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Unformatted text preview: ISyE 2027 Fall 2011 Hmwk 6 Solution November 2, 2011 1. (a) 1 2 ; (b) 2 3 ; (c) 1 2 ; (d) V ar ( X ) = E ( X 2 ) E ( X ) 2 = 1 18 . (e) F ( x ) = if x ≤ , x 2 if 0 ≤ x ≤ 1 1 if x ≥ 1 2. (a) c = 5; (b) 1 e 12 . 5 ; (c) e 12 . 5 ; (d) F ( x ) = ( if x ≤ , e 5 x if 0 ≤ x < ∞ (e) e 5 t ; (f) e 5 s (Memoryless property). 3. mean 0, variance 1. 4. (a) c = 1 50 ; (b) F ( t ) = if t ≤ , t 50 if 0 ≤ t ≤ 50 1 if x ≥ 50 (c) 25; (d) 625 3 ; (e) R = 2 L 4 + 10; (f) E ( R ) = 25 2 + 10 = 22 . 5; (g) V ar ( R ) = 1 4 V ar ( L ) = 625 12 5. (a) 1 / 4 + f/ 2; (b) 1 4 (c) Binomial distribution with n = 56 + 89 and p = 1 4 so P { N = k } = 145 k (1 / 4) k (3 / 4) 145 k for k = 0 , 1 ,.... (d1) Under H , P { N ≥ 56 } = 145 X k =56 P { N = k } = 145 X k =56 145 k (1 / 4) k (3 / 4) 145 k ≈ 2 . 032 * 10 4 from Maple or Mathematica or ... . 1 Since the probability is so small, it seems reasonable to reject the null hypothesis. (d2) Under the Null hypothesis, mean 145 4 = 36 . 25, standard deviation q 145 1 4 3 4 = 5 . 21. Thus, 56 is around 3.79 standard deviation above the mean....
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 Fall '08
 Zahrn
 Standard Deviation, Null hypothesis, Z1, Qk, kPk

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