Comparing Two Sections' 2.4 km Run Times
Two recruit sections complete the same 2.4 km run assessment on the same course. Times are recorded to the nearest 5 seconds. Section A: 11:20, 11:45, 12:10, 12:10, 12:35, 13:00, 13:15, 13:40, 14:05, 15:30. Section B: 11:50, 12:05, 12:20, 12:25, 12:40, 12:50, 13:05, 13:25, 13:45, 13:55. The pass mark is 13:30. One recruit in Section A is returning from injury and is known to be an outlier; the chain of command wants to know which section is performing better overall, which section has the stronger top five, and whether the mean or median should be quoted in the briefing. A retest is planned for the two slowest runners in each section, who are expected to improve their times by 4%. No runner is to be named in the briefing.
1. Convert all times to seconds and calculate the mean and median for each section. Identify the outlier and its effect.
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Section A seconds: 680, 705, 730, 730, 755, 780, 795, 820, 845, 930. Sum = 7,770; mean = 777.0 s = 12:57.0. Median = (755 + 780) ÷ 2 = 767.5 s = 12:47.5. Section B seconds: 710, 725, 740, 745, 760, 770, 785, 805, 825, 835. Sum = 7,700; mean = 770.0 s = 12:50.0. Median = (760 + 770) ÷ 2 = 765.0 s = 12:45.0. The 930 s (15:30) in A is the outlier; it pulls A's mean 9.5 s above its median, whereas B's mean is only 5 s above its median. Without the outlier, A's mean would be (7,770 − 930) ÷ 9 = 760.0 s, faster than B's.
2. Calculate the pass rate at 13:30 and the range for each section. Recommend which section is performing better overall and justify the choice of measure.
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Pass mark 13:30 = 810 s. Section A passes: 680, 705, 730, 730, 755, 780, 795 = 7 of 10 = 70%. Fails: 820, 845, 930. Range = 930 − 680 = 250 s = 4:10. Section B passes: 710, 725, 740, 745, 760, 770, 785, 805 = 8 of 10 = 80%. Fails: 825, 835. Range = 835 − 710 = 125 s = 2:05. Overall, B is the more consistent and higher-passing section: better pass rate, lower mean, lower median and half the spread. A has the faster front end but a long tail. For a briefing on overall performance, quote the median and pass rate; for resourcing remedial training, quote the range and the outlier, because the mean alone hides A's split profile.
3. The two slowest runners in each section improve by 4% on retest. Calculate the new mean time for each section and state whether the ranking or pass rate changes.
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A's slowest: 845 s and 930 s. A 4% improvement means ×0.96: 845 × 0.96 = 811.2 s; 930 × 0.96 = 892.8 s. New sum = 7,770 − 845 − 930 + 811.2 + 892.8 = 7,699.0 s; new mean = 769.9 s = 12:49.9. B's slowest: 825 s and 835 s. 825 × 0.96 = 792.0 s; 835 × 0.96 = 801.6 s. New sum = 7,700 − 825 − 835 + 792.0 + 801.6 = 7,633.6 s; new mean = 763.36 s = 12:43.4. B remains faster on mean, and the gap narrows only slightly. B's retested runners now pass (792.0 and 801.6 are both under 810 s), so B's pass rate rises to 10 of 10 = 100%. A's 811.2 s is still just outside 810 s and 892.8 s also fails, so A stays at 70%. The ranking does not change; A still has the larger spread and the lower pass rate.