Question

Assume a normal distribution and that the average phone call in a certain town lasted 4 min, with a standard deviation of 1 min. What percentage of the calls lasted lessthan 3 min?

Answer 1

The mean is 4min

standard deviation is 1min

the z score is

`z=\frac{x-\bar{x}}{\sigma}`

where

`\begin{gathered} x=3 \\ \bar{x}=4 \\ \sigma=1 \end{gathered}`

then we have

`\begin{gathered} z=(3-4)/(1)=(-1)/(1)=-1 \\ z=-1 \end{gathered}`

The probability the call lasted less than 3 min will be

Therefore, the probability that (z < -1 ) is

[tex]\begin{gathered} Pr(z<-1)=Pr(0Hence, the percentage of the calls that lasted less than 3 min is 16%