Chebyshev's Inequality Consider $X_{1},...,X_{30}$ independent Poisson random variables with mean 1.
I need to find a lower bound for
$$
P(25 \le \sum_{i=1}^{30}X_{i} \le 35)
$$
My first thought was that:
$$
\bar{X}_{30} = \frac{1}{30}\sum_{i=1}^{30}X_{i}
$$
so
$$
P(25 \le \sum_{i=1}^{30}X_{i} \le 35) = P(\frac{25}{30} \le \bar{X}_{30} \le \frac{35}{30}) \ge 1 - \frac{V(\bar{X}_{30})}{(5/30)^{2}}
$$
by Chebyshev's inequality since the mean is 1 and $\frac{25}{30} = 1 - 5/30$ and $\frac{35}{30} = 1 + 5/30$. However, this inequality gives me a nonsense result since $V(\bar{X}_{30}) = \sigma^{2}/30$ with $\sigma = 1$ since it's a Poisson distribution.
Where am I going wrong...?
 A: Revised solution
Ok, after some thought, I realized the following:
The complement of the event in Chebyshevs inequality $P[|Y-30|\geq 5] \leq \frac{30}{25}$ is not $P[|Y-30|\leq 5] \geq 1- \frac{30}{25}$ but $P[|Y-30|< 5] > 1- \frac{30}{25}$ (changes to strict inequality) The discreteness of the Poisson makes this distinction important. Knowing this we can apply Chebyshev directly to your problem. But...we must apply it to $P[|Y-30|\geq 6]$ with the upper bound being $\frac{30}{36} = \frac{5}{6}$. Now, due to the discreteness of the Poisson counts, $P[|Y-30|< 6] = P[|Y-30|\leq 5] >1-\frac{5}{6} = \frac{1}{6}$ Which gives $0.16$. Comparing this to the true value of 0.63, we see how loose/conservative Chebyshev's bound is.
However, Chebyshev's inequality is definitely not the tightest bound out there. Since your RVs are independent, I'd take a look at Chernoff Bounds (also)which are tighter. Note that the sum of independent Poisson RVs is also Poisson (in your case Poisson(30)), so it can be directly applied. 
A: This may not be what you're looking for, but I'll include it anyway:
The sum of independent Poisson-variables is actually very nice to work with: If $X_1,X_2,\ldots,X_n$ are independent Poisson-distributed variables with means $\lambda_1,\ldots,\lambda_n$, then $X_1+\cdots+X_n$ is actually Poisson-distributed with mean $\mu:=\lambda_1+\cdots+\lambda_n$.
I'll give you a proof of this below; there are certainly other proofs as well that may be more to your liking.  But the point of the result is that in your case, $X_1+\cdots+X_{30}$ is Poisson-distributed with mean $30$, so that
$$
P\Bigl(25\leq\sum_{i=1}^{30}X_i\leq 35\Bigr)=\sum_{m=25}^{35}\frac{e^{-30}30^m}{m!}\approx0.685374.
$$
As for a proof of the fact that I mentioned: if you can prove that it is true for two variables, then you can use a simple inductive argument to say that it is true for any finite sum. So: suppose $X_1$ and $X_2$ are independent Poisson variables with means $\lambda_1$ and $\lambda_2$. Then
$$
\begin{align*}
P(X_1+X_2=m)&=\sum_{i=0}^{m}P(X_1=i,\ X_2=m-i)\\
&=\sum_{i=0}^{m}\frac{e^{-\lambda_1}\lambda_1^i}{i!}\cdot\frac{e^{-\lambda_2}\lambda_2^{m-i}}{(m-i)!}\\
&=\frac{e^{-(\lambda_1+\lambda_2)}}{m!}\sum_{i=0}^{m}\binom{m}{i}\lambda_1^i\lambda_2^{m-i}\\
&=\frac{e^{-(\lambda_1+\lambda_2)}}{m!}(\lambda_1+\lambda_2)^m
\end{align*}
$$
by the Binomial Theorem; but, this is precisely the mass function for a Poisson variable with mean $\lambda_1+\lambda_2$.
Inductively, if we have proved that the result is true for $n$ Poisson variables, but we have $n+1$: we know that $X_1+\cdots+X_n$ is Poisson-distributed with mean $\lambda_1+\cdots+\lambda_n$; so, $(X_1+\cdots+X_n)+X_{n+1}$ is a sum of two Poisson variables, and our above argument shows that it is therefore Poisson distributed with mean $\lambda_1+\cdots+\lambda_n+\lambda_{n+1}$.
