STATISTICS AND BUSINESS MATHEMATICS

1 STATISTICS AND BUSINESS MATHEMATICS Regular Annual Examination 2015 Compiled & Solved By: JAHANGEER KHAN Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg. 1 (SECTION A) (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 = { , } (b): For the derivatives in each of the problem. (any Two) (i) ( ) (ii) (iii) SOLUTION (1-b-i): ( ) Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg.

2 2 ( ) ( ) ( )( ) + ( ( )(3) 45 60 30 (2 ) SOLUTION (1-b-ii): ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg. 3 ( ) ( ) ( )( ( ) )*( ) + *( ) ( ) ( ) + *( ) ( ) ( ) ( ) + *( ) ( ) ( ) + *( ) ( ) ( ) + *( )( ) ( ) + * ( ) + * ( ) + * ( ) + X * + * ( ) + Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg.)

3 4 * ( ) + SOLUTION (1-b-iii): ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (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: ( ) ( ) *( ) ( )( )+ ( ) Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg. 5 Hence Solution Set = { , } SOLUTION (2-a-iii): As we know that By substituting values ( ) ( ) ( )( ) ( ) ( ) Hence Solution Set = { , } (b): Find the inverse of the following square matrix A than verify that A-1 x A=I.

4 A=* + SOLUTION (2-b): * + | | | | | | ( )( ) ( )( ) | | | | * + Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg. 6 | | * + [ ] [ ] x [ ] x * +=* + [( )( ) ( )( )( )( ) ( )( )( )( ) ( )( )( )( ) ( )( )] =* + [ ] =* + [ ] =* + [ ] =* + Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg. 7 [ ] =* + * + * + (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 =[ ] (b): Examine maximum and minimum value of the function y =.

5 SOLUTION (3-b): ( ) ( ) ( ) ( ) Now taking ( ) . 0 = 27= = Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg. 8 ( ) ( ) ( ) ( ) Since ( ) > 0, so attains minima at ( )= ( ) ( ) ( )= ( )= ( ) ( ) ( )= ( )= (SECTION B) (a): Calculate , , 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 and cannot be calculated for the given data. Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg.

6 9 Mode = The most repeated value in the data So, Mode = 4 Hence solution set = { , and cannot be calculated, Mode=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= = 2003 2134 x 100= = 2004 2380 x 100= = 2005 2785 x 100= = 2006 2765 x 100= = 2007 2420 x 100= = 2008 2595 x 100= = 2009 2425 x 100= = (c): If an investor buys shares of at a price of per share of at price of per share.

7 Calculate the average price per share. SOLUTION (4-c): Type 1 Shares Type 2 Shares Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg. 10 (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 12 14 26 13 338 14 16 42 15 630 16 18 30 17 510 18 20 08 19 152 Total 120 ----- 1,784 ----- ----- ( ) | | ( ) ( ) Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg.

8 11 (b): In a moderately skewed frequency distribution: Mean = and Median = find Mode. SOLUTION (5-b): Mean = Median = Mode =? ( ) ( ) (c): Given = 20, x = 4 find and y (mean and sd of y) . SOLUTION (5-c): ( ) ( ) Hence Solution Set = { , } (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 * ( ) +* ( ) + ( ) ( )( ) * ( ) ( ) +* ( ) ( ) + * +* + Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg.

9 12 * +* + SOLUTION (6-a-ii): ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) (b): The average is 68 and is 4 of marks of section A. the average is 52 and is 12 marks of section B. which is more consistent? SOLUTION (6-b): Section A Section B Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg. 13 ( ) x ( ) x ( ) ( ) x ( ) x ( ) Conclusion: Since ( )< ( ), it means that section A is more consistent than section B. (SECTION C) (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?

10 SOLUTION (7-a): nPr= 5P3= 5P3= 5P3= x x 5P3= (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 *( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) + ( ) ( ) ( ) ( ) ( ) ( ) Compiled & Solved By: JAHANGEER KHAN STATISTICS AND BUSINESS MATHEMATICS - 2015 Regular Annual Examination 2015 pg. 14 (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.