sam.mid2.06.sol

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online