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Unformatted text preview: Statistics 427: Homework #7 Solutions to Assigned Problems 4.11 (a) The cdf is F ( x ) = x < x 2 4 ≤ x < 2 1 2 ≤ x To find P ( X ≤ 1) we compute F (1) = 1 / 4 = 0 . 25. (b) P (0 . 5 ≤ X ≤ 1) = F (1) F (0 . 5) = 1 / 4 1 / 16 = 3 / 16 = 0 . 1875 . (c) P ( X > . 5) = 1 P ( X ≤ . 5) = 1 1 / 16 = 15 / 16 = 0 . 9375 . (d) The median is the solution to 0 . 5 = ˜ μ 2 / 4 which is ˜ μ = √ 2. (e) The density function is f ( x ) = d dx F ( x ) = d dx x 2 4 = x 2 . To be complete, the pdf is therefore f ( x ) = x 2 ≤ x < 2 o.w. (f) E [ X ] = Z 2 wf ( w ) dw = Z 2 w w 2 dw = Z 2 w 2 2 dw = w 3 6 2 = 8 6 = 4 3 . 1 (g) To compute the variance of X we will first compute E X 2 : E X 2 = Z 2 w 2 f ( w ) dw = Z 2 w 3 2 dw = w 4 8 2 = 16 8 = 2 . The variance is therefore Var [ X ] = E X 2 E [ X ] 2 = 2 16 9 = 2 9 . The standard deviation is σ = p Var [ X ] = √ 2 / 3 . (h) This was done above: E X 2 = $2. 4.13 (a) The pdf is f ( x ) = k x 4 x > 1 x ≤ 1 ....
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This note was uploaded on 04/07/2010 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.
 Spring '08
 Staff
 Statistics

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