HW3%20Solution

HW3%20Solution - STAT113 HW3 Solutions 1. (Chapter 4: 100 )...

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STAT113 HW3 Solutions 09/18/09 1. ( Chapter 4: 100 ) Solution: (a) Two conditions are required for f ( x ) to be a legitimate pdf: i. f ( x ) > 0 , x > 0. ii. Z 0 f ( x ) dx = Z 0 32 / ( x + 4) 3 dx = 32 × 1 2( x + 4) 2 0 =1 (b) Let F ( x ) denote the cdf of X. F ( x ) = Z x 0 f ( t ) dt = Z x 0 32 / ( t + 4) 3 dt = 32 × 1 2( t + 4) 2 0 x = 1 - 16 ( x + 4) 2 ( x > 0) (c) P (2 X 5) = F (5) - F (2) = 0 . 2469 (d) E [ X ] = Z 0 xf ( x ) dx = 32 Z 0 x ( x + 4) 3 dx = 32 Z 0 xd ± - 1 2 ( x + 4) - 2 = 32 ± - 1 2 x ( x + 4) - 2 0 - Z 0 ± - 1 2 ( x + 4) - 2 dx = 4 (e) Let g ( x ) = 100 4 + x ,then E [ g ( x )] = Z 0 g ( x ) f ( x ) dx = 3200 Z 0 1 ( x + 4) 4 dx = 3200 × ± - 1 3 1 ( x + 4) 3 0 16 . 67 2. ( Chapter 4: 106 ) Solution: (a) Let F ( x ) denote the cdf. For 1 x 3: F ( x ) = Z x 1 f ( t ) dt = Z x 1 3 2 t 2 dt = 3 2 ± 1 - 1 x Therefore, F ( x ) = 3 2 ( 1 - 1 x ) if 1 x 3 1 if x > 3 0 o.w.
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HW3%20Solution - STAT113 HW3 Solutions 1. (Chapter 4: 100 )...

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