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Unformatted text preview: 5 + 1.5 = 4 b) Variance σT2 = σA12 + σA22 + 2 ρ σA1 σA2= 2.5 + 1.5 + 2 (0.4) √2.5 √1.5 = 5.55 Standard deviation = √5.55 = 2.36 c) Damages Let D be total damages. D = 5000 A1 + 6000 A2 µD = 5000 µA1 + 6000 µA2 = 5000 x 2.5 + 6000 x 1.5 = $21,500 2/6 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 9 Expected Values; Sampling 3) A random sample of engineering sophomores gave the following grade point averages: 3.01 2.88 2.72 1.94 3.78 3.63 3.21 2.57 3.44 2.49 a) What is the mean of this sample? b) What is the standard deviation of this sample? Ans: a) We know that the sample mean, X = (1/10)* (3.01+2.88+2.72+1.94+3.78+3.63+3.21+2.57+3.44+2.49) = 2.967 b) We also know that the sample variance, S2 = ⦋1/(10 ­1)⦌*⦋(3.01 ­2.967)2+(2.88 ­2.967)2+(2.72 ­2.967)2+(1.94 ­2.967)2+ ...
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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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