Two classes, one test: which class actually did better?
Two JCE classes sat the same 20-mark revision test. Class X, 15 learners: 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 12, 17, 20. Class Y, 15 learners: 8, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 13. Both teachers claim their class did better. The Class X teacher points to the score of 20 and says her class has the strongest mathematicians; the Class Y teacher says her class is clearly ahead because 'more of my learners are in double figures'. The head of department has to write one sentence for the school report saying which class performed better overall and why, and a second sentence about consistency.
1. Calculate the mean and median for each class and explain what the two measures show.
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Class X: total = 7 + 8 + 8 + 8 + 9 + 9 + 9 + 10 + 10 + 10 + 11 + 11 + 12 + 17 + 20 = 159. Mean = 159 ÷ 15 = 10.6. With 15 values the median is the 8th in order = 10.
Class Y: total = 8 + 9 + 9 + 10 + 10 + 10 + 10 + 11 + 11 + 11 + 11 + 12 + 12 + 12 + 13 = 159. Mean = 159 ÷ 15 = 10.6. Median = 8th value = 11.
The means are identical, so 'average score' cannot separate the classes at all. The medians differ by one mark in Y's favour, meaning Y's middle learner outscored X's middle learner. The reason the means still tie is that X's mean is dragged upwards by two extreme scores (17 and 20); when data is skewed like this, the median is the better description of a typical learner and the mean is the more fragile statistic. Any report that quotes only the mean is hiding the difference.
2. Evaluate the two teachers' claims, then write the two report sentences the head of department needs.
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Class X teacher: one learner scoring 20 says nothing about the class; it is a single value out of 15 and is balanced by a 7 and three 8s at the bottom. A highest score measures one learner, not a class.
Class Y teacher: this claim is checkable and survives checking. Learners scoring 10 or more - X: three 10s, two 11s, 12, 17, 20 = 8 of 15; Y: four 10s, four 11s, three 12s, one 13 = 12 of 15. Twelve against eight supports her.
Report sentence 1: 'Class Y performed better overall: the two classes had equal means (10.6), but Class Y had the higher median (11) and 12 of its 15 learners reached double figures against 8 in Class X, so Y's typical learner scored higher.'
Report sentence 2: 'Class Y was also more consistent, with a range of 5 marks (8 to 13) against 13 marks in Class X (7 to 20), whose mean was lifted by two outlying scores of 17 and 20.'
3. Use the combined group of 30 learners for these probability questions: (a) a learner is chosen at random - estimate the probability that the score is at least 12; (b) predict the score of a new learner joining from a class with the same profile as Y, and say how confident the prediction is; (c) which class is more likely to produce a score above 15?
Read the worked answer
(a) Count scores of 12 or more. Class X: 12, 17, 20 → 3. Class Y: 12, 12, 12, 13 → 4. Total 7 out of 30, so P ≈ 7/30 ≈ 0.23 (about 23%). Using the combined group is legitimate because all 30 learners sat the same test under the same conditions.
(b) Best single prediction = 11, Class Y's median. The mean is 10.6, so both measures point at 'about 11'. Confidence is moderate, not high: the estimate rests on 15 learners; the values actually run from 8 to 13, and 8 of the 15 sit at 10 or 11, so 11 is the most crowded region. It is a reasonable forecast, not a guarantee - a prediction of 'between 10 and 12' would be safer.
(c) Class X: scores above 15 are 17 and 20 → 2/15 ≈ 0.13. Class Y: none → 0. So X is far more likely to produce a very high score. This does not make X the better class: X also contains the weakest learners (a 7 and three 8s), which is exactly why its median is lower. Probability of an extreme tells you about the spread, not the standard of the class.