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

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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 .
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2. Let X = number of passengers who do not cancel their reservations. Then X has Binomial distribution,
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This note was uploaded on 03/14/2012 for the course STAT 400 taught by Professor Kim during the Spring '08 term at University of Illinois, Urbana Champaign.

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

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