A taxi company records the waiting time, in minutes, of \(80\) customers calling for a taxi one evening, shown in the histogram below. One bar has been left...

Since the total of all frequencies must equal the number of customers surveyed, the missing frequency is the total minus the sum of the frequencies read from the other bars of the histogram; dividing that frequency by its class width gives its frequency density. The estimated mean for the completed grouped data then uses class midpoints as usual, and is compared directly against the company's threshold.

1. The other three classes have frequency densities of \(3.0\), \(2.0\), \(1.2\) and \(0.4\) on the histogram, corresponding to frequencies of \(3.0\times5=15\) (for \(0\) to \(5\) minutes), \(2.0\times10=20\) (for \(15\) to \(25\) minutes), \(1.2\times15=18\) (for \(25\) to \(40\) minutes) and \(0.4\times20=8\) (for \(40\) to \(60\) minutes), totalling \(15+20+18+8=61\). [1 mark] The missing frequency for \(5<t\le15\) is \(80-61=19\). [1 mark]
2. Frequency density \(=19\div10=1.9\) [1 mark]
3. Using midpoints \(2.5\), \(10\), \(20\), \(32.5\), \(50\): \(fx\) values are \(2.5(15)=37.5\), \(10(19)=190\), \(20(20)=400\), \(32.5(18)=585\), \(50(8)=400\), totalling \(1612.5\) [2 marks]; mean \(=1612.5\div80=20.2\) minutes (1 d.p.) [1 mark]
4. Yes, an extra dispatcher is needed, since the estimated mean waiting time of \(20.2\) minutes is greater than the company's \(15\)-minute threshold. [2 marks]