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;

$z=\frac{x-\mu}{\sigma}$

Where;

$x=5,\text{ }\mu=9\text{ and }\sigma=2$$\begin{gathered} z=\frac{5-9}{2} \\ z=-\frac{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%