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Independent trials, dependent condition

  1. Nov 21, 2006 #1
    A friend gave me this problem; it's been years since I've taken probability, and I'm really rusty. I'm curious how to solve it.

    Suppose you have a normal pdf with mean mu and stdev sigma that represents the distribution of test scores.

    What's the probability that two randomly selected students will have an average higher than k?

    So if I let X1 and X2 be the test scores of two randomly selected students, and if they have values of k1 and k2, then I want to know the probability that:

    k1 + k2 > 2k


    P(X1 + X2 > 2k)

    I'm stumped. How does one approach a problem like that?

  2. jcsd
  3. Nov 21, 2006 #2


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    You need to define a new random variable Y = X1 + X2, then calculate P(Y > 2k). If each X is normally distributed, then Y is, too. Proof. (Even if X's were non-normal, The Central Limit Theorem states that the sum of N independent identical random variables approaches the normal distribution as N goes to infinity.)
  4. Nov 21, 2006 #3
    OK, thanks. Is it correct that the mean of the distribution for Y is the sum of the averages (so mu' = mu + mu) and the stdev of the distribution for Y is the RMS sum of the deviations (so sigma' = sqrt{\sigma^2 + \sigma^2} = \sqrt{2}\sigma?

    Thanks! I can't believe how much I've forgotten :surprised
  5. Nov 21, 2006 #4


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