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HW 2 - Homework Set#2 Solutions IE 336 Spring 2011 1(a Let...

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Homework Set #2 Solutions IE 336 Spring 2011 1. (a) Let X equal number of good chips that appear. Then, X Binomial ( n, p ), where n > 1000 and p = 2 / 5. Since n is sufficiently large, the binomial distribution can be approximated using a normal distribution, i.e. X N ( np, np (1 - p )) . P (4 < X < 8) P 4 - np p np (1 - p ) < Z < 8 - np p np (1 - p ) ! P Z < 8 - np p np (1 - p ) ! - P Z < 4 - np p np (1 - p ) ! Φ 8 - np p np (1 - p ) ! - Φ 4 - np p np (1 - p ) ! 0 Since n is very large, Φ( · ) evaluates very close to 0. (b) E ( X ) = np = 2 5 n (c) Let Y equal the number of bad chips selected. Then, the probability that exactly 3 of the 4 selected chips are bad is equal to: P ( Y = 3) = (1 - p ) 3 · p = 3 5 3 2 5 2. X U ( a, b ). The expected value and the variance are equal to: E ( X ) = a + b 2 = 2 V ar ( X ) = ( b - a ) 2 12 = 1 3 Solving for a and b yields a = 1 and b = 3. 1
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E ( X n ) = Z -∞ x n f ( x ) dx = Z b a 1 b - a x n dx = 1 b - a 1 n + 1 x n +1 b a = 1 b - a 1 n + 1 ( b n +1 - a n +1 ) 10 = 1 3 - 1 1 n + 1 ( 3 n +1 - 1 ) 20 = 3 n +1 - 1 n + 1 20 n + 21 = 3 n +1 There is no easy way to solve for n , therefore this answer will be accepted. However, a numerical computation (e.g. Goal Seek in Excel) yields
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