SOLUTION
The mean is 4min
standard deviation is 1min
the z score is
[tex]z=\frac{x-\bar{x}}{\sigma}[/tex]where
[tex]\begin{gathered} x=3 \\ \bar{x}=4 \\ \sigma=1 \end{gathered}[/tex]then we have
[tex]\begin{gathered} z=\frac{3-4}{1}=\frac{-1}{1}=-1 \\ z=-1 \end{gathered}[/tex]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%
