Homework6 - p ( y ) = ( 1 n y = 1 , 2 ,...,n otherwise ....

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HOMEWORK PROBLEMS #6 SOLUTIONS TO EVEN PROBLEMS 90. Let Y = # of employees tested until three positives are found. Then Y NegBin(3 , 0 . 4) and we want to find P ( Y = 10) = ± 10 - 1 3 - 1 ² 0 . 4 3 0 . 6 10 - 3 = 2 5 × 3 9 5 10 0 . 064 106. Let Y = # of defective machines chosen. Then Y Hypergeometric( N = 10 ,r = 4 ,n = 5), E ( Y ) = 5 × 4 10 = 2 and Var( Y ) = 5 × 4 10 × 6 10 × 5 9 = 2 3 . We want to compute the mean and variance of Z = $50 × Y . So, E ( Z ) = $50 × E ( Y ) = $100 and Var( Z ) = 50 2 × Var( Y ) = 50 2 × 2 3 = 5000 3 1666 . 67. 110. Y Hypergeometric( N = 6 ,r = 2 ,n = 3) (a) P ( Y = 1) = 2 1 ! 4 2 ! 6 3 ! = 0 . 6 (b) P ( Y 1) = 1 - P ( Y = 0) = 1 - 2 0 ! 4 3 ! 6 3 ! = 1 - 0 . 2 = 0 . 8 (c) P ( Y 1) = P ( Y = 0) + P ( Y = 1) = 0 . 2 + 0 . 6 = 0 . 8 Uniform. For a uniform we have
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Unformatted text preview: p ( y ) = ( 1 n y = 1 , 2 ,...,n otherwise . Therefore, by definition, E ( Y ) = n X y =1 yp ( y ) = n X y =1 y 1 n = 1 n n X y =1 y = 1 n n ( n + 1) 2 = n + 1 2 E ( Y 2 ) = n X y =1 y 2 p ( y ) = n X y =1 y 2 1 n = 1 n n X y =1 y 2 = 1 n n ( n + 1)(2 n + 1) 6 = ( n + 1)(2 n + 1) 6 Var( Y ) = E ( Y 2 )-[ E ( Y )] 2 = ( n + 1)(2 n + 1) 6-³ n + 1 2 ´ 2 = 2 n 2 + 3 n + 1 6-n 2 + 2 n + 1 4 = 4 n 2 + 6 n + 2-3 n 2-6 n-3 12 = n 2-1 12 1...
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This note was uploaded on 02/23/2011 for the course MATH 444 taught by Professor Any during the Fall '10 term at Roosevelt.

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