The probability distribution of random variable x is given by:

Find K, mean and variance.

We must have $\sum p_{i}=1$

$0.1+K+0.2+2K+0.3+K=1$

$0.6+4K=1$

$4K=1-0.6=0.4$

$K=0.1$

$\therefore$ Put K=0.1 in the table,

Now,

Mean$=E(x)=\sum p_{i}x_{i}=(-2 \times 0.1)+(-1 \times 0.1)+(0 \times 0.2) + (1 \times 0.2)+(2 \times 0.3)+(3 \times 0.1)$

Mean$=-0.2-0.1+0+0.2+0.6+0.3$

$Mean=E(X)=0.8$

$E(x^{2})=\sum p_{i}x_{i}^{2} = 0.1 (-2)^{2}+0.1 (-1)^{2}+0.2 (0)^{2}+0.2(1)^{2}+0.3(2)^{2}+0.1(3)^{2}$

$E(x^{2})=0.4+0.1+0+0.2+1.2+0.9=2.8$

$\therefore $ Variance$=\sigma^{2}=E(x^{2})-[E(x)]^{2}=2.8-(0.8)^{2}=2.8=0.64=2.16$