Question

The number of electrical outages in a city varies from day to day. Assume that the number of electrical outages ( x ) in the city has the following probability distribution.xf (x)00.8010.1520.0430.01The mean and the standard deviation for the number of electrical outages (respectively) are _____.

Answer 1

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:

$\left|\begin{array}{c|cccc}x&0&1&2&3\\P(x)&0.8&0.15&0.04&0.01\end{array}\right|$

(a)Mean

Expected Value, $\mu =\sum x_iP(x_i)$

=(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

$(x-\mu)^2\\(0-0.26)^2=0.0676\\(1-0.26)^2=0.5476\\(2-0.26)^2=3.0276\\(3-0.26)^2=7.5076$

Standard Deviation $=\sqrt{\sum (x-\mu)^2P(x)}$

$=\sqrt{0.0676*0.8+0.5476*0.15+3.0276*0.04+7.5076*0.01}\\=\sqrt{0.3324}\\=0.5765$

Therefore, the mean and the standard deviation for the number of electrical outages (respectively) are 0.26 and 0.5765 respectively.