Lower bound given expectation and standard deviation. A random variable X with integer values only has mean 3 and standard deviation 2. Under those assumptions, which is the best lower bound for $P[0\leq  X \leq  6]?$. By my calculations, it is $\frac{5}{9}$ however it is not the right answer. Your thoughts please.

You can WLOG assume symmetry about $3$ this way: Suppose $X$ is integer-valued and satisfies $E[X]=3, Var(X)=4$.  Then $Y=6-X$ also satisfies those constraints, as does $$Z  = \left\{ \begin{array}{ll}
X &\mbox{ with prob $1/2$} \\
Y  & \mbox{ with prob $1/2$} 
\end{array}
\right.$$ where the choice is decided by an independent coin flip. Then $Z$ has a symmetric probability mass function about $3$ and also satisfies:
$$P[|Z-3|>3]=P[|X-3|>3]$$
Assuming symmetry makes it easier to prove that you can restrict $X$ to the set $\{-1, ..., 7\}$ (since cases $X\geq 8$ and $X\leq -2$ use up to much variance). Then you can get the best lower bound.
