sam.mid2.06.sol

# sam.mid2.06.sol - Solutions to Sample STAT 200 Midterm Exam...

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Solutions to Sample STAT 200 Midterm Exam II 1. F, T, T, T, T. 2 (16 pts). ii, ii, iv, iv. 3. (a) Let X be the content of a randomly selected bottle. Then X N (298 , 3). We have ¯ X = 4 X i =1 X i N (298 , 3 2 ) = N (298 , 1 . 5) . Thus P (295 < ¯ X < 299) = P ( 295 - 298 1 . 5 < Z < 299 - 298 1 . 5 ) = P ( - 2 < Z < 0 . 67) = 0 . 7486 - 0 . 0228 = 0 . 726 . (b) (10 pts) Let p be the probability that a randomly selected bottle is good. Then p = P ( X 300) = P ( Z 300 - 298 3 ) = P ( Z > 0 . 67) = 0 . 2514 . Let Y be the number of good bottles among the 100 bottles. Then Y follows a binomial distribution with n = 100 and p = 0 . 2514, and E ( Y ) = np = 100 × 0 . 2514 , V ( Y ) = np (1 - p ) = 100 × 0 . 2514 × (1 - 0 . 2514) . Thus, based on the Central Limit Theorem, we have P ( Y 80) P ( Z < 80 - 100 × 0 . 2514 q 100 × 0 . 2514 × (1 - 0 . 2514) ) = P ( Z < 12 . 70) = 1 . 4. (a) Let μ be the population change in stress level. We want to test H 0 : μ = 0 vs. H a : μ < 0 Test statistic

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## This note was uploaded on 05/09/2011 for the course STAT 200 taught by Professor Karim during the Spring '08 term at UBC.

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sam.mid2.06.sol - Solutions to Sample STAT 200 Midterm Exam...

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