At a summer camp an arithmetic test is taken by 250 children. The times taken, to the nearest minute, to complete the test were recorded. The results are summarised in the table. Time taken, in minutes | 1–30 | 31–45 | 46–65 | 66–75 | 76–100 Frequency | 21 | 30 | 68 | 86 | 45
- (a)Draw a histogram to represent this information.[4]
- (b)State which class interval contains the median.[1]
- (c)Given that an estimate of the mean time is 61.05 minutes, state what feature of the distribution accounts for the median and the mean being different.[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2022). The mark or mark type is shown in brackets.
- M1 At least 4 frequency densities calculated
- A1 All heights correct on graph
- B1 Bar ends at 0.5, 30.5, 45.5, 65.5, 75.5, 100.5 (at axis), 5 bars drawn, condone 0 in first bar 0.5 < time axis < 100.5, linear scale with at least 3 values indicated.
- B1 Axes labelled: Frequency density (fd), time (t) and mins (or appropriate title). Linear fd scale, with at least 3 values indicated 0 < fd axis < 8.6
- Class Widths and Frequency Densities:
- 1-30: Width = 30, Frequency = 21, FD = 0.7
- 31-45: Width = 15, Frequency = 30, FD = 2
- 46-65: Width = 20, Frequency = 68, FD = 3.4
- 66-75: Width = 10, Frequency = 86, FD = 8.6
- 76-100: Width = 25, Frequency = 45, FD = 1.8
- B1 Condone 65.5 – 75.5
- 66-75
- B1 Or skewed, ignore nature of skew
- Distribution is not symmetrical
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