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ch3sol - Chapter 3 Common Families of Distributions 3.1 The...

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Chapter 3 Common Families of Distributions 3.1 The pmf of X is f ( x ) = 1 N 1 - N 0 +1 , x = N 0 , N 0 + 1 , . . . , N 1 . Then E X = N 1 x = N 0 x 1 N 1 - N 0 +1 = 1 N 1 - N 0 +1 N 1 x =1 x - N 0 - 1 x =1 x = 1 N 1 - N 0 +1 N 1 ( N 1 +1) 2 - ( N 0 - 1)( N 0 - 1 + 1) 2 = N 1 + N 0 2 . Similarly, using the formula for N 1 x 2 , we obtain E x 2 = 1 N 1 - N 0 +1 N 1 ( N 1 +1)(2 N 1 +1) - N 0 ( N 0 - 1)(2 N 0 - 1) 6 Var X = E X 2 - E X = ( N 1 - N 0 )( N 1 - N 0 +2) 12 . 3.2 Let X = number of defective parts in the sample. Then X hypergeometric( N = 100 , M, K ) where M = number of defectives in the lot and K = sample size. a. If there are 6 or more defectives in the lot, then the probability that the lot is accepted ( X = 0) is at most P ( X = 0 | M = 100 , N = 6 , K ) = ( 6 0 )( 94 K ) ( 100 K ) = (100 - K ) · · · · · (100 - K - 5) 100 · · · · · 95 . By trial and error we find P ( X = 0) = . 10056 for K = 31 and P ( X = 0) = . 09182 for K = 32. So the sample size must be at least 32. b. Now P (accept lot) = P ( X = 0 or 1), and, for 6 or more defectives, the probability is at most P ( X = 0 or 1 | M = 100 , N = 6 , K ) = ( 6 0 )( 94 K ) ( 100 K ) + ( 6 1 )( 94 K - 1 ) ( 100 K ) . By trial and error we find P ( X = 0 or 1) = . 10220 for K = 50 and P ( X = 0 or 1) = . 09331 for K = 51. So the sample size must be at least 51. 3.3 In the seven seconds for the event, no car must pass in the last three seconds, an event with probability (1 - p ) 3 . The only occurrence in the first four seconds, for which the pedestrian does not wait the entire four seconds, is to have a car pass in the first second and no other car pass. This has probability p (1 - p ) 3 . Thus the probability of waiting exactly four seconds before starting to cross is [1 - p (1 - p ) 3 ](1 - p ) 3 .
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3-2 Solutions Manual for Statistical Inference 3.5 Let X = number of effective cases. If the new and old drugs are equally effective, then the probability that the new drug is effective on a case is . 8. If the cases are independent then X binomial(100 , . 8), and P ( X 85) = 100 x =85 100 x . 8 x . 2 100 - x = . 1285 . So, even if the new drug is no better than the old, the chance of 85 or more effective cases is not too small. Hence, we cannot conclude the new drug is better. Note that using a normal approximation to calculate this binomial probability yields P ( X 85) P ( Z 1 . 125) = . 1303. 3.7 Let X Poisson( λ ). We want P ( X 2) . 99, that is, P ( X 1) = e - λ + λe - λ . 01 . Solving e - λ + λe - λ = . 01 by trial and error (numerical bisection method) yields λ = 6 . 6384. 3.8 a. We want P ( X > N ) < . 01 where X binomial(1000 , 1 / 2). Since the 1000 customers choose randomly, we take p = 1 / 2. We thus require P ( X > N ) = 1000 x = N +1 1000 x 1 2 x 1 - 1 2 1000 - x < . 01 which implies that 1 2 1000 1000 x = N +1 1000 x < . 01 . This last inequality can be used to solve for N , that is, N is the smallest integer that satisfies 1 2 1000 1000 x = N +1 1000 x < . 01 . The solution is N = 537. b. To use the normal approximation we take X n(500 , 250), where we used μ = 1000( 1 2 ) = 500 and σ 2 = 1000( 1 2 )( 1 2 ) = 250.Then P ( X > N ) = P X - 500 250 > N - 500 250 < . 01 thus, P Z > N - 500 250 < . 01 where Z n(0 , 1). From the normal table we get P ( Z > 2 . 33) . 0099 < . 01 N - 500 250 = 2 . 33 N 537 . Therefore, each theater should have at least 537 seats, and the answer based on the approx- imation equals the exact answer.
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Second Edition 3-3 3.9 a. We can think of each one of the 60 children entering kindergarten as 60 independent Bernoulli trials with probability of success (a twin birth) of approximately 1 90 . The probability of having
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