HW 2 - Homework Set #2 Solutions IE 336 Spring 2011 1. (a)...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Homework Set #2 Solutions IE 336 Spring 2011 1. (a) Let X equal number of good chips that appear. Then, X Binomial ( n,p ), where n > 1000 and p = 2 / 5. Since n is sufficiently large, the binomial distribution can be approximated using a normal distribution, i.e. X N ( np,np (1 - p )) . P (4 < X < 8) P 4 - np p np (1 - p ) < Z < 8 - np p np (1 - p ) ! P Z < 8 - np p np (1 - p ) ! - P Z < 4 - np p np (1 - p ) ! Φ 8 - np p np (1 - p ) ! - Φ 4 - np p np (1 - p ) ! 0 Since n is very large, Φ( · ) evaluates very close to 0. (b) E ( X ) = np = 2 5 n (c) Let Y equal the number of bad chips selected. Then, the probability that exactly 3 of the 4 selected chips are bad is equal to: P ( Y = 3) = (1 - p ) 3 · p = ± 3 5 ² 3 ± 2 5 ² 2. X U ( a,b ). The expected value and the variance are equal to: E ( X ) = a + b 2 = 2 V ar ( X ) = ( b - a ) 2 12 = 1 3 Solving for a and b yields a = 1 and b = 3. 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
E ( X n ) = Z -∞ x n f ( x ) dx = Z b a 1 b - a x n dx = ± ± ± ± 1 b - a 1 n + 1 x n +1 ± ± ± ± b a = 1 b - a 1 n + 1 ( b n +1 - a n +1 ) 10 = 1 3 - 1 1 n + 1 ( 3 n +1 - 1 ) 20 = 3 n +1 - 1 n + 1 20 n + 21 = 3 n +1 There is no easy way to solve for n , therefore this answer will be accepted. However, a numerical computation (e.g. Goal Seek in Excel) yields
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/05/2011 for the course IE 336 taught by Professor Bruce,s during the Spring '08 term at Purdue.

Page1 / 4

HW 2 - Homework Set #2 Solutions IE 336 Spring 2011 1. (a)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online