The original question: Suppose you are throwing a fair-six-sided die, find the probability that at least 80 rolls are necessary to have the sum exceed 300.
My solution: Let $X_i$ be the result of ith row, since they are iid and $\mathbb{E}X=7/2$ and $\sigma_X=\sqrt{35/12}$, by central limit theorem. $$\mathbb{P}(\geq 80 \ \text{rolls to get} \geq 300)=\mathbb{P}(79 th \text{ rolls receive} < 300 )=\mathbb{P}(\sum_{i=1}^{79}X_i<300)$$ $$\mathbb{P}(\sum_{i=1}^{79}X_i<300)=\mathbb{P}(\overline{X}<\frac{300}{79})\approx \mathbb{P}(Z<\frac{\frac{300.5}{79}-3.5}{\sqrt{\frac{35/12}{79}}})=0.92$$ I used Central Limit Theorem in last approach.
I tried a different approach, let $Y$ be the r.v denoting the number of times to receive at least 300. Then clearly $50\leq Y\leq 300$. We just need $\mathbb{P}(Y\geq 80)$ and $\mathbb{E}Y=\frac{300}{3.5}$, is there a way to find $Var(Y)$ and can this be approximated using normal distribution. I think the intuition is yes, because getting an anticipated number can be seen as a success. This is a almost binomail distribution which can be approximated by normal distrbution. Can anyone help me to solve this problem in this approach.
Using CLT and the necessary Continuity Correction Factor you get
$$\mathbb{P}[Y\le 300]=\mathbb{P}\left[Z\le \frac{300.5-79\cdot\frac{7}{2}}{\sqrt{79\cdot\frac{35}{12}}} \right]=\Phi(1.5811)\approx94.31\%$$