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Unformatted text preview: x y =0 (2 x + 2 y4 xy ) dydx = Z 1 x =0 (2 x 3 + 3 x 22 x + 1) dx = 1 2 . 5. (a) Find the cdf of Y , F Y ( y ) ±rst F Y ( y ) = P ( Y ≤ y ) = P (2 X + 3 ≤ y ) = P ( X ≤ y3 2 ) = ± y3 2 ² 2 The pdf f Y ( y ) = d dy F Y ( y ) = 1 2 ( y3) , 3 < y < 5 . And EY = E (2 X + 3) = Z 1 (2 x + 3)2 xdx = 13 3 . (b) F Y ( y ) = P ( Y ≤ y ) = P ( e X ≤ y ) = P ( X ≤ log y ) = (log y ) 2 and the pdf of Y is f Y ( y ) = 2 log y y , 1 < y < e. and EY = Ee X = Z 1 e x 2 xdx = 2 2...
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 Spring '08
 Staff
 Probability, Standard Deviation, Quartile, FY, xy, dx, 1.2kg

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