HW15Answers(4.5) - Math331 Spring 2008 Instructor David...

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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 ]. E [ X 2 ] = N summationdisplay i =1 i 2 (1 /N ) = (1 /N ) N ( N + 1)(2 N + 1) 6 = ( N + 1)(2 N + 1) 6 . Therefore, V ar ( X ) = E [ X 2 ] - E [ X ] 2 = ( N + 1)(2 N + 1) 6 - ( N + 1) 2 4 = 2 N 2 + 3 N + 1 6 - N 2 + 2 N + 1 4 = 4 N 2 + 6 N + 2 - 3 N 2 - 6 N - 3 12 = N 2 - 1 12 . Then, σ X = radicalbig V ar ( X ) = radicalbig ( N 2 - 1) / 12. 1
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7. We need V ar ( - 3 X + 5) = E [( - 3 X + 5) 2 ] - ( E [ - 3 X + 5]) 2 = E [9 X 2 - 30 X + 25] - ( - 3 E [ X ] + 5) 2 = 9 E [ X 2 ] - 30 E [ X ] + 25 - 9 E [ X ] 2 + 30 E [ X ] - 25 = 9 E [ X 2 ] - 9 . Therefore, we need E [ X 2 ]. Thankfully, we have one more bit of information we haven’t used
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