Answer to Question #320917 in Statistics and Probability for Lyns

Question

Calculate the mean and the variance of the discrete random variable x which one values 12 and 3, given that P(1)=10/33, P(2)=1/3 and P(3)=12/33

Answer 1

The mean:

$\mu=\sum x_i\cdot P(x_i)=\\ =1\cdot\cfrac{10}{33}+2\cdot\cfrac{1}{3}+3\cdot\cfrac{12}{33}=\cfrac{68}{33}\approx2.06.$

The variance:

$\sigma^2=\sum(x_i-\mu)^2\cdot P(x_i),$

$X-\mu=\begin{Bmatrix} 1-\cfrac{68}{33}, 2-\cfrac{68}{33}, 3-\cfrac{68}{33} \end{Bmatrix}=$

$=\begin{Bmatrix} \cfrac{-35}{33},\cfrac{-2}{33}, \cfrac{31}{33} \end{Bmatrix},$

$\sigma^2=\\ =\begin{pmatrix} \cfrac{-35}{33}\end{pmatrix}^2\cdot\cfrac{10}{33}+\begin{pmatrix} \cfrac{-2}{33}\end{pmatrix}^2\cdot\cfrac{1}{3}+\begin{pmatrix} \cfrac{31}{33}\end{pmatrix}^2\cdot\cfrac{12}{33}=\\ =2.01.$