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# HW4sol - IE 300/GE 331 Homework 4 Sol ASSIGNMENT 4-8 4-15...

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IE 300/GE 331 Homework 4 Sol ASSIGNMENT: 4-8, 4-15, 4-36, 4-42, 4-62, 4-82, 4-102, 4-180 (from the textbook, 5 th edition) Problem Conversion Guide: Problem number in 5 th Edition Corresponding Problem Number in 4 th Edition 4-8 4-6 4-15 4-15 4-36 4-30 4-42 4-36 4-62 4-54 4-82 4-70 4-102 4-88 4-180 4-150 4-8 The probability density function of the time to failure of an electronic component in a copier (in hours) is g1858(g1876) =g1857 g2879 g3299 g3117g3116g3116g3116 /1000 for g1876>0 . Determine the probability that (a) A component lasts more than 3000 hours before failure. (b) A component fails in the interval from 1000 to 2000 hours. (c) A component fails before 1000 hours. (d) Determine the number of hours at which 10% of all components have failed. a) 05 . 0 1000 ) 3000 ( 3 3000 3000 1000 1000 = = - = = - - - e e dx e X P x x b) 233 . 0 1000 ) 2000 1000 ( 2 1 2000 1000 2000 1000 1000 1000 = - = - = = < < - - - - e e e dx e X P x x c) 6321 . 0 1 1000 ) 1000 ( 1 1000 0 1000 0 1000 1000 = - = - = = < - - - e e dx e X P x x d) 10 . 0 1 1000 ) ( 1000 / 0 0 1000 1000 = - = - = = < - - - x x x e e dx e x X P x x . Then, e x - = / . 1000 09 , and x = - 1000 ln 0.9 = 105.36. 4-15 Determine the cumulative function for the distribution in Exercise 4-1. 4-1 Suppose that g1858(g1876)=g1857 g2879g3051 for 0<g1876 . Since we have:

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IE 300/GE 331 Homework 4 Sol g1858(g1876)={ 0 g1876≤0 g1857 −g1876 g1876>0 We have g1832(g1876)= g1842(g1850≤g1876)=g1516 g1858(g1873)g1856g1873=g1829−g1857 g2879g3051 g1858g1867g1870 g1876>0 g3051 g2879g2998 , where C is a constant. Since when x-> +∞, F(x)=1, we have C=1, Hence, g1832(g1876)={ 0 g1876≤0 1−g1857 −g1876 g1876>0 4-36 The probability density function of the weight of packages delivered by a post office is g1858(g1876)= 70/(69g1876 g2870 ) for 1<g1876<70 pounds. (a) Determine the mean and the variance of weight. (b) If the shipping cost is \$2.50 per pound, what is the average shipping cost of a package? (c) Determine the probability that the weight of a package exceeds 50 pounds. (a) 3101 . 4 1 70 | ln 69 70 69 70 ) ( ) ( 70 1 70 1 = = = = x dx x dx x xf X E 70 69 70 ) ( ) ( 70 1 70 1 2 2 = = = dx dx x f x X E g1848g1853g1870(g1850)= 51.4230 18.5770 70 ) ( ) ( 2 2 = - = - EX X E (b) 2.5*4.3101=10.7753 (c) 0058 . 0 ) ( ) 50 ( 70 50 = = dx x f X P 4-42
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