Finding mean, variance of $X^2$ and $3X$ given the distribution of $X$ If a random variable $X$ has the distribution:

Compute the mean, the variance, and the standard deviation of $X$.


*

*$X^2$?

*$3X$?


I found the mean, variance and SD of $X$ but I don't know to to compute those of $X^2$ and $3X$. Is there a specific formula or do I just have to multiply the $X$?
 A: General formula for mathematical expectation for discrete variables: $$\mathbb {E} [X]=\sum \limits _{i=1}^{\infty }x_{i}\,p_{i},$$ where $X$ is your random variable, 
$x_i$ is i-value of random variable, $p_i$ is probability of i-value of random variable.
$X^2$ is a new random variable with new distribution, but it's generated by $X$ random variable, which has distribution: $$\mathbb{P}(x = 0) = \frac{1}{5}; \mathbb{P}(x = 1) = \frac{1}{10} + \frac{1}{10}  = \frac{1}{5};
 \mathbb{P}(x = 2) = \frac{1}{10} + \frac{1}{5} = \frac{3}{5}$$
We have 3 values, bcs $(-1)^2 = (1^2) = 1$, $(0^2) = 0$ and $(-2^2) = (2^2) = 2.$ 
Similarly, $3X$ has distribution: $$\mathbb{P}(x = -6) = \frac{1}{2}; \mathbb{P}(x = -3) = \frac{1}{5}; \mathbb{P}(x = 0) = \frac{1}{5};     \mathbb{P}(x = 3) = \frac{1}{10}; \mathbb{P}(x = 6) = \frac{1}{10}$$
Thus, $$\mathbb {E} [X^2]=\sum \limits _{i=1}^{3}x_{i}\,p_{i} = 0\cdot\frac{1}{5} + 1 \cdot \frac{1}{5}+2 \cdot\frac{3}{5} = \frac{7}{5}$$
Calculate the same way: $$\mathbb {E} [3X] = ?$$
General formula for variance :$$\mathbb D[X]=\mathbb {E}\left[{\big (}X-\mathbb {E}[X]{\big )}^{2}\right] =\mathbb {E}[X^{2}]-\left(\mathbb {E}[X]\right)^{2}$$
General furmula for SD: $$\sigma ={\sqrt {\mathbb D[X]}}$$.
Can you finish it yourself?
