Every cumulative frequency question from 2 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 9.6. 2 Extended-tier questions. Each question is labelled with its question paper code (for example 0580/21/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Statistics past paper questions.
0580 Cumulative Frequency past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q14
The cumulative frequency diagram shows the time 60 students spend watching television on one day.
(a) Find an estimate for (i) the median (ii) the interquartile range.
(b) Use the diagram to complete the frequency table: 0 < t ≤ 1: 7; 1 < t ≤ 2: …; 2 < t ≤ 3: …; 3 < t ≤ 4: 9; 4 < t ≤ 5: 4; 5 < t ≤ 6: 2.
Show solution
- There are 60 students, so the median is at a cumulative frequency of 60 ÷ 2 = 30.
- Read across from 30 to the curve, then down to the time axis.
- Median ≈ 2.5 hours
- Lower quartile: read across from 60 ÷ 4 = 15. It is about 1.8 hours.
- Upper quartile: read across from 3 × 15 = 45. It is about 3 hours.
- IQR = 3 − 1.8 = 1.2 hours
- Read the cumulative frequency at the end of each class from the graph.
- At t = 2 the CF is about 18. The first class has 7, so 1 < t ≤ 2 has 18 − 7 = 11
- At t = 3 the CF is about 45, so 2 < t ≤ 3 has 45 − 18 = 27
- Check: 7 + 11 + 27 + 9 + 4 + 2 = 60 ✓
May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q12
The time spent on the internet by each of 120 adults is recorded for one day. The cumulative frequency diagram shows this information.
(a) Use the cumulative frequency diagram to find an estimate of the interquartile range.
(b) 70% of the adults spent less than k hours on the internet. Use the cumulative frequency diagram to find an estimate of the value of k.
Show solution
- There are 120 adults.
- Lower quartile: read across from 120 ÷ 4 = 30 to the curve, then down: about 4.2 hours.
- Upper quartile: read across from 3 × 30 = 90 to the curve, then down: about 5.6 hours.
- Interquartile range = upper quartile − lower quartile = 5.6 − 4.2 = 1.4 h
- 70% of 120 = 0.7 × 120 = 84.
- Read across from 84 to the curve, then down: k ≈ 5.4
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