CLT problem.

**juanma101285** · December 11th 2011, 10:39 AM

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 $p(X/120>6.5)=1-p((X/120-6)/(1/2)<1) \approx 1 - \Phi (1)= \Phi(-1)$ .

**emakarov** · December 12th 2011, 08:39 AM

Originally Posted by juanma101285

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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