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3502(0809)midterm-sol

3502(0809)midterm-sol - STAT 3502(Winter 2009 SOLUTIONS TO...

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STAT 3502 (Winter 2009) SOLUTIONS TO MIDTERM TEST 1. 10 · 9 · 8 · 7 · 6 10 5 = 0 . 3024 . 2. Let us introduce the events: G = { student graduates } and S = { student studies } . By assumption P ( S ) = 0 . 75 , P ( G | S ) = 0 . 85 , P ( G | S ) = 0 . 35 . (a) By the Law of Total Probability P ( G ) = P ( G | S ) P ( S ) + P ( G | S ) P ( S ) = (0 . 85)(0 . 75) + (0 . 35)(1 - 0 . 75) = 0 . 725 . (b) According to the Bayes’ Theorem, using the result of part (a), P ( S | G ) = P ( G | S ) P ( S ) P ( G | S ) P ( S ) + P ( G | S ) P ( S ) = P ( G | S ) P ( S ) P ( G ) = (0 . 85)(0 . 75) 0 . 725 = 0 . 879 . 3. Because each engine is assumed to fail or function independently of what happens with the other engines, it follows that the number X of engines remaining operative is a binomial r.v. Hence, the probability that a four-engine plane makes a successful flight is P ( X 2) = 4 2 p 2 (1 - p ) 2 + 4 3 p 3 (1 - p ) 1 + 4 4 p 4 (1 - p ) 0 = 6 p 2 (1 - p ) 2 + 4 p 3 (1 - p ) + p 4 = 3 p 4 - 8 p 3 + 6 p 2 . 4. Let X be the number of lines of code that have a mistake. Then X Bin(20 , 0 . 02) , where p = 0 . 02 < 0 . 1 and np = (20)(0 . 02) = 0 . 4 10 . Therefore the Poisson approximation can safely be used and we get (a) P ( X = 0) e - 0 . 4 (0 . 4) 0 0!

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3502(0809)midterm-sol - STAT 3502(Winter 2009 SOLUTIONS TO...

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