solution_homework6

solution_homework6 - SOLUTIONS TO HOMEWORK 6 (CHAPTER 5)...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SOLUTIONS TO HOMEWORK 6 (CHAPTER 5) Sections 5.1, 5.2: 4, 9, 12, 22, 30, 31 4. a. P 1 (0) = P(X 1 = 0) = p(0,0) + p(0,1) + p(0,2) + p(0,3) = 0.08 + 0.07 + 0.04 + 0.00 = .19 P 1 (1) = P(X 1 = 1) = p(1,0) + p(1,1) + p(1,2) + p(1,3) = .30, etc. The marginal pmf of X 1 x 1 0 1 2 3 4 p 1 (x 1 ) .19 .30 .25 .14 .12 The expected number of customers in line at the express checkout is E(X 1 ) = 0 0.19 + 1 0.30 + 2 0.25 + 3 0.14 + 4 0.12 = 1.7 b. P 2 (0) = P(X 2 = 0) = p(0,0) + p(1,0) + p(2,0) + p(3,0) + p(4,0) = .19, etc The marginal pmf of X 2 x 2 0 1 2 3 p 2 (x 2 ) .19 .30 .28 .23 c. p(4,0) = 0, yet p 1 (4) = .12 > 0 and p 2 (0) = .19 > 0 , so p(x 1 , x 2 ) p 1 (x 1 ) p 2 (x 2 ) for every (x 1 , x 2 ), and the two variables are not independent. 9. a. Since ) , ( y x f is a legitimate joint pdf, ! ! ! ! + = = " " # " " # 30 20 30 20 2 2 ) ( ) , ( 1 dxdy y x K dxdy y x f ! ! ! ! ! !...
View Full Document

This note was uploaded on 04/07/2010 for the course ECE 351 taught by Professor Staff during the Spring '10 term at Ohio State.

Page1 / 3

solution_homework6 - SOLUTIONS TO HOMEWORK 6 (CHAPTER 5)...

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

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