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Thread: 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 $\displaystyle X/120 \approx N(6, 1/4)$ 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 $\displaystyle X$ be the total number of turns needed. Then $\displaystyle X \backsim NB(120,1/6)$ is the sum of $\displaystyle 120$ independent RVs with distribution $\displaystyle Geom(1/6)$. The CLT gives $\displaystyle X/120 \approx N(6,1/4)$, so $\displaystyle p(X/120>6.5)=1-p((X/120-6)/(1/2)<1) \approx 1 - \Phi (1)= \Phi(-1)$.
    Last edited by mr fantastic; Dec 11th 2011 at 11:12 AM. 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 $\displaystyle X/120 \approx N(6, 1/4)$ comes from.
    For Geom(p), the mean $\displaystyle \mu=1/p$ and the variance $\displaystyle \sigma^2=\frac{1-p}{p^2}$ (Wikipedia). For $\displaystyle p = 1/6$, we have $\displaystyle \mu=6$ and $\displaystyle \sigma^2=30$. By CLP, $\displaystyle X/120\sim N(\mu,\sigma^2/120)$, i.e., $\displaystyle X/120\sim N(6,1/4)$.
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