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solution_homework6

solution_homework6 - SOLU TIONS TO HOM EWOR K 6(C HAPTER 5...

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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 ! ! ! ! ! ! + = + = 30 20 2 30 20 2 30 20 30 20 2 30 20 30 20 2 10 10 dy y K dx x K dxdy y K dydx x K 000 , 380 3 3 000 , 19 20 = ! " # $ % & ( = K K b. P(X < 26 and Y < 26) = ! ! ! = + 26 20 2 26 20 26 20 2 2 12 ) ( dx x K dxdy y x K 3024 . 304 , 38 4 26 20 3 = = K Kx c. I II 2 + = x y 2 ! = x y 20 20 30 30 III
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P( | X – Y | 2 ) = !! III region dxdy y x f ) , ( = !! !! " " II I dxdy y x f dxdy y x f ) , ( ) , ( 1 = ! ! ! ! " + " " 30 22 2 20 28 20 30 2 ) , ( ) , ( 1 x x dydx y x f dydx y x f = ! ! ! !
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