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