quiz solution09fall

# quiz solution09fall - Corr ( W,Z ) = Cov ( W,Z ) / p V ar (...

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

ISyE 6413(Fall 2009): “ASSESS YOUR BACKGROUND” Quiz Solution Problem 1. 1. We have 26 letters and 10 digits, so the total number is 26 5 * 10 = 118813760 . 2. 26 5 * 5 = 59406880 . 3. 25 5 * 10 = 97656250 . Problem 2. The event that the sample will contain no more than 2 defective units can be decom- posed into 3 disjoint events: 1. sample contains 0 defective unit, with a probability of (4 / 5) 4 = 256 / 625 . 2. sample contains 1 defective unit, with a probability of (1 / 5) * (4 / 5) 3 = 64 / 625 . 3. sample contains 2 defective units, with a probability of (1 / 5) 2 * (4 / 5) 2 = 16 / 625 . The ﬁnal answer is the sum of the three, which is 336 / 625 . Problem 3. Cov ( W,Z ) = E ( W - E ( W ))( Z - E ( Z )) = E ( X + Y - E ( X ) - E ( Y ))(2 X - E (2 X )) = 2 E ( X + Y - 6)( X - 2) = 2 E ( X 2 - 2 X + XY - 2 Y - 6 X + 12) = 2 E ( X 2 ) - 2 E ( X ) + E ( XY ) - 2 E ( Y ) - 6 E ( X ) + 12 = 229 - 2 * 2 + E ( X ) E ( Y ) - 2 * 4 - 6 * 2 + 12 = 229 - 4 + 2 * 4 - 2 * 4 - 6 * 2 + 12 = 50 V ar ( W ) = E ( W 2 ) - ( E ( W )) 2 = E ( X 2 + Y 2 + 2 XY ) - ( E ( X ) + E ( Y )) 2 = E ( X 2 ) + E ( Y 2 ) + 2 E ( XY ) - (2 + 4) 2 = 29 + 52 + 2 * 2 * 4 - 36 = 61 V ar ( Z ) = E ( Z 2 ) - ( E ( Z )) 2 = E (4 X 2 ) - (2 EX ) 2 = 4 * 29 - 4 * 4 = 100 By deﬁnition,we have

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.

Unformatted text preview: Corr ( W,Z ) = Cov ( W,Z ) / p V ar ( W ) V ar ( Z ) = 50 / 61 * 100 = 0 . 64 1 Problem 4. 1. P ( X > = 750 , 000) = P (( X-500 , 000) / 150 , 000 > = (750 , 000-500 , 000) / 150 , 000) = P ( Z > = 5 / 3) = 0 . 0478 where Z is a standard normal distribution variable and the probability can be found from the normal distribution table or using standard statistical software. 2. Let X be the amount of water supplied. We then have P ( X > = X ) = P (( X-500 , 000) / 150 , 000 > = ( X-500 , 000) / 150 , 000) = P ( Z > = ( X-500 , 000) / 150 , 000) = 0 . 01 Again from the standard normal distribution table or use software, we have ( X-500 , 000) / 150 , 000 = 2 . 33 . Thus X = 849500 . 2...
View Full Document

## This note was uploaded on 12/25/2011 for the course ISYE 6413 taught by Professor Staff during the Spring '08 term at Georgia Institute of Technology.

### Page1 / 2

quiz solution09fall - Corr ( W,Z ) = Cov ( W,Z ) / p V ar (...

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

View Full Document
Ask a homework question - tutors are online