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# set4solutions - IE 300 GE 331 Discussion Problem Set 4...

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IE 300 / GE 331 Discussion Problem Set 4 4-5 (a), (e), (f) a) The distribution of X is symmetric around 0. Therefore, (0 ) 0.5. P X < = e) ( < 0 > 0.5) = ( 1 < < 1) = 1. f) 1 1 2 3 3 ( ) 1.5 0.5 0.5 0.5 0.05, 0.9655. x x P x X x dx x x x < = = = - = = 4-6 (a), (b), (d) Let X denote the time to failure a) /1000 /1000 3 3000 3000 ( 3000) /1000 0.05. x x P X e dx e e - - - = = - = = b) 2000 2000 /1000 /1000 1 2 1000 1000 (1000 2000) /1000 0.233. x x P X e dx e e e - - - - < < = = - = - = d) /1000 /1000 /1000 0 0 ( ) /1000 1 0.10, 105.36. x x x x x P X x e dx e e x - - - < = = - = - = = 4-14 (c), (d) c) Let X denote the waiting time until a customer arrives. = ( < 40) = / /10 = / | = 1 = 0.9817. Out of five, the number of customers that arrive in 40 minutes is binomial: 1 5 0 (1 ) 5 1 (1 ) = 0.9999994. d) ( ) = / /10 = / | = 1 . (15 ≤ ≤ 30) = (30) (15) = 0.1733. PDF created with pdfFactory Pro trial version www.pdffactory.com

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