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ieor_hw1_sol - IEOR 165 Engineering Statistics Quality...

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IEOR 165: Engineering Statistics, Quality Control and Forecasting, Spring 2008 Homework 1 Solution Chapter 6 Question 4 (a)Since winning a single game is sufficient for winning P (Win) = 1 - P (Lost all) = 1 - (37 / 38) 34 = 0 . 596 Let W be the amount earned in a single game E [ W ] = 1 38 · 35 + 37 38 · ( - 1) = - 0 . 0526 V ar [ W ] = 1 38 · 35 2 + 37 38 - ( - 0 . 0526) 2 = 33 . 21 Then we can use normal approximation P (Profit > 0) = P (Profit > 0 . 5) 1 - Φ 0 . 5 + 0 . 0526 n 33 . 21 n ! (b) n = 1000 P (Profit > 0 . 5) = 1 - Φ(0 . 29) = 0 . 386 (c) n = 100000 P (Profit > 0 . 5) = 1 - Φ(2 . 89) = 0 . 002 Question 8 Let T be the lifetime of 12 batteries: P ( T < 52) Φ 52 - 5 · 12 1 . 5 12 ! = Φ( - 1 . 5396) = 0 . 0618 Question 11 We can approximate the sample mean by a normal random variable with mean E [ X ] and variance V ar ( X ) /n . Then the probability is 1 - Φ(25 n/ 80) (a) 0.2676 (b) 0.1056 (c) 0.0307 (d) 0.0062 Question 17 Using the software we obtain 0.5831. With the approximations (a) P ( X 10) = 10 X i =1 e - 10 10 i i ! = 0 . 5830 (b) P ( X < 10 . 5) = Φ 10 . 5 - 100 · 0 . 1 100 · 0 . 1 · 0 . 9 ! = 0 . 5636 1
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Question 18 Note that ( n - 1) S 2 2 is distributed with χ 2 with n - 1 degrees of freedom. Since the sample is of size n
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