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. $$\begin{array} { c c }
x & f ( x ) \
\hline 0 & 0.80 \
1 & 0.15 \
2 & 0.04 \
3 & 0.01
\end{array}$$ The mean and the standard deviation for the number of electrical outages (respectively) are _____.
A) 2.6 and 5.77
B) 0.26 and .577
C) 3 and .01
D) 0 and .8

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. $$\begin{array} { c c } 
x & f ( x ) \
\hline 0 & 0.80 \
1 & 0.15 \
2 & 0.04 \
3 & 0.01
\end{array}$$ The mean and the standard deviation for the number of electrical outages (respectively) are _____.
A) 2.6 and 5.77
B) 0.26 and .577
C) 3 and .01
D) 0 and .8

Quizplus · Accepted Answer

The mean (μ) of a probability distribution is calculated as $\sum [x \cdot f(x)]$, and the standard deviation (σ) is the square root of the variance, which is $\sum [(x - \mu)^2 \cdot f(x)]$. For this distribution, $\mu = (0 \cdot 0.80) + (1 \cdot 0.15) + (2 \cdot 0.04) + (3 \cdot 0.01) = 0.26$. The variance is $(0 - 0.26)^2 \cdot 0.80 + (1 - 0.26)^2 \cdot 0.15 + (2 - 0.26)^2 \cdot 0.04 + (3 - 0.26)^2 \cdot 0.01 = 0.3322$, and the standard deviation is the square root of 0.3322, which is approximately 0.577. Therefore, the mean is 0.26 and the standard deviation is approximately 0.577.

The Number of Electrical Outages in a City Varies from Day