Is There A Way To Calculate Expected value From Variance And Vice Versa? If I am given the value of one (just the value), can I calculate the value of the other?
 A: In general, no. Given the variance, i.e. $\mathrm{Var}(X)<\infty$, all we can deduce about $\mathbb E[X]$ is that it is finite as well (for intuition, consider that $\mathrm{Var}(X) = \mathbb E[(X-\mathbb E[X])^2]$). Given the mean, we don't even know if $\mathbb E[|X|^2]$ (and hence $\mathrm{Var}(X)$) exists.
For example, let $X$ be a discrete random variable with $$\mathbb P(X=n) = \frac1{n^3\zeta(3)},\, n=1,2,3,\ldots.$$
Then $$\mathbb E[X] = \sum_{n=1}^\infty \frac1{n^2\zeta(3)} = \frac1{\zeta(3)}\sum_{n=1}^\infty \frac1{n^2} = \frac{\zeta(2)}{\zeta(3)}<\infty,$$
but
$$\mathbb E[X^2] = \sum_{n=1}^\infty \frac1{n\zeta(3)}= \frac1{\zeta(3)}\sum_{n=1}^\infty \frac1n = \infty.$$
A: As Ethan Bolker mentions in his hint, for every $\mu$ and $\sigma^2 \geq 0$ there is a random variable with mean $\mu$ and variance $\sigma^2$, namely a normal random variable with these parameters. So the parameters are completely independent.
More generally, it is natural to ask whether there are any constraints on the moment sequence $m_i = \mathbb{E}[X^i]$, where $i = 0, 1, 2, \ldots$. Clearly $m_0 = 1$, and the fact that the variance is non-negative is equivalent to the inequality $m_2 \geq m_1^2$, but are there any other constraints? This is the Hamburger moment problem.
A: Let $a > 0$ and let $X_a$ be the discrete random variable with $P(X_a = a) = 1/2$ and $P(X_a = -a) = 1/2$.  Then
\begin{align*}
    E(X_a) &= 0 \\
    \operatorname{Var}(X_a) &= a^2
\end{align*}
Also, as David points out in the comments, adding a constant to a variable changes the expected value but not the variance.  So these two statistics are independent.
A: No.  For example, for a fair die toss, the expected value is 3.5, and the variance is 2.92.
Now draw an extra pip on every face.  The expected value increases by one, but the variance stays the same.
