Two Classes Compared: Missing Data, Averages and a Random Draw
At the end of term the mathematics department compared two Form 2 classes on a test marked out of 100.
- Class 2A: 26 students, mean 62.
- Class 2B: 25 students, mean 58.
The head teacher announced that 2A was 'clearly the stronger class'. The 2B teacher objected: three of her students were away at a netball tournament on the day of the test and were entered as 0, which she says is not a mathematics score. The record also shows that 20 of the 26 students in 2A scored 50 or more, and 19 of the 25 students in 2B scored 50 or more.
Two students are to be chosen at random from the 51 to represent the school at a district quiz. The head teacher wants the pair drawn only from students who scored 50 or more, because 'a weak pair will embarrass us', but the department insists the draw must be random from all 51.
1. Recalculate 2B's mean with the three absentees removed, and assess the head teacher's claim.
Read the worked answer
Recover the total. Mean = total ÷ 25, so 2B's total = 25 × 58 = 1,450 marks.
Remove the three zeros. The three absent students contributed 0 marks, so the 22 students who actually sat still hold all 1,450 marks. New mean = 1,450 ÷ 22 = 65.9 (1 d.p.).
Assess the claim. Comparing like with like - students who sat the test - 2B's mean of 65.9 is higher than 2A's 62, so the head teacher's 'clearly the stronger class' is not supported; on this evidence the claim reverses.
Caveats that a good answer adds. (i) A zero recorded for absence measures attendance, not mathematics, so including it makes the 58 meaningless as a measure of attainment. (ii) Simply deleting the zeros flatters 2B - the 22 remaining students are not a random sample, and the three students may in fact be weak. (iii) The comparison now rests on 22 students against 26, and no median or spread is given, so the honest statement is that the classes are close, with 2B's attending students slightly ahead on the mean, pending the resit.
2. Two students are drawn at random, without replacement, from all 51. Find the probability that both scored 50 or more, and the probability that at least one scored below 50.
Read the worked answer
Counts. Scored 50+: 20 + 19 = 39. Below 50: 51 − 39 = 12.
Both at 50 or more. First draw: 39/51. With that student removed, 38 of the remaining 50 are at 50+: P = (39/51) × (38/50) = 1,482 ÷ 2,550 = 0.581 (58.1%).
At least one below 50. This is the complement: 1 − 0.581 = 0.419 (41.9%). Equivalently, 1 − (12/51)(11/50) = 1 − 132/2,550 = 1 − 0.0518 = 0.948 is the chance of at least one at or above 50 - do not confuse the two complements.
Judgement. The head teacher's hope fails about 2 times in 5: a random pair contains at least one student below 50 in roughly 42% of draws. Conditioning the draw on the 50+ group would raise the 'both strong' chance to 1 (certainty by construction) but it is no longer a random sample of the 51, which is what the department requires. The fair compromise is to state the odds in advance and accept them, or to redefine the selection rule openly (e.g. top 39 by score) rather than to redraw quietly until a strong pair appears - redrawing changes the probabilities to something no one has calculated.
3. Using the proportions scoring 50 or more, write the single most defensible sentence for the school report comparing the two classes, and justify your wording.
Read the worked answer
Calculate the proportions.
- 2A: 20 ÷ 26 = 0.769 (76.9%).
- 2B: 19 ÷ 25 = 0.760 (76.0%).
Compare. The difference is 0.9 percentage points - under one student in each class. On this measure the classes are effectively indistinguishable.
Suggested sentence. 'Of the students who sat the test, 2B's mean was 65.9 compared with 62 in 2A, and the proportion reaching 50 or more was almost identical (76.9% in 2A, 76.0% in 2B); three 2B students were absent and will resit.'
Why this wording. (i) It names the group - only students who sat - because 2B's published mean of 58 includes three zeros that are missing data, not results. (ii) It gives two measures, a mean (sensitive to every mark) and a threshold proportion (sensitive to how many reached a standard), because a single average can be dragged by a few extreme scores. (iii) It reports the absenteeism openly, so readers can judge the comparison rather than being handed a conclusion. (iv) It avoids 'clearly stronger', which the numbers do not support in either direction. A sentence quoting only 62 against 58 would be arithmetically true and statistically misleading.