Answer:
Therefore, the mean and the standard deviation for the number of electrical outages (respectively) are 0.26 and 0.5765 respectively.
Step-by-step explanation:
Given the probability distribution table below:
[tex]\left|\begin{array}{c|cccc}x&0&1&2&3\\P(x)&0.8&0.15&0.04&0.01\end{array}\right|[/tex]
(a)Mean
Expected Value, [tex]\mu =\sum x_iP(x_i)[/tex]
=(0*0.8)+(1*0.15)+(2*0.04)+(3*0.01)
=0+0.15+0.08+0.03
Mean=0.26
(b)Standard Deviation
[tex](x-\mu)^2\\(0-0.26)^2=0.0676\\(1-0.26)^2=0.5476\\(2-0.26)^2=3.0276\\(3-0.26)^2=7.5076[/tex]
Standard Deviation [tex]=\sqrt{\sum (x-\mu)^2P(x)}[/tex]
[tex]=\sqrt{0.0676*0.8+0.5476*0.15+3.0276*0.04+7.5076*0.01}\\=\sqrt{0.3324}\\=0.5765[/tex]
Therefore, the mean and the standard deviation for the number of electrical outages (respectively) are 0.26 and 0.5765 respectively.
