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# hw5sol - Engineering 0020 Probability and Statistics for...

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Engineering 0020: Probability and Statistics for Engineers 1 Spring, 2007; Solutions: Homework #5 Chapter 4.4, 4.5 1. Based on data for Question 81, p139. a Let X be the time between aircraft arrivals. X ~ exp( λ =8) μ x = E(X) = 1/8 hr. = 7.5 minutes b P(X<5) = 1 – e -(8)(5/60) = 0.4866 c P(X>30) = e -(8)(0.5) = 0.0183 d P(X<5 or X>10) = P(X<5) + P(X>10) = 0.4866+ e -(8)(10/60) =0.7502 e P(X>30 | last arrival occurred 30 min. ago) = P(X>30) = 0.0183 This follows from the "Memoryless" property of the Exponential Distribution. 86. Compute the sample and Normal percentiles as shown below: i ( i-0.5 )/ n X i Z i 1 0.025 47.1 -1.96 2 0.075 68.1 -1.44 3 0.125 68.1 -1.15 4 0.175 90.8 -0.93 5 0.225 103.6 -0.76 6 0.275 106 -0.60 7 0.325 115 -0.45 8 0.375 126 -0.32 9 0.425 146.6 -0.19 10 0.475 229 -0.06 11 0.525 240 0.06 12 0.575 240 0.19 13 0.625 278 0.32 14 0.675 278 0.45 15 0.725 289 0.60 16 0.775 289 0.76 17 0.825 367 0.93 18 0.875 385.9 1.15 19 0.925 392 1.44 20 0.975 505 1.96 A plot of X i vs. Z i is shown on the next page:

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Normal Probability Plot 0 100 200 300 400 500 600 -3.00 -2.00 -1.00 0.00 1.00 2.00 3.00 z percentile observed data The plot looks reasonably straight and an assumption of Normality seems to be a plausible one. The intercept (approx. 200+ ??) would be an estimate of the mean and the slope would estimate the std. dev.
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hw5sol - Engineering 0020 Probability and Statistics for...

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