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Unformatted text preview: (0 failures) λ t = (0.0002 hr ­1) (3000 hr) = 0.6 e −0.6 0.6 P (x; λ t) = = 0.548 0! P (failure) = 1 ­ 0.548 = 0.451 b) 10 aircraft 0, 1 or 2 have failures This is binomial (n trials p probability) ȹ 10 ȹ P (X =0) = ȹ ȹ 0.4510 (1 ­0.451) 10 = 0.002 ȹ 0 ȹ ȹ Ⱥ ȹ 10 ȹ P (X =1) = ȹ ȹ 0.4511 (1 ­0.451) 9 = 0.020 ȹ 1 ȹ ȹ Ⱥ ȹ 10 ȹ P (X =0) = ȹ ȹ 0.4512 (1 ­0.451) 8 = 0.075 ȹ 2 ȹ ȹ Ⱥ Total = 0.002 + 0.020 + 0.075 = 0.097 c) Need P (failure) to be 0.01 So, P (0 failure) = 0.99 0 e − λt (λt ) = 0.99 0! e − λt = 0.99 λt = 0.01 t =50 hrs 4/6 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 6 Lots of Distributions Comments on HW grading on Problem 3 a) This question is asking about the probability of one or more failures of an aircraft every 3000 flight hours, NOT just one failure. If you find out parameters for Poisson distribution and plug in the equation, you will get 5 points even...
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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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