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Homework 4 Solutions _

Homework 4 Solutions _ - 5.28 ~(0,1 Z N Using the z-table...

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Homework 4 Solutions Note: Solutions are incomplete if the solution to the first part is a good representation of the solutions to the other parts. 5.4 2 ,0 1 ( 0, ) 2 ,1 x othe x x rw f e x is x < < < = - .8 .8 2 2 2 1 1 2 2 .2 .2 1 2 .8) (.8) (.2 (.2) .3 X d x P x x - = = = = 1 1.2 1 1.2 2 2 1 1 2 2 .6 1 .6 1 2 1 1 2 1 2 2 2 (.6 1.2) (2 ) 2.4 (1.2) 2 . ) 2 (1 . 32 5 6 .18 . X xdx x dx x x x P = + - = + - = - + - - + = + = 5.6 2 ( ) , 1 k f x x x = -∞ < < ∞ + Find k. 2 1 1 k dx x -∞ = + must be true. 1 1 1 2 2 2 tan lim tan lim tan 1 x x k dx k x k x k x k k k x π π π - - - →∞ →-∞ -∞ -∞ = = - = + = + kπ=1, so k=1/π.
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5.14 2 ,0 1 ( 0, ) 2 ,1 x othe x x rw f e x is x < < < = - Find µ and σ 2 . 2 2 1 2 1 3 0 1 1 1 8 1 1 3 3 1 1 2 2 3 2 1 3 3 3 0 0 ( (2 ) (2 2 ) 0 ) 4 1 1 dx x xdx dx x x d xf x x x x dx xx x x μ -∞ + - + - + - = = = - + - = - + = = We know that 2 2 2 ( ) EX EX σ = - , so 1 2 1 2 16 1 2 1 1 2 1 4 3 4 2 2 3 2 3 4 3 4 3 3 4 0 1 0 1 4 ( ) (2 ) 0 4 1.167 EX x f x dx x dx x x dx x x x -∞ = = + - = + - = - + - - + = Then, 2 2 1.167 1 .167
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Unformatted text preview: 5.28 ~ (0,1) Z N Using the z-table in the text, find the value of z such that .25 of the area lies BELOW. 0.25 .50 0.25 .67 .67 z z z-= -= = (Then do .5 lies below, then do .25 lies ABOVE since the score will be positive.) 5.46 ~ ( , ) ( .025,.025) X Unif a b Unif =-1 20, .025 .025 ( ) .025 .025 0, x f x otherwise =-< < =- - a. ] .015 .015 .01 .01 (.01 .015) 20 20 .1 P X dx x < < = = = ∫ b. ] .012 .012 .012 .012 ( .012 .012) 20 20 .48 P X dx x---< < = = = ∫ Bonus Question will be given at class....
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