solution3 - ---dy e y . We see that the value of K must be...

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EL630: Homework3 Solutions 1) } , , , { } 3 { hhht hhth hthh thhh X = = } , , , , , , , , , , { } 2 { tthh thth thht htth htht hhtt ttth ttht thtt httt tttt X = } , , , , { } 2 { ttth ttht thtt httt tttt X = < ϕ = = } 5 { X . 2) i. Yes ii. No iii. Yes 2) i. No, because 1 ) ( x f X on the interval 5 5 < < - x , so its total area exceeds 1. ii. Yes. To find K note that p K p K k X P k k k - = = = = = 1 ) ( 0 0 . In order that the sum be 1, we must have p K - = 1 . iii. Yes. To find K use the transformation of variables x y ln = . Then dx x dy 1 = and K dy e K dx e x K dx x f y x X π μ 2 ) ( 2 / ) ( 0 2 / ) (ln 0 2 2 = = = - - - - - where the last equality follows from the fact that for a standard normal distribution, 1 2 1 2 / ) ( 2 =
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Unformatted text preview: ---dy e y . We see that the value of K must be 2 1 = K . 3) From the data given for 1000 men, 001 . = p . So 10 ) 001 . )( 10000 ( = = = np . Then, 0103 . ! 10 ) 3 ( 3 10 = = =-k k k e X P . 4) Since X Y for every outcome, Y implies that X , and so } { } { X Y . Therefore, } { } { X P Y P , or ) ( ) ( X Y F F ....
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This note was uploaded on 04/24/2011 for the course ECE 6303 taught by Professor Voltz during the Spring '11 term at NYU Poly.

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solution3 - ---dy e y . We see that the value of K must be...

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