Which Batch Gets the Trophy: Mean Versus Median
Two coaching batches of 12 students each appeared for the same 100-mark mathematics mock test. The marks were recorded in ascending order, without names, to keep the comparison anonymous.
Batch A: 30, 58, 61, 63, 64, 65, 66, 68, 70, 72, 75, 78 Batch B: 55, 58, 60, 62, 63, 64, 65, 66, 68, 70, 72, 74
The institute can award its 'Best Batch' rolling trophy to only one group, and the director wants it to go to the batch with the stronger typical performance. The head tutor argues for Batch A because its students hold the top two ranks. The accounts clerk, who only glances at the totals, says Batch B is ahead. Both groups were taught by the same teacher, and one student in Batch A missed three weeks of class after an illness. There are no marks below 30 and none above 78 in either list. Give exact fractions or round to two decimal places.
1. Calculate the mean mark of each batch and state which batch the clerk's method favours.
Read the worked answer
Batch A sum = 30 + 58 + 61 + 63 + 64 + 65 + 66 + 68 + 70 + 72 + 75 + 78 = 770 Mean = 770 / 12 = 64.17
Batch B sum = 55 + 58 + 60 + 62 + 63 + 64 + 65 + 66 + 68 + 70 + 72 + 74 = 777 Mean = 777 / 12 = 64.75
So Batch B is ahead by 0.58 marks on the mean. The clerk is right about the totals, but the margin is tiny - about 1% of the score - and a single mark moved anywhere in Batch A would reverse it.
2. Calculate the median of each batch. Explain why the median reverses the ranking given by the mean, using the shape of the two lists.
Read the worked answer
Both lists are already in ascending order with 12 values, so the median is the average of the 6th and 7th values.
- Batch A: (65 + 66)/2 = 65.5
- Batch B: (64 + 65)/2 = 64.5
Now Batch A is ahead by 1.0 mark.
Why the reversal: Batch A is skewed low by the single score of 30, which sits far below the rest of the list (the next lowest is 58). The mean uses every value, so that one score drags it down by roughly (64 − 30)/12 ≈ 2.8 marks per student. The median only counts positions, so one extreme value cannot move it. Batch B is compact and nearly symmetric (range 19, from 55 to 74), so its mean and median sit close together; Batch A has a range of 48.
The mean answers 'what is the total divided among all?' The median answers 'what does a typical student score?' When a list has one outlier, they disagree, and the choice of average becomes a judgement, not a fact.
3. Recommend which batch should receive the trophy, and state how the picture changes if the absent Batch A student's mark is revised from 30 to 60 after re-evaluation.
Read the worked answer
Recommendation: award the trophy to Batch A, but record the reasoning. Two of the three central measures (median 65.5 vs 64.5, and top ranks 78 and 75 vs 74 and 72) favour A, and A's low mean is explained by one documented absence rather than by weak teaching. Batch B, however, is the more consistent group and its mean is higher, so it should be recognised for consistency rather than for 'typical performance'.
After revision to 60:
- New sum = 770 − 30 + 60 = 800; mean = 800/12 = 66.67, now clearly above Batch B's 64.75.
- The revised list is 58, 60, 61, 63, 64, 65, 66, 68, 70, 72, 75, 78, so the 6th and 7th values are still 65 and 66; median stays 65.5.
- The mean rose by 2.5 marks, the median did not move at all, because 60 is still below the middle of the list.
This confirms that the median is resistant to extreme values while the mean is not - the deciding factor in which average to quote.