Finding mean and variance - Discrete Probability 
So I think I know how to get the variance for these, first we need to find $E(X)$ right? So $E(X)$ for $a$ would be:
$$-1 * \frac{1}{4} + 0 * \frac{1}{2} + 1 * \frac{1}{4} = 0$$
So to find the variance, we need $E((x - E(x))^{2})$

So is it $(-1 - 0)^{2} + (0-0)^2 + (1-0)^2$ which is just $2$?

 A: When you compute the expectation of $E(X-E[X])^2$, remember to compute the weighted sum. 
The variance is 
$$Var(X)=\frac14(-1-0)^2+ \frac12 (0-0)^2+\frac14(1-0)^2=\frac12$$
Alternatively, you can also use the formula, $$Var(X)=E(X^2)-E(X)^2$$
Here $E(X^2)=\frac14(-1)^2+\frac12(0)^2+\frac14(1)^2=\frac12$
Hence $$Var(X)=\frac12-0^2=\frac12$$
A: The mean you have calculated correct. But variance is wrong. You can use formula :
Var(x)=((sigma)(px^2))-(mean^2)
Which comes out to be ((-1)^2)×(1/4) + ((0)^2)×(1/2) + ((1)^2)×(1/4)) -  0 which is 1/2
A: You can calulate the values in the table:
a) $P(X=-1)=1/4, P(X=0)=1/2, P(X=1)=1/4$:
$$\begin{array}{c|c|c}
X&P(X)&XP(X)&[X-\mathbb E(X)]^2\cdot P(X)&X^2P(X^2)\\
\hline
-1&1/4&-1/4&1/4&1/4\\
0&1/2&0&0&0\\
1&1/4&1/4&1/4&1/4\\
\hline
&1&\mathbb E(X)=0&Var (X)=1/2&\mathbb E(X^2)=1/2\\
\end{array}$$
Hence:
$$\mathbb E(X)=\sum XP(X)=0\\
Var (X)=\mathbb E(X-\mathbb E(X))^2=\sum[X-\mathbb E(X)]^2\cdot P(X)=1/2\\
Var (X)=\mathbb E(X^2)-(\mathbb E(X))^2=\sum X^2P(X^2)-\left[\sum XP(X)\right]^2=1/2-0^2=1/2.$$
