CEE 202 HW 4 Solutions

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Unformatted text preview: can be found using either PDF or CDF i) using PDF ∫ 0.1 −∞ 0.1 6(1 − t ) 5 dt = ∫ 6(1 − t ) 5 dt = −(1 − 0.1) 6 + 1 = 0.469 0 ii) using CDF F (0.1) = −(1 − 0.1) 6 + 1 = 0.469 c) Find the probability of y>0.5 This value can be found using either PDF or CDF i) using PDF ∫ ∞ 5 1 6(1 − t ) 5 dt = ∫ 6(1 − t ) 5 dt = −(1 − 1) 6 + (1 − 0.5) 6 = 0.0156 0.5 ii) using CDF ( ) 1 − F (0.5) = 1 − − (1 − 0.5) 6 + 1 = 1 + (1 − 0.5) 6 − 1 = 0.0156 d) For continuous distribution ∞ E ( X ) = ∫ xf ( x)dx −∞ ∞ 1 −∞ 0 E ( y ) = ∫ y ⋅ 6(1 − y )5 dy = ∫ y ⋅ 6(1 − y )5 dy =0.143 4/8 CEE 202 – Engineering Risk & Uncertainty Homework 4 Spring 2011 (Bond) Probability Distributions, Expected Values Therefore, 14.3% of the budget Mr.Vapor should expect to spend on pollution controls e) var iance, σ 2 = E ( X 2 ) − µ 2 For continuous distribution ∞ E ( X 2 ) = ∫ x 2 f ( x)dx −∞ 1 ∞ E ( y 2 ) = ∫ y 2 ⋅ 6(1 − y ) 5 dy = ∫ y 2 ⋅ 6(1 − y ) 5 dy =0.0357 0 −∞ 2 2 µ y = 0.143 = 0.0204 σ 2 = 0.0357 − 0.0204 = 0.0153 5/8 CEE 202 – Engineering Risk &a...
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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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