quiz 4 solution

quiz 4 solution - c=-0.5 c It wouldn’t change the value...

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Solution to Quiz4 1. a. It’s discrete, because the possible values are distinct and seperate. b. 0 + 120(0 . 2) + 150(0 . 05) + 200(0 . 03) + 450(0 . 02) = 46 . 5 c. 120 2 (0 . 2) + 150 2 (0 . 05) + 200 2 (0 . 03) + 450 2 (0 . 02) - 46 . 5 2 = 7092 . 75 d. σ x = 7092 . 75 = 84 . 2 e. No, because the price is greater than the mean. 2. a. It’s continuous, because the possible values form a whole interval. b. Since R 0 - 1 x 2 + 1 2 dx = 1 4 > 1 16 ,c must be less than 0 and greater than -1. hence, we try to solve R c - 1 x 2 + 1 2 dx = 1 16 . Then we get the equation 4 c 2 +8 c +3 = (2 c +1)(2 c +3) = 0. The solution to this equation is c=-0.5 or -1.5, but -1.5 is less than -1 hence not our answer. Therefore
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Unformatted text preview: c=-0.5. c. It wouldn’t change the value of c since X is a continuous random variable. 3. a. It’s continuous, because the possible values form a whole interval. b. E ( X ) = R ∞ x . 5 e-. 5 x dx = R ∞-xde-. 5 x =-xe-. 5 x | ∞ + R ∞ e-. 5 x dx = 0 + e . 5 x-. 5 | ∞ = 2 c. R ∞ 10 . 5 e-. 5 x dx =-e-. 5 x | ∞ 10 = e-5 = 0 . 0067. d. P ( X < 20 | X > 10) = P (10 <X< 20) P ( X> 10) = R 20 10 . 5 e-. 5 x dx R ∞ 10 . 5 e-. 5 x dx = e-5-e-10 e-5 = 0 . 9933 1...
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This note was uploaded on 04/29/2010 for the course MATH 218 at USC.

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