0005 ex n p 4000 x 00005 2 ey 10000 ex

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Unformatted text preview: 1/6 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 6 Lots of Distributions c) Cost is a random variable Y = 10,000 X E(Y) = 10,000 E(X) = 40,000 d) Let’s call the new probability p’, the new expected value X’, and the new cost Y’ The new probability would be p’ = 0.0005, E(X’) = n p’ = 4000 x 0.0005 = 2, E(Y) = 10,000 E(X) = 20,000 The program is worth it. Because after spending $17,000, the expected cost of repair will be reduced to $20000 from $40,000. So the total cost is (20000+17000) = $37000 which is less than previous cost of repair 40, 000. 2/6 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 6 Lots of Distributions 2) (10 points) To avoid detection at customs, a traveler obtains 18 fluffy teddy bears and fills 6 of them with illegal substances. If a customs official s...
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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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