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Math Help - CLT problem.

  1. #1
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    CLT problem.

    Hi,

    I have the following problem that comes with the solution, but I don't understand where X/120 \approx N(6, 1/4) comes from. Thanks a lot for your help!

    The problem and the solution are here:

    Problem:
    I roll a fair die until I get a six. I do this 120 times. Estimate the probability that I need more than an average of 6.5 rolls each time.

    Solution:
    Let X be the total number of turns needed. Then X \backsim NB(120,1/6) is the sum of 120 independent RVs with distribution Geom(1/6). The CLT gives X/120 \approx N(6,1/4), so p(X/120>6.5)=1-p((X/120-6)/(1/2)<1) \approx 1 - \Phi (1)= \Phi(-1).
    Last edited by mr fantastic; December 11th 2011 at 12:12 PM. Reason: Title.
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  2. #2
    MHF Contributor
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    Re: CLT problem.

    Quote Originally Posted by juanma101285 View Post
    I have the following problem that comes with the solution, but I don't understand where X/120 \approx N(6, 1/4) comes from.
    For Geom(p), the mean \mu=1/p and the variance \sigma^2=\frac{1-p}{p^2} (Wikipedia). For p = 1/6, we have \mu=6 and \sigma^2=30. By CLP, X/120\sim N(\mu,\sigma^2/120), i.e., X/120\sim N(6,1/4).
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