hw3sol - ORIE 3500/5500 Fall Term 2008 Assignment...

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ORIE 3500/5500 Fall Term 2008 Assignment 3-Solution 1. We have p X ( r ) = a + b · r, r = 1 , . . . , 5 and E ( X ) = 11 / 3. So 5 X r =1 ( a + b · r ) = 1 5 a + 15 b = 1 5 X r =1 r ( a + b · r ) = 11 / 3 15 a + 55 b = 11 / 3 . (a) Solving the above two equations, a = 0 and b = 1 / 15. (b) E ( X 2 ) = 5 X r =1 r 2 ( a + b · r ) = 5 r =1 r 3 15 = 225 / 15 = 15 . So var ( X ) = E ( X 2 ) - [ E ( X )] 2 = 15 - (11 / 3) 2 = 14 / 9. 2. (a) Let X be the number of students interviewed by the first investi- gator who owns a car. Then X Binomial (5 , 3 / 4). So P ( X 4) = 1 - P ( x = 5) = 1 - (3 / 4) 5 = 781 / 1024 . (b) Let p be the probability that an investigator finds at least one student who does not own a car. Then p = 781 / 1024. So N Binomial (20 , p ), p N ( r ) = ± 20 r ² (781 / 1024) r (243 / 1024) 20 - r , r = 0 , 1 , . . . , 20 . (c) E ( N ) = 20 p = 20 · 781 / 1024 = 3905 / 256 = 15 . 2539.
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This note was uploaded on 12/04/2008 for the course STSCI 6000 taught by Professor Turnbull during the Fall '08 term at Cornell University (Engineering School).

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hw3sol - ORIE 3500/5500 Fall Term 2008 Assignment...

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