Solve for the mean and the variance of the discrete random variable x wich can take only the values 2,4,5 and 9 given that P(2) =9/20, P(4) =1/20, P(5) = 1/5 and P(9)= 3/10
The mean:
μ=∑xi⋅P(xi)==2⋅920+4⋅120+5⋅15+9⋅310=4.8.\mu=\sum x_i\cdot P(x_i)=\\ =2\cdot\cfrac{9}{20}+4\cdot\cfrac{1}{20}+5\cdot\cfrac{1}{5}+9\cdot\cfrac{3}{10}=4.8.μ=∑xi⋅P(xi)==2⋅209+4⋅201+5⋅51+9⋅103=4.8.
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={2−4.8,4−4.8,5−4.8,9−4.8}=X-\mu=\begin{Bmatrix} 2-4.8, 4-4.8, 5-4.8, 9-4.8 \end{Bmatrix}=X−μ={2−4.8,4−4.8,5−4.8,9−4.8}=
={−2.8,−0.8,0.2,4.2},=\begin{Bmatrix} -2.8, -0.8, 0.2, 4.2 \end{Bmatrix},={−2.8,−0.8,0.2,4.2},
σ2=(−2.8)2⋅920+(−0.8)2⋅120+0.22⋅15+4.22⋅310=8.86.\sigma^2=(-2.8)^2\cdot \cfrac{9}{20}+(-0.8)^2\cdot \cfrac{1}{20}+0.2^2\cdot \cfrac{1}{5}+4.2^2\cdot \cfrac{3}{10}=8.86.σ2=(−2.8)2⋅209+(−0.8)2⋅201+0.22⋅51+4.22⋅103=8.86.
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