# hw5 - Solutions to Homework 5 Chapter 5 5.9 a âˆ âˆ âˆ âˆ =...

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Unformatted text preview: Solutions to Homework 5 Chapter 5 5.9 a. âˆ« âˆ« âˆ« âˆ« + = = âˆž âˆž- âˆž âˆž- 30 20 30 20 2 2 ) ( ) , ( 1 dxdy y x K dxdy y x f âˆ« âˆ« âˆ« âˆ« âˆ« âˆ« + = + = 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 = â‡’ â‹… = K K b. P(X < 26 and Y < 26) = âˆ« âˆ« âˆ« = + 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. P( | X â€“ Y | â‰¤ 2 ) = âˆ«âˆ« III region dxdy y x f ) , ( âˆ«âˆ« âˆ«âˆ«-- II I dxdy y x f dxdy y x f ) , ( ) , ( 1 âˆ« âˆ« âˆ« âˆ«- +-- 30 22 2 20 28 20 30 2 ) , ( ) , ( 1 x x dydx y x f dydx y x f = (after much algebra) .3593 d. f x (x) = 30 20 3 2 30 20 2 2 3 10 ) ( ) , ( y K Kx dy y x K dy y x f + = + = âˆ« âˆ« âˆž âˆž- = 10Kx 2 + .05, 20 â‰¤ x â‰¤ 30 e. f y (y) is obtained by substituting y for x in (d); clearly f(x,y) â‰  f x (x) â‹… f y (y), so X and Y are not independent....
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hw5 - Solutions to Homework 5 Chapter 5 5.9 a âˆ âˆ âˆ âˆ =...

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