Exam2-solution

Exam2-solution - Math 218 Exam2 Solutions 1. (25) Assume...

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Unformatted text preview: Math 218 Exam2 Solutions 1. (25) Assume the scores of an exam in a class are normally distributed with mean 80 and variance 25. Now a student is randomly selected from that class. (i) What is the probability that his exam score is more than 90? Solution. Let Y = exam score of the selected student. Then Y N (80 , 25). Let Z = Y- 80 5 . Then Z N (0 , 1). P ( Y 90) = P Y- 80 5 90- 80 5 = P ( Z 2) = 1 2- . 4772 = 0 . 0228 . (ii) Find the value k such that with probability 90% his score is less than k . Solution. For the Y,Z in part (i), we have . 9 = P ( Y k ) = P Z k- 80 5 . Then P Z k- 80 5 = 0 . 9- 1 2 = 0 . 4 . By the normal table, we get k- 80 5 = 1 . 28 . Then k = 80 + (1 . 28) 5 = 86 . 4 . 2. (30) Assume students come to my office hours in Math Center randomly over time at a constant expected rate 3 persons per 30 minutes. (i) What is the probability that I have exactly 4 students within 30 minutes?...
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This note was uploaded on 06/29/2008 for the course MATH 218 taught by Professor Haskell during the Spring '06 term at USC.

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Exam2-solution - Math 218 Exam2 Solutions 1. (25) Assume...

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