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

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators