Let $X_1,\cdots, X_{65}$ be a sequence of i.i.d. discrete random variables such that for each $i \in$ {$1,\cdots, 65$}, $\mu_{X_{i}} = 0.7$ and $\sigma_{X_i} = 0.1$. Then, with $Y=X_1+\cdots+X_{65}$, approximate $\mathbb{P}\left(Y>46\right)$.
I've been struggling with this question all afternoon. I've calculated $\mu_Y = \sqrt{0.65}$ and $\sigma_Y = 45.5$ but the best I've been able to do (via Markov's inequality) is show that $\mathbb{P}\left(Y>46\right)$ less than about 0.99, which is pretty useless.
If anyone has any ideas about how to get the bounds any tighter, I'd appreciate the input.
I think that you have to use the fact that $$\mathbb P(Y>46)=\mathbb P\left( \frac{Y-\mathbb E[Y]}{\sqrt{\operatorname{Var}(Y)}}>\frac{46-\mathbb E[Y]}{\sqrt{\operatorname{Var}(Y)}}\right)$$ and use the fact that $\frac{Y-\mathbb E[Y]}{\sqrt{\operatorname{Var}(Y)}}$ has a distribution close to that of a standard normal random variable $N$. Therefore, the wanted approximation is $$ \mathbb P(Y>46) \approx \mathbb P\left( N>\frac{46-\mathbb E[Y]}{\sqrt{\operatorname{Var}(Y)}}\right). $$ It remains to simplify $\frac{46-\mathbb E[Y]}{\sqrt{\operatorname{Var}(Y)}}$ and use the table of the distribution function of a standard normal distribution.