The waiting times of 60 callers are grouped in classes of width 10 minutes from 0 to 60, with the…

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The waiting times of $60$ callers are grouped in classes of width $10$ minutes from $0$ to $60$ , with the frequencies $4, 10, 15, 14, 11$ and $6$ . The company has promised that the slowest $20\%$ of the callers will get a free call back. To keep the promise, it must know the waiting time that a caller has to exceed to be in this group. Assume that the callers are spread evenly through each class, and find this waiting time.

Need a reminder? The key ideaShow

Read a cumulative frequency curve at the height $\dfrac{N}{2}$ for the median, at $\dfrac{N}{4}$ and $\dfrac{3N}{4}$ for the quartiles, and at $\dfrac{p}{100}N$ for the $p$ th percentile. With straight segments, a reading at the height $H$ is $l + \dfrac{H - cf}{f} \times h$ in the class where the running total reaches $H$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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