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Stats problem involving the Bienayme-Chebyshev inequality

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Question 3 here: http://www.stat.washington.edu/peter/395/samplemidterm.pdf [Broken]

    Solution to it here: http://www.stat.washington.edu/peter/395/prmt.sln/sln.html [Broken]

    By the way, I could use help soon since this is the practice exam for an exam I'm taking 7 hours from now!!!!

    2. Relevant equations

    Bienayme-Chebyshev inequality: P(|X-E(X)|>t) ≤ Var(X) / t2

    3. The attempt at a solution

    So I'm just confused at how he gets the expected value. He multiplies the chance of choosing a first pair of students to be boy and girl by 100 .... which doesn't make sense since with each successive choosing of pairs the probability of choosing boy and girl depends on whether the last pair choice happened to be boy and girl.

    Also, on the solution to problem 1 he says "Assuming the density is defined on x>1 ...." How would I have known this on an exam????
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 11, 2013 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    E(sum) = sum E, and this is true even if the terms are dependent. The expected value for each pair is the same as for the first pair, because, for example, the expectation for pair 17 is obtainable by laying out all 200 students in a line (a random permutation), then picking the first two, then the next two, etc until reaching students 33 and 34 to get the 17th pair. The gender characteristics of the pair occupying positions 33 and 34 are probabilistically the same as for those occupying positions 1 and 2---the random permutation does not distinguish between those pairs.

    This can be hard to believe at first, I know, but if you think about it it will start to make sense.
     
    Last edited by a moderator: May 6, 2017
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