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Unformatted text preview: Math331, Spring 2008 Instructor: David Anderson Section 4.5 Homework Answers Homework: pgs. 182 #’s 3, 4, 5, 7, 8. 3. We first need E [ X ]. E [ X ] = 3 * (7 / 28) 2 * (6 / 28) 1 * (5 / 28) + 0 * (4 / 28) + 1 * 3 / 28 + 2 * 2 / 28 + 3 * 1 / 28 = 1 . Next we need E [ X 2 ]: E [ X 2 ] = 9 * (7 / 28) + 4 * (6 / 28) + 1 * (5 / 28) + 0 * (4 / 28) + 1 * 3 / 28 + 4 * 2 / 28 + 9 * 1 / 28 = 4 . Therefore, Var( X ) = E [ X 2 ] E [ X ] 2 = 4 1 = 3. 4. We first need to determine the probability mass function. We can determine this via the jumps in the distribution function. We see p ( 3) = 3 / 8, p (0) = 3 / 8, and p (6) = 1 / 4. Therefore, E [ X ] = 3 * 3 / 8 + 0 * 3 / 8 + 6 * 1 / 4 = 9 / 8 + 6 / 4 = 3 / 8 . E [ X 2 ] = 9 * 3 / 8 + 0 * 3 / 8 + 36 * 1 / 4 = 99 / 8 . Therefore, Var( X ) = E [ X 2 ] E [ X ] 2 = 99 / 8 9 / 64 = 783 / 64 and σ X = radicalbig 783 / 64 = √ 87(3 / 8). 5. We have calculated in class that E [ X ] = ( N + 1) / 2. We need E [ X 2 ]....
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This note was uploaded on 03/31/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at University of Wisconsin.
 Spring '08
 Anderson
 Math, Probability

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