CEE 202 Homework 9 solution

# 01 288 272 194 378 363 321 257 344 249 a

This preview shows page 1. Sign up to view the full content.

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

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+ ...
View Full Document

## This document was uploaded on 03/06/2014 for the course CEE 202 at University of Illinois, Urbana Champaign.

Ask a homework question - tutors are online