The question is: What’s the variance of random variable that takes values from -3 to 3 with probabilities set by PMF
$ P \quad -3 \quad -2 \quad -1 \qquad 0 \quad\ 1 \quad\ 2 \quad 3$
$ X \qquad \frac{1}{14} \qquad \frac{1}{7} \qquad \frac{1}{7} \quad\ \frac{2}{7} \quad \frac{2}{7} \quad 0 \quad \frac{1}{14} $
and let me show you what I did on and keep got wrong answers. I was already found the $\mathbb{E}X = -1/7$ then I subtract the $\mathbb{E}X$ from $X$
$ P \quad -3 \quad -2 \quad -1 \qquad 0 \quad\ 1 \quad\ 2 \quad 3$
$ X \qquad \frac{1}{14} \qquad \frac{1}{7} \qquad \frac{1}{7} \quad\ \frac{2}{7} \quad \frac{2}{7} \quad 0 \quad \frac{1}{14} $
$\mathbb{E} \qquad\ \frac{3}{14} \qquad \frac{2}{7} \qquad \frac{2}{7} \quad\ \frac{3}{7} \quad \frac{3}{7} \quad \frac{1}{7} \quad \frac{3}{14}$
then I squared the values (for shorthand I am going to write $S$ for that row)
$ P \quad -3 \quad -2 \quad -1 \qquad 0 \quad\ 1 \quad\ 2 \quad 3$
$ X \qquad \frac{1}{14} \qquad \frac{1}{7} \qquad \frac{1}{7} \quad\ \frac{2}{7} \quad \frac{2}{7} \quad 0 \quad \frac{1}{14} $
$\mathbb{E} \qquad\ \frac{3}{14} \qquad \frac{2}{7} \qquad \frac{2}{7} \quad\ \frac{3}{7} \quad \frac{3}{7} \quad \frac{1}{7} \quad \frac{3}{14}$
$S \qquad \frac{9}{196} \qquad \frac{4}{49} \quad\ \frac{4}{49} \quad\ \frac{9}{49} \quad \frac{9}{49} \quad \frac{1}{49} \quad \frac{27}{196}$
Then I've multiplied with the corresponding probabilities as $M$
$ P \quad -3 \quad -2 \quad -1 \qquad 0 \quad\ 1 \quad\ 2 \quad 3$
$ X \qquad \frac{1}{14} \qquad \frac{1}{7} \qquad \frac{1}{7} \quad\ \frac{2}{7} \quad \frac{2}{7} \quad 0 \quad \frac{1}{14} $
$\mathbb{E} \qquad\ \frac{3}{14} \qquad \frac{2}{7} \qquad \frac{2}{7} \quad\ \frac{3}{7} \quad \frac{3}{7} \quad \frac{1}{7} \quad \frac{3}{14}$
$S \qquad \frac{9}{196} \qquad \frac{4}{49} \quad\ \frac{4}{49} \quad\ \frac{9}{49} \quad \frac{9}{49} \quad \frac{1}{49} \quad \frac{9}{196}$
$M \qquad \frac{-27}{196} \qquad \frac{-8}{49} \quad\ \frac{-4}{49} \quad\ 0 \quad \frac{9}{49} \quad \frac{2}{49} \quad \frac{27}{196}$
and I summed all the values and get $\frac{-1}{49}$ but this isn't acceptable. Where do I taking a wrong turn? I couldn't get around. many thanks in advance.
The easiest way is to use the following formula
$$\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}^2[X]$$
thus you have
$$\mathbb{E}[X]=-3\cdot\frac{1}{14}-2\cdot\frac{2}{14}-1\cdot\frac{2}{14}+1\cdot\frac{4}{14}+3\cdot\frac{1}{14}=-\frac{2}{14}$$
$$\mathbb{E}[X^2]=9\cdot\frac{1}{14}+4\cdot\frac{2}{14}+1\cdot\frac{2}{14}+1\cdot\frac{4}{14}+9\cdot\frac{1}{14}=\frac{32}{14}$$
that means
$$\mathbb{V}[X]=\frac{32}{14}-\frac{1}{49}=\frac{111}{49}$$