Forty Mock Scores: Mean, Median and a Principal's Claim
A mathematics teacher in a 40-student SSS 3 class recorded the following scores in a 70-mark mock test.
| Score (%) | 0–9 | 10–19 | 20–29 | 30–39 | 40–49 | 50–59 | 60–69 |
|---|---|---|---|---|---|---|---|
| Number of students | 4 | 7 | 10 | 9 | 6 | 3 | 1 |
The school's pass mark is 30%. The principal's report states that 'the class average is about 29 and, since most students scored near the average, well over half the class passed'. A parent challenges the claim and asks for the mean, the median and the actual pass rate. The teacher also needs the probability that a student picked at random from the class scored below 20, so that a remedial class can be planned. The examination officer insists that grouped data be handled with class midpoints and the median formula, and that probabilities be given as fractions in simplest form and as decimals.
1. Calculate the mean score and identify the modal class. Comment on how much the mean can be trusted.
Read the worked answer
Use class midpoints 4.5, 14.5, 24.5, 34.5, 44.5, 54.5, 64.5.
| Midpoint (x) | f | fx |
|---|---|---|
| 4.5 | 4 | 18.0 |
| 14.5 | 7 | 101.5 |
| 24.5 | 10 | 245.0 |
| 34.5 | 9 | 310.5 |
| 44.5 | 6 | 267.0 |
| 54.5 | 3 | 163.5 |
| 64.5 | 1 | 64.5 |
| Total | 40 | 1,170.0 |
Mean = Σfx ÷ Σf = 1,170 ÷ 40 = 29.25%.
Modal class: 20–29, with frequency 10 (the largest in the table).
Comment. The mean is a grouped estimate: it assumes every score sits at its class midpoint, so the true mean may differ slightly. It is also pulled downwards by the four very low scores, and it is close to the modal class, so it describes where the mass of the class sits. As the principal's own figure of 'about 29' it is accurate - but it says nothing about how many students cleared 30%.
2. Use the median formula to find the median score and interpret it against the 30% pass mark.
Read the worked answer
Cumulative frequencies: 4, 11, 21, 30, 36, 39, 40.
Locate the median. N/2 = 40/2 = 20. The cumulative frequency first reaches 20 in the class 20–29 (it goes from 11 to 21), so the median class is 20–29.
Apply the formula. Median = L + [(N/2 − cf) ÷ f] × c
- L = lower class boundary = 19.5
- cf = cumulative frequency before the median class = 11
- f = frequency of the median class = 10
- c = class width = 10
Median = 19.5 + [(20 − 11) ÷ 10] × 10 = 19.5 + 9 = 28.5%.
Interpretation. Half the class scored below 28.5%, which is under the 30% pass mark. That means at least half the class failed, and possibly more, since some of those between 28.5 and 30 also failed. The median is the more honest summary here: it sits below the pass mark even though the mean sits just below it too, and it directly answers 'how did the middle student do?'.
3. Find the probability that a randomly chosen student passed and the probability that a randomly chosen student scored below 20. Then assess the principal's claim.
Read the worked answer
Pass (30% and above). Frequencies from 30–39 upwards: 9 + 6 + 3 + 1 = 19. P(pass) = 19/40 = 0.475, i.e. 47.5%.
Below 20. Frequencies 0–9 and 10–19: 4 + 7 = 11. P(below 20) = 11/40 = 0.275, i.e. 27.5%. That is 0.275 × 40 = 11 students for the remedial class.
Check. Passing (19), failing (21) and the two groups are exhaustive: 19 + 21 = 40 ✓, and 0.475 + 0.525 = 1 ✓.
Assessing the claim. The first half is defensible - the mean is 29.25, so 'about 29' is fair. The second half is false: 47.5% passed, which is fewer than half, not 'well over half'. The principal has confused a measure of central tendency with the proportion of students above a fixed threshold; a class can have a mean of 29 and still fail more than half its members when the scores are spread out. The correct summary is: mean 29.25%, median 28.5%, pass rate 47.5%, with 27.5% scoring below 20%.