STATISTICS AND BUSINESS MATHEMATICS - City of …cityofcommercepk.weebly.com/uploads/5/6/7/6/56761669/statistics... · Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS

STATISTICS AND BUSINESS MATHEMATICS

B.com-1 Private Annual Examination 2015

Compiled & Solved By: JAHANGEER KHAN

B.com-1


STATISTICS AND BUSINESS MATHEMATICS - 2015 B.com1 Private Annual Examination 2015

pg. 1

(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. 2

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

( )


( ) ( )

( )

( )

( )

( ) ( )

Q.2 (b): Find the derivative of the function (i); √ (ii)

SOLUTION (2-b-i): √

( )

( )



pg. 3

( )

( )

( )

( )

( )

( )

√

SOLUTION (2-b-ii):

( )

( )( ) ( )( )

( )

( )

Q.3 (a): Solve the equation by Cramer’s rule: ,

SOLUTION (3-a):

|

|

( )( ) ( )( )



pg. 4

|

|

( )( ) ( )( )

|

|

( )( ) ( )( )

As we know that



Q.3 (b): Given two matrices

*

+ , *

+ Show that ( )

SOLUTION (3-b): *

+ *

+

[( )( ) ( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( ) ( )( )

]

*

+

*

+

( ) *

+

( ) *

+



pg. 5

*

+

*

+

*

+ *

+

[( )( ) ( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( ) ( )( )

]

*

+

*

+

( )

*

+ *

+

(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 12 18 27 20 17 6

SOLUTION (4-b): 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. 6

(

)

(

( ) )

(

)

(

)

( )


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. 7

( ) | |

( )

( )

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. 8

√

(

)

√ (

)

√ ( )

√

√

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. 9

( )

( )( ) ( )

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. 10

√

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. 11

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 x 3

( ) ( )

Q.7(c): Two unbiased coins are tossed once: Find the probability of (i) No head (ii) Two tails

SOLUTION (7-c-i):



pg. 12

* + ( ) * + ( )

( ) ( ) ( )

( )

( )

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. 13

√ ( )

√

( √

) ( ) ( √

)

( ) ( ) (

)

( ) ( ) ( ) ( ) ( )

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. 14

( ) ( ) 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.

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

E ( ) = µ

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. 15

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

E ( ) = µ

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. 16

( ) ( )

( )

( ) ( ) ( )

( )

( )

Q.10(a): A random sample of size 16 was found to have and . Test the hypotheses against at

SOLUTION (10-a):

1. H0 : 2. HA : 3. α 1% OR 0.01 4. Use –test:

√ ⁄

√ ⁄

5. Critical Value:

6. Decision/Conclusion: Since | | > | | so reject H0 and accept HA. 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. 17

1. H0 : 2. HA : 3. α 0.05 4. Use –test:

√( )

( )

√( )( )

( )( )

√( )( ) ( )( )

√

√

√

( )

√

( )

√

√

( )



pg. 18

5. Critical Value:

6. Decision/Conclusion:

Since | | > | | so reject H0 and accept HA. it means that .

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STATISTICS AND BUSINESS MATHEMATICS - City of …cityofcommercepk.weebly.com/uploads/5/6/7/6/56761669/statistics... · Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS