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Unformatted text preview: Statistics Assignment #3 4.24 Bridge network reliability (a) P (x > 1) for a1 = 0.85 P (x > 1) for a2 = 0.60 Assuming they are independent events, P (x>1) = 0.85 * 0.60 = 0.51 (b) p(1) for a1 = 0.10 p (1) for a3 = 0.90 p(1) for a6 = 0.25 Therefore p(1) = 0.10 * 0.90 * 0.25 = 0.0225 4.38 Likelihood of flood damage (a) Probability Distribution p(x) x (b) Loss x Sample Point Probability 300,000 Measurable amoun of rain 0.30 No Rain 0.70 E(x) = (300,000)(0.3) + (0)(0.7) = $90,000 Therefore the firms expected losses due to flood are $90,000. This however is only gives an idea of the central tendency and is not actually a possible value of x, because either it will rain and the company will lose $300,000, or it will not rain and the company will not lose anything. 4.60 Parents who condone spanking n = 3 , p = 0.6 X ~ Bin (3, 0.6) (a) P (X=0) =( n ) p^x * q^(nx) x = ( 3 ) 0.6^0 * 0.4^3 = 0.064 (b) P (X 1) = 1 P(X=0) = 1 0.064 = 0.936 (c) mean = = np = 3 x 0.6 = 1.8 standard deviation = = sqrt (npq) = sqrt (3 x 0.6 x 0.4) = 0.85 These results show the central tendency and variation of the sample results. Of course these These results show the central tendency and variation of the sample results....
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This note was uploaded on 08/13/2008 for the course MATH 203 taught by Professor Dr.josecorrea during the Fall '08 term at McGill.
 Fall '08
 Dr.JoseCorrea
 Statistics

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