Assignment4

Assignment4 - ASSIGNMENT#4 SOLUTIONS(1 Y ∼ U(15 30(a P...

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Unformatted text preview: ASSIGNMENT #4 SOLUTIONS (1) Y ∼ U(15 , 30) (a) P ( Y > 20) = 30- 20 30- 15 = 2 3 . (b) P ( Y < 25 | Y > 20) = P (20 < Y < 25) P ( Y > 20) = 25- 20 30- 15 2 3 = 1 2 . (2) Y ∼ N(3315 , 575 2 ), 0 . 90 = P ( Y ≤ c ) = P Z ≤ c- 3315 575 = 1- P Z > c- 3315 575 , so c- 3315 575 = 1 . 28 and c = 4051. (3) Y ∼ N(166 , 400). (a) P (162 < Y < 190) = P 162- 166 20 < Z < 190- 166 20 = P (- . 2 < Z < 1 . 2) = 1- . 4207- . 1151 = . 4642. (b) P (175 ≤ Y ≤ 185) = P (0 . 45 ≤ Zle . 95) = 0 . 3264- . 1711 = 0 . 1553. (4) Let Y = # of bad checks in the time interval [0 ,t ], then Y ∼ Poisson ( t 4 ) , let X = time until the first bad check arrives, then P ( X > t ) = P ( Y = 0) = e- t/ 4 , so X ∼ Exp ( 1 4 ) . (a) P ( X > 3) = e- 3 / 4 ≈ . 47. (b) P ( X < 3 | X > 1) = P (1 < X < 3) P ( X > 1) = P ( X > 1)- P ( X > 3) P ( X > 1) = e- 1 / 4- e- 3 / 4 e- 1 / 4 = 1- e- 1 / 2 ≈ . 39. (5) For- 1 < t < 1, we have that m ( t ) = E ( e tY ) = Z ∞-∞ e ty 1 2 e-| y | dy = 1 2 Z-∞ e ty +...
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This note was uploaded on 02/23/2011 for the course MATH 444 taught by Professor Any during the Fall '10 term at Roosevelt.

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