STATISTICS AND BUSINESS...

STATISTICS AND BUSINESS MATHEMATICS

B.com-1 Regular Annual Examination 2015

Compiled & Solved By: JAHANGEER KHAN

B.com-1


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

pg. 1

(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(

) + 15(

) =15(1)

( ) ( )

Comparing

with

Where

Hence Solution Set = { ,

}

Q.1 (b): For the derivatives

in each of the problem. (any Two)

(i) ( ) (ii)

√ (iii) √

SOLUTION (1-b-i): ( )



pg. 2

( )

( )

( )( ) + ( ( )(3)

45

60

30 (2 )

SOLUTION (1-b-ii):

√

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )



pg. 3

( ) ( ) ( )(

( )

)

*( ) +

*( ) ( )

( )

+

*( )

( )

( )

( )

+

*( )

( )

( )

+

*( )

( )

( )

+

*( )( )

( )

+

*

( )

+

*

( )

+

*

( )

+ X *

+

*

( )

+



pg. 4

*

( )

+

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. 5

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. 6

| |

*

+

[

]

[

]

x

[

] x *

+=*

+

[(

) ( ) (

)( ) (

) ( ) (

)( )

(

) ( ) (

)( ) (

) ( ) (

)( )

] =*

+

[

] =*

+

[

] =*

+

[

] =*

+



pg. 7

[

] =*

+

*

+ *

+

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. 8

( )

( )

( )

(

)

Since (

) > 0, so attains minima at

(

)= (

) (

)

(

)=

(

)= ( ) ( )

(

)=

(

)=

(SECTION – B)

Q.4 (a): Calculate A.M, G.M, HM and Mode for the given frequency distribution. 0 1 2 3 4 5

2 2 4 6 8 3

SOLUTION (4-a): 0 2 0 1 2 2 2 4 8 3 6 18 4 8 32 5 3 15

Total 25 75

Note: Since data contains value zero so G.M and H.M cannot be calculated for the given data.



pg. 9

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 2001 2002 2003 2004 2005 2006 2007 2008 2009 production 2046 1776 2134 2380 2785 2765 2420 2595 2425

SOLUTION (4-b): Year Production Link Relatives Chain Indices

2001 2046

x 100=100%

= 100%

2002 1776

x 100=86.80%

= 86.80%

2003 2134

x 100=120.16%

=104.30%

2004 2380

x 100=111.53%

=116.33%

2005 2785

x 100=117.02%

=136.13%

2006 2765

x 100=99.28%

=135.15%

2007 2420

x 100=87.52%

=118.28%

2008 2595

x 100=107.23%

=126.83%

2009 2425

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. 10

Q.5(a): For the following frequency distribution:

C – B 10 – 12 12 – 14 14 – 16 18 – 20 18 – 20 f 14 26 42 08 08

Find Mean Deviation from Mean.

SOLUTION (5-a): C – B | | | |

10 – 12 14 11 154 0.87 3.87 54.18 12 – 14 26 13 338 1.87 1.87 48.62 14 – 16 42 15 630 0.13 0.13 5.46 16 – 18 30 17 510 2.13 2.13 63.9 18 – 20 08 19 152 4.13 4.13 33.04

Total 120 ----- 1,784 ----- ----- 205.2

( ) | |

( )

( )



pg. 11

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, ơx = 4 find and ơy (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 63 65 66 67 67 68 Heights of sons 66 68 65 67 69 70

(i) Find the Karl Pearson Coefficient of Correlation. (ii) Find the equation of the regression line of son on father.

SOLUTION (6-a-i):

63 66 3969 4356 4158 65 68 4225 4624 4420 66 65 4356 4225 4290 67 67 4489 4489 4489 67 69 4489 4761 4623 68 70 4624 4900 4760

396 405 26152 27355 26740

√* ( ) +* ( ) +

( ) ( )( )

√* ( ) ( ) +* ( ) ( ) +

√* +* +



pg. 12

√* +* +

√

√

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. 13

( )

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):

nPr=

5P3=

5P3=

5P3= x x

5P3=

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. 14

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

9

14

10

12

7

13

11

12

88

ơ√

( )

ơ√

(

√ ) ( ) (

√

)

(

) ( ) (

)

( )( ) ( )



pg. 15

( ) ( )

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. 16

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. H0 : Boy and girl are equally likely.

2. HA : Boy and girl are not equally likely.

3. α =0.05

4. Use χ2 –test:

χ2cal= ( )

( ) ( ) ( ) ∑

( )

Boy 210 0.5 200 100 0.5 Girl 190 0.5 200 100 0.5

Total N=400 1 400 ----- 1

χ2cal=1

5. Critical Value:

d.f=n-1=2-1=1

χ2tab= χ20.05; 1=3.841

6. Decision/Conclusion:

Since | | < |

| so accept H0 and reject HA. 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 ( )

0 3 ( ) ( ) 0.04

1 3 ( ) ( ) 0.22

2 3 ( ) ( ) 0.44

3 3 ( ) ( ) 0.30

Total ----- 1



pg. 17

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 = 50

( )

( )

( ) ( )

( )

( )

√

1. H0 : µ=0.45 2. HA : µ<0.45 3. α 4. Use –test:

√ ⁄

√ ⁄

⁄



pg. 18

5. Critical Value:

6. Decision/Conclusion: Since | | > | | so reject H0 and accept HA. it means that population

mean is less than 0.45. 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. 19

SOLUTION (10-a-ii):

ơ ( ≥ )

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, 2 2 2. 2, 4 3 3. 2, 10 6 4. 4, 2 3 5. 4, 4 4 6. 4, 10 7 7. 10, 2 6 8. 10, 4 7 9. 10,10 10

Total ----- 49

E ( ) = µ

Verified

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