The game costs 5 to enter and pays 125 for each head

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Unformatted text preview: snowfall? ! ∞ 45 −∞ 10 E( X ) = ∫x⋅ f ( x )dx = ∫x( −0.00095x + 0.0545)dx 45 ∫ −0.00095 x = 2 + 0.0545 xdx 10 € € 45 = −0.000316 x 3 + 0.0272 x ]10 = 26.32 – 2.40 = 23.9 45 ∞ € 2 σ= ∫x 2 2 ⋅ f ( x ) dx − µ = ∫x 2 ( −0.00095 x + 0.0545) dx − (23.9 2 ) 10 −∞ 45 = ∫ −0.00095 x 3 + 0.0545 x 2 dx − (571.2) 10 € € € 45 = −0.0002375 x 4 + 0.0182 x 3 ]10  ­571.2 = 681.5 ­15.79 ­571.2 = 94.5 5/7 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) σ2 = 94.5 P(X) = Homework 5 Expected Values, More Distributions σ = 9.72 b) What should a manager budget for average annual snow removal? Y = cost, X = inches, Y = 1000 + 400 X E(Y) = 1000 + 400 E(X) = 1000 + 400 (23.9) = $10,560 c) What is the standard deviation of the expected cost for snow removal? σY = 400 σX = 400 (9.72) = $3,890 d) What is the probability that the actual cost in a year will exceed 150% of the average budgeted amount? It’s tempting to work with values of Y, but remember that Y has an offset from X. While we have easy rules for E(Y) and σY, we don’t have easy rules for probabili...
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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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