hw6sol - HW6 Solution Section 4.3 1. (5 . 5) = 4 . 5 3 . 5...

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Unformatted text preview: HW6 Solution Section 4.3 1. (5 . 5) = 4 . 5 3 . 5 2 . 5 1 . 5 . 5 = 52 . 34 . Section 4.5 1. (a) Z 1 Ax 3 (1- x ) 2 dx = A Z 1 x 3 (1- 2 x + x 2 ) dx = A Z 1 ( x 3- 2 x 4 + x 5 ) dx = A 1 4- 2 5 + 1 6 = A 60 . Since A 60 = 1, that is A = 60. (b) E X = R 1 60 x x 3 (1- x ) 2 dx = 4 7 E X 2 = R 1 60 x 2 x 3 (1- x ) 2 dx = 5 14 Var X = 3 98 . (c) This is a Beta Distribution with a = 4 , b = 3. Therefore, A = (4 + 3) (4)(3) = (7) (4)(3) = 6! 3!2! = 60 and E X = a a + b = 4 7 , Var X = ab ( a + b ) 2 ( a + b + 1) = 4 3 (4 + 3) 2 (4 + 3 + 1) = 3 98 Section 5.1 ( Note: Some of the numerical answers are different from the solution in Text- book. As I computed all the results from the Table I, and the solution is using statistical package. In the final exam, please bring and use the Table I.) 1. (a) .9099, (b) .5871, (c) .6521, (d) .4249, (e) .2960, (f) .13, (g) -0.58, (h) .40 3 = 10 , 2 = 2 , = 2....
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This note was uploaded on 04/16/2010 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.

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hw6sol - HW6 Solution Section 4.3 1. (5 . 5) = 4 . 5 3 . 5...

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