The distances driven on 100 delivery rounds are grouped like this: 40 rounds under 5 km, 36 rounds from 5 km up to 8 km, and 24 rounds from 8 km up to 20 km...
Question 1 Report
The distances driven on 100 delivery rounds are grouped like this: 40 rounds under 5 km, 36 rounds from 5 km up to 8 km, and 24 rounds from 8 km up to 20 km.
- Work out the frequency density for each of the three classes. (2)
- Calculate an estimate of the mean distance driven. (2)
Answer Details
The three classes have different widths, so a histogram needs frequency density, and an estimated mean needs each class represented by its midpoint.
(a)
- Frequency density is frequency divided by class width. The class from 0 to 5 km has width 5: \(\frac{40}{5} = 8\). The class from 5 to 8 km has width 3: \(\frac{36}{3} = 12\). [1]
- The class from 8 to 20 km has width 12: \(\frac{24}{12} = 2\). All three are measured in rounds per km. [1]
(b)
- The midpoints are \(\frac{0+5}{2} = 2.5\), \(\frac{5+8}{2} = 6.5\) and \(\frac{8+20}{2} = 14\) km. [1]
- Mean \(= \frac{40 \times 2.5 + 36 \times 6.5 + 24 \times 14}{100} = \frac{100 + 234 + 336}{100} = \frac{670}{100} = 6.7\) km. [1]
Notice how the densities rank differently from the frequencies: the widest class has the most spread-out data and so the shortest bar, even though it contains 24 rounds. Drawing bars of height 40, 36 and 24 would badly misrepresent the distribution.
The mean of 6.7 km is higher than the median class because the long 8 km to 20 km class pulls it up; with grouped data the answer is always an estimate, since the individual distances within each class are unknown.
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