The following table lists the relative frequency distribution of the number of calls coming into a call center each hour. Each call takes five minutes to process. What is the mean and standard deviation of the number of minutes the operators are answering questions each hour?

x 2 3 4 5 6 7
P(x) 0.07 0.11 0.23 0.31 0.18 0.1

Solution

The mean of the number of minutes the operators are
answering questions each hour is

$$ \begin{aligned} E(X)&= \sum x*p(x)\\ &= 2*0.07+3*0.11+4*0.23+5*0.31+6*0.18+7*0.1\\ &= 4.72 \end{aligned} $$

To find $V(x)$ let us find $E(X^2)$.

$$ \begin{aligned} E(X^2)&= \sum x^2*p(x)\\ &= 4*0.07+9*0.11+16*0.23+25*0.31+36*0.18+49*0.1\\ &= 24.08 \end{aligned} $$

Variance of $X$ is

$$ \begin{aligned} V(X) &= E(X^2) - [E(X)]^2 \\ &= 24.08 - 4.72^2\\ &= 1.8016 \end{aligned} $$

The standard deviation of the number of minutes the operators are answering questions each hour is

$$ \begin{aligned} sd(X) &= \sqrt{V(X)}\\ &= \sqrt{1.8016}\\ &=1.3422 \end{aligned} $$

Further Reading