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Unformatted text preview: M362K Probability Homework Solutions Homework 11: Due November 24 Chapter 7 theoretical exercises 50, 54 Chapter 8, problems 2, 3, 7, 11, 15, 16, theoretical exercise 9. Theoretical exercise 7.50 Since Ψ( t ) = log( M ( t )), Ψ ′ ( t ) = M ′ ( t ) /M ( t ), and Ψ ′′ ( t ) = M ( t ) M ′′ ( t ) − M ′ ( t ) 2 M ( t ) 2 . Evaluating at t = 0 gives Ψ ′′ (0) = 1 · E ( X 2 ) − ( E ( X )) 2 1 2 = V ar ( X ). This result can also be understood using the Taylor series log(1 + x ) = x x 2 / 2 + x 3 / 3 + ··· . Since M ( t ) = 1 + tE ( X ) + t 2 E ( X 2 ) / 2+higher powers of t , we have Ψ( t ) = ( tE ( X ) + t 2 E ( X 2 ) / 2) t 2 E ( X ) 2 / 2+higher powers of t . Thus Ψ ′ (0) = E ( X ) and Ψ ′′ (0) = E ( X 2 ) E ( X ) 2 = V ar ( X ). Theoretical exercise 7.54 Cov ( Z,Z 2 ) = E ( Z 3 ) E ( Z ) E ( Z 2 ) = 0 × 1 = 0. Problem 8.2 (a) Let X be the score of a random student. Since E ( X ) = 75, P ( X > 85) ≤ P ( X ≥ 85) ≤ 75 / 85, by Markov’s inequality. However, we can do much better using Chebyshev’s inequality. P ( X > 85) ≤ P (  X 75  > 10) ≤ V ar ( X ) / (10) 2 = 1 / 4. (b) By the same reasoning, the probability of being between 65 and 85 is at least 3/4. (c) If we average n scores, then ¯ X n has a mean of 75, and a variance of 25 /n . The probability of being within 5 of 75 is then at least 1....
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 Spring '11
 Berg
 Central Limit Theorem, Probability, Standard Deviation, Variance, Probability theory, Theoretical Exercise

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