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10_18ans - STAT 400 1 Fall 2011 Examples for(Answers...

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STAT 400 Examples for 10/18/2011 ( Answers ) Fall 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, n = 100, p = 0.85. Normal approximation: μ = 100 × 0.85 = 85 , σ 2 = 100 × 0.85 × 0.15 = 12.75 . σ = 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 . A fair 6-sided die is rolled 180 times. The sum of the outcomes is likely to be around __________, give or take __________ or so.
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