Exam2-solution

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

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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?...
View Full Document

## This note was uploaded on 06/29/2008 for the course MATH 218 taught by Professor Haskell during the Spring '06 term at USC.

### Page1 / 3

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

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

View Full Document
Ask a homework question - tutors are online