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# hw6sol - HW6 Solution Section 4.3 1(5.5 = 4.5 3.5 2.5 1.5...

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HW6 Solution Section 4.3 1. Γ(5 . 5) = 4 . 5 × 3 . 5 × 2 . 5 × 1 . 5 × 0 . 5 × π = 52 . 34 . Section 4.5 1. (a) Z 1 0 Ax 3 (1 - x ) 2 dx = A Z 1 0 x 3 (1 - 2 x + x 2 ) dx = A Z 1 0 ( 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 0 60 x · x 3 (1 - x ) 2 dx = 4 7 E X 2 = R 1 0 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 = Γ(4 + 3) Γ(4)Γ(3) = Γ(7) Γ(4)Γ(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 μ = 10 , σ 2 = 2 , σ = 2. (a) P ( X 10 . 34) = P ( Z 10 . 34 - 10 2 ) = Φ( . 24) = . 5948 . (b) P ( X 11 . 98) = 1 - Φ(1 . 4) = 1 - . 9192 = 0 . 0808 (c) P (7 . 67 X 9 . 90) = P ( - 1 . 65 Z ≤ - 0 . 07) = 0 . 4721 - 0 . 0495 = 0 . 4216 (d) P (10 . 88 X 13 . 22) = Φ(2 . 28) - Φ( . 62) = . 9887 - . 7324 = . 2563 . (e) P ( | X - 10 | ≤ 3) = P (7 X 13) = Φ(2 . 12) - Φ( - 2 . 12) = . 9830 - . 0170 = . 9660 . 1

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(f) First find z such that P ( Z z ) = 0 . 81. That is Φ( z ) = 0 . 81, z = . 88. Then, x = μ + σ ( . 88) = 10 + 2( . 88) = 11 . 24. (g) First find z such that P ( Z z ) = 0 . 04. That is Φ( z ) = 0 . 96, z = 1 . 75. Then, x = μ + σ (1 . 75) = 10 + 2(1 . 75) = 12 . 47. (h) To find
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