Solution_Assign3_Fall_2010

# Solution_Assign3_Fall_2010 - Solutions for Exercise 4 1 a...

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Solutions for Exercise 4 1. a) c=1/30 since 1 = 3 X =0 c ( X 2 + 4) = 30 C. b) c=1/10 since 1 = 2 x =0 c ( 2 x )( 3 3 - x ) = c [( 2 0 )( 3 3 ) + ( 2 1 )( 3 2 ) + ( 2 2 )( 3 1 )] = 10 c. 2. a) P(T=5)=F(5)-F(4)=3/4-1/2=1/4. b) P ( T > 3) = 1 - F (3) = 1 - 1 / 2 = 1 / 2 . c) P (1 . 4 < T < 6) = F (6) - F (1 . 4) = 3 / 4 - 1 / 4 = 1 / 2 . 3. F ( x ) = 0 , x < 0; 0 . 41 , 0 x < 1; 0 . 78 , 1 x < 2; 0 . 94 , 2 x < 3; 0 . 99 , 3 x < 4; 1 , x 4. 4. a) Area = R 3 1 1 / 2 dx = x/ 2 | 3 1 = 1 b) P (2 < x < 2 . 5) = R 2 . 5 2 1 / 2 dx = x/ 2 | 2 . 5 2 = 1 / 4 . c) P ( x 1 . 6) = R 1 . 6 1 1 / 2 dx = x/ 2 | 1 . 6 1 = 0 . 3 . 5. a) 1 = k R 1 0 xdx = 2 k 3 x 3 / 2 | 1 0 = 2 k/ 3 . Therefore k=3/2. b) F ( x ) = 3 2 R x 0 tdt = t 3 / 2 | x 0 = x 3 / 2 . F (0 . 3 < x < 0 . 6) = F (0 . 6) - F (0 . 3) = (0 . 6) 3 / 2 - (0 . 3) 3 / 2 = 0 . 3004 . 6. a) 3 x =1 3 y =1 f ( x, y ) = c 3 x =1 3 y =1 xy = 36 c = 1 . Hence c=1/36. b) 3 x =1 3 y =1 f ( x, y ) = c 3 x =1 3 y =1 | x - y | = 15 c = 1 . Hence c=1/15. 7. a) P ( X 2 , Y = 1) = f (0 , 1) + f (1 , 1) + f (2 , 1) = 1 / 30 + 2 / 30 + 3 / 30 = 1 / 5 . b) P ( X > 2 , Y 1) = f (3 , 0) + f (3 , 1) = 3 / 30 + 4 / 30 = 7 / 30 . c) P ( X > Y ) = f (1 , 0) + f (2 , 0) + f (3 , 0) + f (2 , 1) + f (3 , 1) + f (3 , 2) = 1 / 30 + 2 / 30 + 3 / 30 + 3 / 30 + 4 / 30 + 5 / 30 = 3 / 5 . d) P ( X + Y = 4) = f (2 , 2) + f (3 , 1) = 4 / 30 + 4 / 30 = 4 / 15 . 8. a)

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Solution_Assign3_Fall_2010 - Solutions for Exercise 4 1 a...

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