Assignment4 - ASSIGNMENT#4 SOLUTIONS(1 Y U(15 30(a P(Y > 20...

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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 ( - 0 . 2 < Z < 1 . 2) = 1 - 0 . 4207 - 0 . 1151 = 0 . 4642. (b) P (175 Y 185) = P (0 . 45 Zle 0 . 95) = 0 . 3264 - 0 . 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 0 . 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 0 . 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 0 -∞ e ty + y dy + Z 0 e ty - y dy = 1 2 1 t + 1 e ( t +1) y 0 -∞ + 1 t - 1 e ( t - 1) y 0 ! = 1 2 1 t + 1 - 1 t - 1 = 1 1 - t 2 . (6) (a) P ( 3 4 Y 1 1 , 0 Y 2 1 2 ) = Z 1 3 / 4 Z 1 / 2 0 3 y 1 dy 1 dy 2 = Z 1 3 / 4 3 y 2 1 2 1 / 2 0 dy 2 = 3 8 y 2 1 3 / 4 = 3 32 . (b) P ( Y 1 + Y 2 1) = Z 1 / 2 0 Z 1 - y 2 y 2 3 y 1 dy 1 dy 2 = Z 1 / 2 0 3 y 2 1 2 1 - y 2 y 2 dy 2 = 3 2 Z 1 / 2 0 (1 - 2 y 2 ) dy 2 = 3 2 ( y 2 - y 2 2 ) 1 / 2 0 = 3 8 . (7) (a) Y Exp( β ), then for y > 0, f ( y ) = 1 β e - y/β and S ( y ) = e - y/β , so h ( y ) = f ( y ) S ( y ) = 1 β , which is constant (does not depend on y ). (b) Note that h ( y ) = - S 0 ( y ) S ( y ) = - d dy ln[ S ( y )]. Therefore, if
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