CEE 202 Homework 9 solution

# 221 1091 therefore p x 64 p z1091 01379 using

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Unformatted text preview: (n − 1) s 2 We know, χ 2 = 2 σ So, 71.42= 50*s2/4.22; s =5.02 For 95% probability range the sample standard deviation will lie between 3.379 and 5.02. e) 90% probability that the sample mean lies between 64 and 66 ksi Z = (X (bar)  ­ µ)/( σ/√n) For P = 0.05 in lower tail from table, z =  ­1.645 4/6 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 9 Expected Values; Sampling  ­1.645 = (64 ­65)/(4.2/√n); n =47.7 ~48 For P = 0.95 in upper tail from table, z = 1.645 1.645= (66 ­65)/(4.2/√n); n =47.7~48 To ensure a 90% probability that the sample mean lies between 64 and 66 ksi, the sample number will be 48 5/6 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 9 Expected Values; Sampling 5) Repeat each step of Problem 4, but now assume that only 21 sampl...
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## This document was uploaded on 03/06/2014 for the course CEE 202 at University of Illinois, Urbana Champaign.

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