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
The mean:
μ=∑xi⋅P(xi)==1⋅1033+2⋅13+3⋅1233=6833≈2.06.\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.μ=∑xi⋅P(xi)==1⋅3310+2⋅31+3⋅3312=3368≈2.06.
The variance:
σ2=∑(xi−μ)2⋅P(xi),\sigma^2=\sum(x_i-\mu)^2\cdot P(x_i),σ2=∑(xi−μ)2⋅P(xi),
X−μ={1−6833,2−6833,3−6833}=X-\mu=\begin{Bmatrix} 1-\cfrac{68}{33}, 2-\cfrac{68}{33}, 3-\cfrac{68}{33} \end{Bmatrix}=X−μ={1−3368,2−3368,3−3368}=
={−3533,−233,3133},=\begin{Bmatrix} \cfrac{-35}{33},\cfrac{-2}{33}, \cfrac{31}{33} \end{Bmatrix},={33−35,33−2,3331},
σ2==(−3533)2⋅1033+(−233)2⋅13+(3133)2⋅1233==2.01.\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.σ2==(33−35)2⋅3310+(33−2)2⋅31+(3331)2⋅3312==2.01.
