# HW_5 - Solutions for Homework#5 7 a 0.030 b 0.025 0.015...

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Unformatted text preview: Solutions for Homework #5 7. a. 0.030 b. 0.025+0.015+0.050+0.030=0.12 c. P(exactly one car)=0.05+0.03+0.02=0.07 P(exactly one bus)=0.015+…+0.03=0.3 d. P(Y=1,X=3,4,5)+P(Y=2)=(0.09+0.06+0.03)+(0.01+0.02+0.05+0.06+0.04+0.02)=0.38 e. yes, since ( , ) ( ) ( ) X Y p x y p x p y = ⋅ for all x and y. 12. a. ( 1 ) ( ) [ ] , x y x xy X xy x x f x xe dy xe e dy e xe e x x ∞ ∞- +---- ∞- = = = = <- ∫ ∫ 3 3 3 ( 3) [ ] x x P X e dx e e ∞-- ∞- = =- = ∫ b. ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) 2 2 ( ) [ ] 1 [ ( ) ] (i nt egr al by par t s) 1 1 1 [ ] , (1 ) (1 ) x y x y Y x y x y x y e f y xe dx x dx y e e x dx y y e y y y- + ∞ ∞- +- +- + ∞ ∞- + ∞- ′ = = +-- =- + + = + - = < + + ∫ ∫ ∫ Since ( , ) ( ) ( ) X Y f x y f x f y ≠ , they are not independent. c. P(at least one >3)=1-P(X<3 and Y<3) P(X<3,Y<3)=-- +--------- =- = =- =- =- + =--- ∫ ∫ ∫ ∫ ∫ 3 3 3 ( 1 ) 3 3 3 3 4 12 4 3 3 [ ] 1 ( ) ( ) 1 [ / 4] 1 ( ) 4 4 xy x y x x x x x x x e xe dydx xe dx x e xe dx e e dx x...
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## This note was uploaded on 01/06/2011 for the course STAT 511 taught by Professor Bud during the Fall '08 term at Purdue.

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HW_5 - Solutions for Homework#5 7 a 0.030 b 0.025 0.015...

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