Question

Assume a normal distribution and that the average phone call in a certain town lasted 9 minutes, with a standard deviation of 2 minutes. What percentage of the calls lasted less than 5 minutes?

Answer 1

Given the sample mean of the distribution as 9 minutes.

Also, the sample deviation as 2 minutes;

Given the z-score formula;

`z=(x-\mu)/(\sigma)`

Where;

`x=5,\text{ }\mu=9\text{ and }\sigma=2`
`\begin{gathered} z=(5-9)/(2) \\ z=-(4)/(2) \\ z=-2 \end{gathered}`

Now, from the Z-table;

`P(x<-2)=0.02275`

Thus, the percentage of the calls lasted less than 5 minutes is 2.3%