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There are two Certified Public Accountants in a particular

Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye ISBN: 9780321629111 32

Solution for problem 115E Chapter 3

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 115E

There are two Certified Public Accountants in a particular office who prepare tax returns for clients. Suppose that for a particular type of complex form, the number of errors made by the first preparer has a Poisson distribution with mean value ?1, the number of errors made by the second preparer has a Poisson distribution with mean value ?2, and that each CPA prepares the same number of forms of this type. Then if a form of this type is randomly selected, the function gives the pmf of X = the number of errors on the selected form. a.? ?Verify that ?p?(?x?; ?1, ?2) is in fact a legitimate pmf ( ? 0and sums to 1). b.?? hat is the expected number of errors on the selected form? c.? hat is the variance of the number of errors on the selected form? d. ?How does the pmf change if the first CPA prepares 60% of all such forms and the second prepares 40%?

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Answer: Step1: There are two certified public accountants in a particular office who prepare tax returns for clients. The pmf is given by 1 x 2 x 1 e 1 1e 2 P(x; ,1 )2 2 x! + 2 x! x = 0,1,2…. Given, the pmf of ‘x’ is the number of errors on the selected form. Step2: a). To verify that P(x; ,1) i2 in fact a legitimate pmf ( 0and sums to 1). Therefore, 0.5 1 x 2 x P(x; ,1 ) 2 x!( e 1e 2) 0 Here, e 1,e 2> 0 and a1d ar2 positive. And x! is always positive (x 0) P(x; ,1 ) 2 1 P(x, ) 1 1 P(x, )2 x=0 2 x=0 2 x=0 = 1 + 1 ( where, P(x, )= 1) 2 2 x=0 = 1. Therefore, P(x; , 1 =21. Step3: b). Here we need to find the expected number of errors on the selected form. 1 1 E(x) = P(x; , )1= 2 2 P(x, )1+ 2 P(x, ) 2 x=0 x=0 x=0 = 1 ( + ) 2 1 2 Therefore, E(x) = 1 (1+ )2 . 2 Step4: c). Here we need to find the variance of the number of errors on the selected form. Therefore, V(x) = E(x ) - [E(x)] 2 Where, E(x ) = x P(x; , ) 1 2 2 1 2 1 2 E(x ) = 2 x P(x, ) 1 2 x P(x, ) 2 x=0 x=0 For poisson distribution, var (y,) = = E(y ) [E(y)] 2 Where, E(y )= 2 + 2 Since, E(x ) = 1 ( + ) + ( + ) 2 2 1 1 2 2 2 1 ( + ) + ( + ) 2 1 2 Therefore, V(x) = 2 1 1 2 2 2 - [2( 1 )]2 Step 5: d). To find the pmf change if the first CPA prepares 60% of all such forms and the second prepares 40%. The pmf is given by 1 x 2 x P(x; , )= 0.6 e 1 + 0.4 e 2 . 1 2 x! x!

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Chapter 3, Problem 115E is Solved
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Textbook: Probability and Statistics for Engineers and the Scientists
Edition: 9
Author: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye
ISBN: 9780321629111

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