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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 32x3 x 4 . 0 x<4 E[X] =
 xf (x)dx =
4 x 32x3 dx 1 4 = 32
4 x2 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) = 3e3x , x 0. 3 e2x dx =  e2x 2 This is the density for an exponential random variable with parameter 3. We have E[eX ] =
0 ex 3e3x 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.
 Spring '08
 Anderson
 Math, Probability

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