Having trouble identifying this probability distribution and its parameters For $y = \frac{1}{4}\text{ }x\text{ }e^{-x/2}$, my initial hunch was that it is a normal distribution but I wasn't able to figure out what the mean and variance would be. What is it? What are its mean and variance?
 A: As a previous answer has suggested, it appears the support for this distribution is:
$$x\in[0,\infty)$$
To verify it as a density, we can integrate across the support and the area should be equal to 1:
$$\begin{align}
\frac{1}{4}\int_{0}^{\infty}xe^{-x/2}\,dx&=\frac{1}{4}\Big[-2xe^{-x/2}\Big]_{0}^{\infty}+\frac{1}{4}\int_{0}^{\infty}2e^{-x/2}\,dx\\
&=0+\frac{1}{4}\Big[-4e^{-x/2}\Big]_{0}^{\infty}\\
&=1
\end{align}$$
Now, as to what density this is, note that the $\text{Gamma}(\alpha,\beta)$ density has the form:
$$f(x)=\frac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}$$
If we let $\alpha=2$ and $\beta=0.5$, this leads to:
$$f(x)=\tfrac{1}{4}xe^{-x/2}$$
Thus, your distribution is that of a $\text{Gamma}(2,0.5)$. This verifies the mean and variance calculations:
$$\begin{align}
E[Y]&=\frac{\alpha}{\beta}=4\\
\text{Var}(Y)&=\frac{\alpha}{\beta^{2}}=8
\end{align}$$
This also confirms the support we defined above. One way that such a Gamma distribution can arise, for example, is if we have:
$$X_{i}\sim \text{Exp}(\lambda=0.5)$$
where all $X_{i}$ are independent and
$$Y=\sum_{i=1}^{2}X_{i}$$
then
$$Y\sim\text{Gamma}(2,0.5)$$
As to why you might think it is a normal distribution, you should note that the equation for $y$ does not follow the form of a normal density at all. Further inspection of the plot of $y$ shows that the density is skewed, which should tell you immediately that it cannot be normal.

A: Evidently the range of this RV is $(0, \infty)$, because that's how you get the density to integrate to 1. 
The mean is $$\frac{1}{4} \int_{0}^{\infty} x^2 {\rm exp}(x/2) dx = 4$$
This is easy to solve using integration by parts. 
Through a similar (but more tedious) calculation, you can find that the second moment is 
$$\frac{1}{4} \int_{0}^{\infty} x^3 {\rm exp}(x/2) dx = 24$$
which means the variance is $24 - 4^2 = 8$. 
