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The long-distance calls made by South Africans are normally distributed with a mean of 16.3 minutes and a standard deviation of 4.2 minutes for 1500 south Africans what is the expected number of callers whose calls last less than 15 minutes?

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The question provides the following parameters:


\begin{gathered} \mu=16.3 \\ \sigma=4.2 \end{gathered}

For 15 minutes, the z-score is calculated using the formula:


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

At x = 15:


z=(15-16.3)/(4.2)=-0.3

The probability is calculated using the formula:


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}

The expected number of callers will be calculated using the formula:


\begin{gathered} E=xP(x) \\ At\text{ }x=1500 \\ E=1500*0.38 \\ E=570 \end{gathered}

Therefore, the expected number of callers whose calls last less than 15 minutes is 570 callers.

User Lee Treveil
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