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

2. ## Re: CLT problem.

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., $X/120\sim N(6,1/4)$.