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Unformatted text preview: Math331, Spring 2008 Instructor: David Anderson Section 11.3 Homework Answers Homework: pg. 484  485, #’s 1, 2, 3, 4, 5, 16 ( k > 0). 1. Let X be the life of a given bill. We know that E [ X ] = 22, where time is in months. Thus, by Markov’s inequality P ( X > 60) ≤ E [ X ] 60 = 22 60 = 0 . 3667 . 2. By Markov, E [ X ] ≥ P ( X ≥ 2)2 = (1 P ( X < 2))2 = (1 3 / 5) × 2 = 4 / 5 . 3. By Markov’s inequality, P ( X ≥ 11) ≤ E [ X ] 11 = 5 11 = . 4545 . By Chebyshev’s inequality, P ( X ≥ 11) = P ( X 5 ≥ 6) . Note that X 5 can not be less than 6 because X is nonnegative. Therefore, the above equality is actually P ( X ≥ 11) = P ( X 5 ≥ 6) = P (  X 5  ≥ 6) ≤ V ar ( X ) 6 2 = E [ X 2 ] E [ X ] 2 36 = 42 25 36 = 17 / 36 = . 4722 . You could also use that for any X P ( X 5 ≥ 6) ≤ P ( { X 5 ≥ 6 } ∪ { X 5 ≤  6 } = P (  X 5  ≥ 6) to arrive at the same conclusion. In this case, the inequalities give similar results....
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This note was uploaded on 06/24/2008 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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