# HW 21 Answers (6.3) - Math331, Spring 2008 Instructor:...

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Unformatted text preview: Math331, Spring 2008 Instructor: David Anderson Section 6.3 Homework Answers Homework: pgs. 254 - 255, #'s 1, 2, 4. 1. Differentiating shows that the density of X is given by f (x) = Therefore, the expected value is 32x-3 x 4 . 0 x<4 E[X] = - xf (x)dx = 4 x 32x-3 dx 1 4 = 32 4 x-2 dx = 32 = 8. The variance does not exist because E[X ] = 4 2 x 32 x dx = -2 2 4 32dx = , and so V ar(X) = E[X 2 ] - E[X]2 does not exist. 2. We are given that f (x) = Therefore, 2 6(x - 1)(2 - x) if 1 < x < 2 0 else E[X] = - xf (x)dx = 1 2 x 6(x - 1)(2 - x)dx =- 1 6x3 - 18x2 + 12x dx 2 x=1 3 = - x4 + 6x3 - 6x2 2 3 = . 2 The second moment of X is 2 E[X 2 ] = 1 x2 6(x - 1)(2 - x)dx = 23 . 10 Therefore, V ar(X) = 23/10 - 9/4 = 1/20. Thus, the standard deviation is X = 1 V ar(X) = . 20 1 4. We have f (x) = 3e-3x , x 0. 3 e-2x dx = - e-2x 2 This is the density for an exponential random variable with parameter 3. We have E[eX ] = 0 ex 3e-3x dx = 3 0 x=0 3 = . 2 2 ...
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## This homework help was uploaded on 03/31/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at Wisconsin.

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HW 21 Answers (6.3) - Math331, Spring 2008 Instructor:...

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