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

This preview shows pages 1–2. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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).

### Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online