The question provides the following parameters:
[tex]\begin{gathered} \mu=16.3 \\ \sigma=4.2 \end{gathered}[/tex]For 15 minutes, the z-score is calculated using the formula:
[tex]z=\frac{x-\mu}{\sigma}[/tex]At x = 15:
[tex]z=\frac{15-16.3}{4.2}=-0.3[/tex]The probability is calculated using the formula:
[tex]P(X<15)=Pr(z<-0.3)=Pr(z<0)-Pr(0From tables, we have:[tex]\begin{gathered} Pr(z<0)=0.5 \\ Pr(0Therefore, the probability is given to be:[tex]\begin{gathered} P(X<15)=0.5-0.1179 \\ P(X<15)=0.38 \end{gathered}[/tex]The expected number of callers will be calculated using the formula:
[tex]\begin{gathered} E=xP(x) \\ At\text{ }x=1500 \\ E=1500\times0.38 \\ E=570 \end{gathered}[/tex]Therefore, the expected number of callers whose calls last less than 15 minutes is 570 callers.
