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

1
Expert's answer
2022-03-25T07:00:56-0400

The mean:

μ=xiP(xi)==2920+4120+515+9310=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=(xiμ)2P(xi),\sigma^2=\sum(x_i-\mu)^2\cdot P(x_i),

Xμ={24.8,44.8,54.8,94.8}=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)2920+(0.8)2120+0.2215+4.22310=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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