Damaged Frequency Column: Rebuilding a Table from the Median
A class teacher tabulates the marks of 40 students in a practice test out of 70. The table has class intervals 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70 with frequencies 4, 6, a, 10, b, 6, 4. The registers for two columns were damaged, so a and b are unknown. From her memory of the result sheet the teacher is certain that the median is exactly 34. She must reconstruct the table, decide whether the mean and median tell the same story, and answer a parent who writes: 'Since the median is 34, more than half the class must have scored above 45.' The school also wants the probability that a student picked at random from this class scored 50 or more, for a scholarship shortlist. Grouped data must be read with its limits in mind: the teacher may not invent individual marks, and every conclusion must be defensible from the table alone.
1. Find a and b using the total frequency and the given median of 34.
Read the worked answer
Total frequency: 4 + 6 + a + 10 + b + 6 + 4 = 40 → a + b = 10.
Cumulative frequencies: 4, 10, 10 + a, 20 + a, 20 + a + b, 26 + a + b, 30 + a + b = 40 ✔ (consistent with a + b = 10).
Median 34 lies in the class 30–40, where l = 30, h = 10, f = 10, and the cumulative frequency before the class is 10 + a. With N/2 = 20:
Median = l + [(N/2 − cf)/f] × h = 30 + [(20 − (10 + a))/10] × 10 = 30 + (10 − a) = 40 − a.
Setting 40 − a = 34 gives a = 6, hence b = 4.
Check the class is right: cf before 30–40 is 16 (< 20) and cf up to it is 26 (≥ 20) ✔. The reconstructed frequencies are 4, 6, 6, 10, 4, 6, 4, summing to 40.
2. Compute the mean from the reconstructed table and say whether it agrees with the median.
Read the worked answer
Using class midpoints 5, 15, 25, 35, 45, 55, 65 with frequencies 4, 6, 6, 10, 4, 6, 4:
| midpoint | frequency | product |
|---|---|---|
| 5 | 4 | 20 |
| 15 | 6 | 90 |
| 25 | 6 | 150 |
| 35 | 10 | 350 |
| 45 | 4 | 180 |
| 55 | 6 | 330 |
| 65 | 4 | 260 |
Σfx = 20 + 90 + 150 + 350 + 180 + 330 + 260 = 1,380. Mean = 1,380 ÷ 40 = 34.5.
Interpretation: mean 34.5 and median 34 nearly coincide, so the distribution is close to symmetric about the mid-30s, with the modal class 30–40 (frequency 10). The slight excess of the mean over the median comes from the small high-scoring tail (10 students at 50 and above), which pulls the mean up by about half a mark. Mean and median tell the same story here.
3. Answer the parent's claim and find the probability that a randomly chosen student scored 50 or more.
Read the worked answer
Parent's claim is false. Students scoring above 45 are the 6 in 50–60 and the 4 in 60–70, i.e. 10 for certain, plus however many of the 4 students in 40–50 scored above 45 - at most 4 more. So the number above 45 lies between 10 and 14 out of 40, never more than 20. The median of 34 says only that about half the class scored below 34; it says nothing about the 45 mark. The claim confuses 'half below the median' with 'half above any chosen mark'.
Probability: P(score ≥ 50) = (6 + 4)/40 = 10/40 = 0.25, i.e. 25%.
Caveat worth stating: this probability is exact because the classes 50–60 and 60–70 are whole classes at the top. By contrast, the number above 45 is only bounded (10 to 14) because 45 cuts across the 40–50 class, and grouped data do not reveal how the 4 marks inside that class are spread.