Homework7 - HOMEWORK PROBLEMS#7 SOLUTIONS TO EVEN PROBLEMS...

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HOMEWORK PROBLEMS #7 SOLUTIONS TO EVEN PROBLEMS Chapter 3 122. Y Poisson(7). We use tha table in the back of the book for parts (a) and (b): (a) P ( Y 3) = 0 . 082 (b) P ( Y 2) = 1 - P ( Y < 2) = 1 - P ( Y 1) = 1 - 0 . 007 = 0 . 993 (c) P ( Y = 5) = 7 5 5! e - 7 0 . 13 128. Let Y = # cars that arrive during a one minute phone call. Then λ = 80 cars 1 hour = 80 cars 1 hour 1 hour 60 minutes = 4 3 cars/minute. We want P ( Y 1) = 1 - P ( Y < 1) = 1 - P ( Y = 0) = 1 - e - 4 / 3 0 . 74 130. Let Y = # cars that arrive at entrance I in one hour, and X = # cars that arrive at entrance II in one hour. We are given that Y Poisson(3) and X Poisson(4), and they are independent. We want (we can add because they are disjoint events, we can multiply because they are independent) P ( Y + X = 3) = P ( Y = 0 , X = 3) + P ( Y = 1 , X = 2) + P ( Y = 2 , X = 1) + P ( Y = 3 , X = 0) = P ( Y = 0) P ( X = 3) + P ( Y = 1) P ( X = 2) + P ( Y = 2) P ( X = 1) + P ( Y = 3) P ( X = 0) = 3 0 e - 3 0! 4 3 e - 4 3! + 3 1 e - 3 1! 4 2 e - 4 2! + 3 2 e - 3 2! 4 1 e - 4 1! + 3 3 e - 3 3! 4 0 e - 4 0! = 343 6 e - 7 0 . 052 148. m ( t ) = pe t 1 - qe t , so E ( Y ) = m 0 (0) = pe t (1 - qe t ) 2 t =0 = p (1 - q ) 2 = 1 p E ( Y 2 ) = m 00 (0) = pe t (1 + qe t ) (1 - qe t ) 3 t =0 = p (1 + q ) p 3 = 1 + q p 2 Var( Y ) = E ( Y 2 ) - [ E ( Y )] 2 = 1 + q p 2 - 1 p 2 = q p
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