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Stat400Lec16(Ch5.7)_ans

# Stat400Lec16(Ch5.7)_ans - STAT 400(Chapter 5.7 Spring 2011...

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STAT 400 (Chapter 5.7) Spring 2011 1. Binomial distribution, n = 25, p = 0.50. Normal approximation: mean = n p = 25 0.50 = 12.5 . n p ( 1 p ) = 25 0.50 0.50 = 6.25 . SD = 25 . 6 = 2.5 . a) P(X = 17) = PMF @ 17 = 0.0322 . b) P(X = 17) = P(16.5 X 17.5) = 5 . 2 5 . 12 5 . 17 Z 5 . 2 5 . 12 5 . 16 P = P(1.60 Z 2.00) = 0.9772 0.9452 = 0.0320 .

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c) P(X 11) = 1 CDF @ 10 = 1 0.2122 = 0.7878 . d) P(X 11) = P(X 10.5) = 5 . 2 5 . 12 5 . 10 Z P = P(Z 0.80) = 1 0.2119 = 0.7881 . e) P(10 X 14) = CDF @ 14 CDF @ 9 = 0.7878 0.1148 = 0.6730 . f) P(10 X 14) = P(9.5 X 14.5) = 5 . 2 5 . 12 5 . 14 Z 5 . 2 5 . 12 5 . 9 P = P( 1.20 Z 0.80) = 0.7881 0.1151 = 0.6730 .
2. Let X = number of passengers who do not cancel their reservations. Then X has Binomial distribution, n = 100, p = 0.85. Normal approximation: = 100 0.85 = 85 , 2 = 100 0.85 0.15 = 12.75 . = 3.57 . P ( X 92 ) = P ( X 92.5 ) = 57 . 3 85 5 . 92 Z P = P ( Z 2.10 ) = 0.9821 . Binomial: P ( X 92 ) = 0.9878 . 3. Binomial distribution, n = 180, p = 1 / 6 .

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