HW 2 Solutions

# HW 2 Solutions - E 243 2016 Fall HW#2 Solution Due Sept...

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E 243- 2016 Fall HW #2 Solution Due: Sept 26 th Total Points -100 3-30. An assembly consists of two mechanical components. Suppose that the probabilities that the first and second components meet specifications are 0.95 and 0.98, respectively. Assume that the components are independent. Determine the probability mass function of the number of components in the assembly that meet specifications. (10 points) Sol: X = number of components that meet specifications 𝑃(? = 0) = (0.05)(0.02) = 0.001 𝑃(? = 1) = (0.05)(0.98) + (0.95)(0.02) = 0.068 𝑃(? = 2) = (0.95)(0.98) = 0.931 3-52. The thickness of wood paneling (in inches) that a customer orders is a random variable with the following cumulative distribution function: (2+2+2+2+2 points) Determine the following probabilities: (a) 𝑃(? ≤ 1/18) (b) 𝑃(? ≤ 1/ 4) (c) 𝑃(? ≤ 5 /16) (d) 𝑃(? > 1/ 4) (e) 𝑃(? ≤ 1/ 2) Sol: The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; ?(1/8) = 0.2, ?(1/4) = 0.7, ?(3/8) = 0.1 a) 𝑃(? 1/18) = 0 b) 𝑃(? 1/4) = 0.9 c) 𝑃(? 5/16) = 0.9 d) 𝑃(? > 1/4) = 0.1 e) 𝑃(? 1/2) = 1 3-68. Trees are subjected to different levels of carbon dioxide atmosphere with 6% of them in a minimal growth condition at 350 parts per million (ppm), 10% at 450 ppm (slow growth), 47% at 550 ppm (moderate growth), and 37% at 650 ppm (rapid growth). What are the mean and standard deviation of the carbon dioxide atmosphere (in ppm) for these trees in ppm? (10 points) Sol: μ = E(X) = 350(0.06) + 450(0.1) + 550(0.47) + 650(0.37) = 565 V(X)= 4 1 2 ) )( ( i i x x f = 6875 𝜎 = √𝑉(?) = 82.92

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