tut_5_Chap_5_sol - CollegeofEngineering,QatarUniversity

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College of Engineering, Qatar University Page 1of 4 Probability and Statistics, GENG200 2B Tutorial - 5 Question Sections 5-1 Page 163: 1) 5.1 2) 5-2 3) 5-8 SOLUTION: 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 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 e) 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 V(X)=E(X 2 )-[E(X)] 2 =[1 2 (1/4)+1.5 2 (3/8)+2.5 2 (1/4)+3 2 (1/8)]-1.8125 2 =0.4961 V(Y)= E(Y 2 )-[E(Y)] 2 =[1 2 (1/4)+2 2 (1/8)+3 2 (1/4)+4 2 (1/4)+5 2 (1/8)]-2.875 2 =1.8594 f) marginal distribution of X x f(x) 1 ¼ 1.5 3/8 2.5 ¼ 3 1/8 g) ) 5 . 1 ( ) , 5 . 1 ( ) ( 5 . 1 X XY Y f y f y f = and ) 5 . 1 ( X f = 3/8. Then, y ) ( 5 . 1 y f Y 2 (1/8)/(3/8)=1/3 3 (1/4)/(3/8)=2/3 h) ) 2 ( ) 2 , ( ) ( 2 Y XY X f x f x f = and ) 2 ( Y f = 1/8. Then, x ) ( 2 y f X 1.5 (1/8)/(1/8)=1 i) E ( Y | X =1.5) = 2(1/3)+3(2/3) =2 1/3 j) Since f Y|1.5 (y) f Y (y), X and Y are not independent 5-2 Let R denote the range of (X,Y). Because 1 ) 6 5 4 5 4 3 4 3 2 ( ) , ( = + + + + + + + + = c y x f R , 36c = 1, and c = 1/36 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
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tut_5_Chap_5_sol - CollegeofEngineering,QatarUniversity

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