exam1-sol - Stat 217 First Exam Solutiosn Spring 2008 1...

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Unformatted text preview: Stat 217 First Exam Solutiosn Spring 2008 1 Short Answers (a) Compute the expected value directly, as an average of value weighted by probabilities: E ( V ) = 4 X k =0 k P { V = k } = 0 + 1 1 2 + 2 1 4 + 3 1 9 + 4 1 8 = 11 6 . (b) The total probability must be 1: 1 = 4 X k =0 P { U = k } = 5 X j =1 jc = 15 c. So, c = 1 / 15. (c) Let N be the number of heads that come up in 1000 flips; this is a sum of independent indicators with E ( N ) = 500. Apply Markovs inequality: P { N > 600 } 500 600 = 5 6 . Or apply Hoeffdings inequality: P { N > 600 } = P { N > E ( N ) + 1000 . 1 } e- 2(1000) 2 (0 . 1) 2 1000 = e- 20 . The second bound is much sharper, but either is fine for the exam. (d) By logic, for X to be 16 it must be greater than 15 but not greater than 16. That is, 1 { X =16 } = 1 { X> 15 }- 1 { X> 16 } so, taking expected values yields Stat 217 First Exam Solutiosn Spring 2008 P { X = 16 } = P { X > 15 } - P { X > 16 } = 1 3- 1 6 = 1 6 ....
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exam1-sol - Stat 217 First Exam Solutiosn Spring 2008 1...

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