ch05 - CHAPTER 5 Section 5-1 5-1. First, f(x,y) 0. Let R...

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CHAPTER 5 Section 5-1 5-1. First, f(x,y) 0. Let R denote the range of (X,Y). Then, = + + + + = R y x f 1 ) , ( 8 1 4 1 4 1 8 1 4 1 5-2 a) P(X < 2.5, Y < 3) = f(1.5,2)+f(1,1) = 1/8+1/4=3/8 b) P(X < 2.5) = f (1.5, 2) + f (1.5, 3) +f(1,1)= 1/8 + 1/4+1/4 = 5/8 c) P(Y < 3) = f (1.5, 2)+f(1,1) = 1/8+1/4=3/8 d) P(X > 1.8, Y > 4.7) = f ( 3, 5) = 1/8 5-3. E(X) = 1(1/4)+ 1.5(3/8) + 2.5(1/4) + 3(1/8) = 1.8125 E(Y) = 1(1/4)+2(1/8) + 3(1/4) + 4(1/4) + 5(1/8) = 2.875 5-4 a) marginal distribution of X x f(x) 1 1/4 1.5 3/8 2.5 1/4 3 1/8 b) ) 5 . 1 ( ) , 5 . 1 ( ) 5 . X XY Y f y f = ) 5 . 1 ( X f ( 1 y f and = 3/8. Then, y ) ( 5 . 1 y f Y 2 (1/8)/(3/8)=1/3 3 (1/4)/(3/8)=2/3 c) ) 2 ( ) 2 , ( ) Y XY X f x f x = ( 2 f and = 1/8. Then, ) 2 ( Y f x ) ( 2 y f X 1.5 (1/8)/(1/8)=1 d) E ( Y | X =1.5) = 2(1/3)+3(2/3) =2 1/3 e) Since f Y|1.5 (y) f Y (y), X and Y are not independent 5-5 Let R denote the range of (X,Y). Because , 36c = 1, and c = 1/36 1 ) 6 5 4 5 4 3 4 3 2 ( ) , ( = + + + + + + + + = c y x f R 5-6. a) 4 / 1 ) 4 3 2 ( ) 3 , 1 ( ) 2 , 1 ( ) 1 , 1 ( ) 4 , 1 ( 36 1 = + + = + + = < = XY XY XY f f f Y X P b) P(X = 1) is the same as part a. = 1/4 c) 3 / 1 ) 5 4 3 ( ) 2 , 3 ( ) 2 , 2 ( ) 2 , 1 ( ) 2 ( 36 1 = + + = + + = = XY XY XY f f f Y P d) 18 / 1 ) 2 ( ) 1 , 1 ( ) 2 , 2 ( 36 1 = = = < < XY f Y X P 5-1
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5-7. [] [ ] () 167 . 2 6 / 13 3 2 1 ) 3 , 3 ( ) 2 , 3 ( ) 1 , 3 ( 3 ) 3 , 2 ( ) 2 , 2 ( ) 1 , 2 ( 2 ) 3 , 1 ( ) 2 , 1 ( ) 1 , 1 ( 1 ) ( 36 15 36 12 36 9 = = × + × + × = + + + + + + + + = XY XY XY XY XY XY XY XY XY f f f f f f f f f X E 639 . 0 ) 3 ( ) 2 ( ) 1 ( ) ( 36 15 2 6 13 36 12 2 6 13 36 9 2 6 13 = + + = X V 639 . 0 ) ( 167 . 2 ) ( = = Y V Y E 5-8 a) marginal distribution of X x ) 3 , ( ) 2 , ( ) 1 , ( ) ( x f x f x f x f XY XY XY X + + = 1 1/4 2 1/3 3 5/12 b) ) 1 ( ) , 1 ( ) ( X XY X Y f y f y f = y fy YX 1 (2/36)/(1/4)=2/9 2 (3/36)/(1/4)=1/3 3 (4/36)/(1/4)=4/9 c) ) 2 ( ) 2 , ( ) Y XY X f x f x = ( Y f and 3 / 1 ) 2 , 3 ( ) 2 , 2 ( ) 2 , 1 ( ) 2 ( 36 12 = = + + = XY XY XY Y f f f f x fx XY 1 (3/36)/(1/3)=1/4 2 (4/36)/(1/3)=1/3 3 (5/36)/(1/3)=5/12 d) E ( Y | X =1) = 1(2/9)+2(1/3)+3(4/9) = 20/9 e) Since f XY (x,y) f X (x)f Y (y), X and Y are not independent. 5-9. = R y x f and y x f 1 ) , ( 0 ) , ( 5-10. a) 8 3 4 1 8 1 ) 1 , 5 . 0 ( ) 2 , 1 ( ) 5 . 1 , 5 . 0 ( = + = + = < < XY XY f f Y X P b) 8 3 ) 1 , 5 . 0 ( ) 2 , 1 ( ) 5 . 0 ( = + = < XY XY f f X P c) 8 7 ) 1 , 5 . 0 ( ) 1 , 5 . 0 ( ) 2 , 1 ( ) 5 . 1 ( = + + = < XY XY XY f f f Y P d) 8 5 ) 2 , 1 ( ) 1 , 5 . 0 ( ) 5 . 4 , 25 . 0 ( = + = < > XY XY f f Y X P 5-2
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5-11. 4 1 8 1 2 1 4 1 8 1 8 1 8 1 2 1 4 1 8 1 ) ( 2 ) ( 1 ) ( 1 ) ( 2 ) ( ) ( 1 ) ( 5 . 0 ) ( 5 . 0 ) ( 1 ) ( = + + = = + + = Y E X E 5-12 a) marginal distribution of X x ) ( x f X -1 1/8 -0.5 ¼ 0.5 ½ 1 1/8 b) ) 1 ( ) , 1 ( ) ( X XY X Y f y f y f = y fy YX () 2 1/8/(1/8)=1 c) ) 1 ( ) 1 , ( ) Y XY f x f x = ( Y X f x fx XY 0.5 ½/(1/2)=1 d) E ( X | Y =1) = 0.5 e) no, X and Y are not independent 5-13. Because X and Y denote the number of printers in each category, 4 0 , 0 = + Y X and Y X 5-14. a) The range of (X,Y) is 3 2 1 0 y x 12 3 The problem needs probabilities to total one. Modify so that the probability of moderate distortion is 0.04. 5-3
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x,y f xy (x,y) 0,0 0.857375 0,1 0.1083 0,2 0.00456 0,3 0.000064 1,0 0.027075 1,1 0.00228 1,2 0.000048 2,0 0.000285 2,1 0.000012 3,0 0.000001 b) x f x (x) 0 0.970299 1 0.029403 2 0.000297 3 0.000001 c) E(X)=0(0.970299)+1(0.029403)+2(0.000297)+3*(0.000001)=0.03 (or np=3*0.01) d) ) 1 ( ) , 1 ( ) ( 1 X XY Y f y f y f = , f x (1) = 0.029403 y f Y|1 (x) 0 0.920824 1 0.077543 2 0.001632 e) E(Y|X=1)=0(.920824)+1(0.077543)+2(0.001632)=0.080807 g) No, X and Y are not independent because, for example, f Y (0) f Y|1 (0). 5-4
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5-15 a) The range of (X,Y) is 4 0 , 0 + Y X and Y X . X is the number of pages with moderate graphic content and Y is the number of pages with high graphic output out of 4. x=0 x=1 x=2 x=3 x=4 y=4 5.35x10 -05 000 0 y=3 0.00184 0.00092 0 0 0 y=2 0.02031 0.02066 0.00499 0 0 y=1 0.08727 0.13542 0.06656 0.01035 0 y=0 0.12436 0.26181 0.19635 0.06212 0.00699 b) x=0 x=1 x=2 x=3 x=4 f(x) 0.2338 0.4188 0.2679 0.0725 0.0070 c) E(X)= 2 . 1 ) 0070 . 0 ( 4 ) 7248 . 0 ( 3 ) 2679 . 0 ( 2 ) 4188 . 0 ( 1 ) 2338 . 0 ( 0 ) ( 4 0 = = + + + = i i x f x d) ) 3 ( ) , 3 ( ) ( 3 X XY Y f y f y f = , f x (3) = 0.0725 y f Y|3 (y) 0 0.857 1 0.143 2 0 3 0 4 0 e) E(Y|X=3) = 0(0.857)+1(0.143) = 0.143 f) V(Y|X=3) = 0 2 (0.857)+1 2 (0.143)- 0.143 2 = 0.123 g) No, X and Y are not independent 5-5
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5-16 a) The range of (X,Y) is 4 0 , 0 + Y X
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This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue.

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ch05 - CHAPTER 5 Section 5-1 5-1. First, f(x,y) 0. Let R...

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