to find. (iii) The standard deviation is positive square root of. (iv) The limit of coefficient of correlation is. (v) The base year index is always. SOLUTION (4-a): (i) Statistics mainly concerned with sample data. (ii) Ogive is used to find Median. (iii) The standard deviation is positive square root of Variance. (iv) The limit of coefficient of correlation is. (v) The base year index is always 100%. Q.4(b): Find Mean, mode and comments on the shape of distribution SOLUTION (4-b): 12 18 27 20 17 6 12 5 60 12 18 15 270 30 27 25 675 57 20 35 700 77 17 45 765 94 6 55 330 100 Total 100 ----- 2800 ----- pg. 5
Hence Solution Set = {, } Comment: Since Mean > Mode so, the distribution is skewed to the right Q.5(a): Calculate Mean absolute deviation about mean: 32, 28, 47, 63, 75, 39, 10, 60, 96 SOLUTION (5-a): 32-18 18 28-22 22 47-3 3 63 13 13 75 25 25 39-11 11 10-40 40 60 10 10 96 46 46 450 0 188 pg. 6
Q.5(b): Find the variance of the numbers; 1, 2, 3, 4, 5. SOLUTION (5-b): 1 1 2 4 3 9 4 16 5 25 15 55 Q.5(c): Find the coefficient of variation: 3 5 7 9 11 2 5 10 2 1 SOLUTION (5-c): 3 2 9 6 18 5 5 25 25 125 7 10 49 70 490 9 2 81 18 162 11 1 121 11 121 35 20 ----- 130 916 pg. 7
Q.6(a): Find Regression line y on x. X 6 2 10 4 8 y 9 11 5 8 7 SOLUTION (6-a): 6 9 36 54 2 11 4 22 10 5 100 50 4 8 16 32 8 7 64 56 30 40 220 214 pg. 8
Q.6(b): Calculate Karl Pearson coefficient of correlation: x 1 2 3 4 5 6 y 6 4 3 5 4 2 SOLUTION (6-b): 1 6 1 36 6 2 4 4 16 8 3 3 9 9 9 4 5 16 25 20 5 4 25 16 20 6 2 36 4 12 21 24 91 106 75 * +* + * +* + * +* + * +* + pg. 9
Q.6(c): Shift the base year from 1991 to 1995: Year 1991 1992 1993 1994 1995 Price Index % 100 103 105 107 110 SOLUTION (6-c): 1991 100 1992 103 1993 105 1994 107 1995 110 (SECTION C) Q.7(a): Find the number of permutations of the words (i) KARIM (ii) ACCOUNTANT SOLUTION (7-a-i): n 5 5 5 pg. 10
Compiled & Solved By: JAHANGEER KHAN SOLUTION (7-a-ii): n 10 10 10 10 Q.7(b): From a group of 5 men and 3 women. How many combinations are possible with 2 men and 1 woman? SOLUTION (7-b): n ( ) 2 men and 1 woman from 5 men and 3 women can be selected into following ways: 5 x3 ( Q.7(c): ) ( ) Two unbiased coins are tossed once: Find the probability of (i) No head (ii) Two tails SOLUTION (7-c-i): pg. 11
* + * + SOLUTION (7-c-ii): * + * + Q.8(a): A random sample of 36 students selected from different colleges showed an average number of marks 62 with S.D of 5. Construct 95% confidence interval for the average marks of all students. SOLUTION (8-a): (95% Confidence Interval) 5 pg. 12
( ) ( ) Q.8(b): A plant manufactures 8% of defective items. A sample of 6 is selected. What is the probability that: (i) At most two are defective (ii) all are defective. SOLUTION (8-b-i): n 6 6 6 SOLUTION (8-b-i): pg. 13
n 6 Q.9(a): Draw all possible samples each of size 2 from population 9, 11, 15, 21. Show that E=µ. SOLUTION (9-a): If Repetition is not Allowed. E = µ S.no Samples Sample Mean 1. 9, 11 10 2. 9, 15 12 3. 9, 21 15 4. 11, 15 13 5. 11, 21 16 6. 15, 21 18 Total ----- 84 Verified If Repetition is Allowed. S.no Samples Sample Mean 1. 9, 9 9 2. 9, 11 10 3. 9, 15 12 4. 9, 21 15 5. 11, 9 10 6. 11, 11 11 7. 11, 15 13 8. 11, 21 16 9. 15, 9 12 pg. 14
E = µ Compiled & Solved By: JAHANGEER KHAN 10. 15, 11 13 11. 15, 15 15 12. 15, 21 18 13. 21, 9 15 14. 21, 11 16 15. 21, 15 18 16. 21, 21 21 Total ----- 224 Verified Q.9(b): A random variable follows Poisson Dist. With mean is 1.5 Find: (i) (ii) SOLUTION (9-b-i): SOLUTION (9-b-ii): pg. 15
Q.10(a): A random sample of size 16 was found to have and. Test the hypotheses against at SOLUTION (10-a): 1. H 0 : 2. H A : 3. α 1% OR 0.01 4. Use test: 5. Critical Value: 6. Decision/Conclusion: Since > so reject H 0 and accept H A. it means that population mean is not equal to 32. Q.10(b): Given two random samples of size with and Test the Hypotheses against at. SOLUTION (10-b): pg. 16
1. H 0 : 2. H A : 3. α 0.05 4. Use test: ( ) pg. 17
5. Critical Value: 6. Decision/Conclusion: Since > so reject H 0 and accept H A. it means that. pg. 18