Question #317473

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

Expert's answer

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.



The variance:

σ2=∑(xi−μ)2⋅P(xi),\sigma^2=\sum(x_i-\mu)^2\cdot P(x_i),

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}=

={−2.8,−0.8,0.2,4.2},=\begin{Bmatrix} -2.8, -0.8, 0.2, 4.2 \end{Bmatrix},

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.



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