Dr. Hackney STA Solutions pg 28

Dr. Hackney STA Solutions pg 28 - 6 1 94 K-1 100 K By trial...

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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 x = N 0 x 1 N 1 - N 0 +1 = 1 N 1 - N 0 +1 ± N 1 X x =1 x - N 0 - 1 X x =1 x ! = 1 N 1 - N 0 +1 ² N 1 ( N 1 +1) 2 - ( N 0 - 1)( N 0 - 1 + 1) 2 ³ = 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 ² N 1 ( N 1 +1)(2 N 1 +1) - N 0 ( N 0 - 1)(2 N 0 - 1) 6 ³ 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 ) + (
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Unformatted text preview: 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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