Find the mean and standard deviation of the following probability distribution: `x_i : \ -2 \ \ -1\ \ 0\ \ 1\ \ 2` `p_i :\ \ 0. 1\ \ 0. 2 \ \ 0. 4\ \ 0. 2 \ \ 0. 1`

Answer

`Mean=0` and Standard Deviation`=1.09`

Explanation

The mean of the probability distribution is:


`Mean = E(X)=Sigma x_i cdot p_i`


`=-2 times 0.1+(-1) times 0.2+0 times 0.4+1 times 0.2+2 times 0.1`


`=-0.2-0.2+0+0.2+0.2=0`


`rArr E(X)=0`


Now,


`E(X^2)=Sigma x_i^2 times p_i`


`= (-2)^2 times 0.1+ (-1) ^2 times 0.2+ 0^2 times 0.4+ 1^2 times 0.2+2^2 times 0.1`


...