Question about multiple poisson distributions

Question

Suppose $X_1, X_2, X_3, ...X_n$ ~iid Poisson(3), and $\bar{X}_m = (X_1+X_2+X_3+...+X_m)/m$, $\bar{X}_m$ being the mean of the first m terms of X.

a. Find the exact value of $P(2.6<=\bar{X}_2<=3.1)$

b. Estimate $P(2.6<=\bar{X}_{50}<=3.1)$

Answer 1

For a) because the R.V.'s are integer valued, the question reduces to $P(X_1+X_2=6)=e^{-6}\sum_{k=0}^6 \frac{3^k}{k!}\frac{3^{6-k}}{(6-k)!}=\frac{e^{-6}3^6}{6!}2^6=\frac{e^{-6}6^6}{6!}=0.1606$. Here $2^6=\sum_{k=0}^6\binom{6}{k}$.

For b) (as noted in comment) use the normal approximation with mean = $3$ and variance = $\frac{3}{50}$ for $\bar{X}_{50}$. This is the probability $P(-1.633\lt Y \lt .4082)=0.8186$ where $Y$ (normal) has mean $0$ an variance $1$.