Every histograms question from 7 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 9.7. 7 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 Histograms past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q22
The table shows the lengths of 145 pieces of wire.
| length (L cm) | 70 < L ≤ 90 | 90 < L ≤ 100 | 100 < L ≤ 115 | 115 < L ≤ 120 |
|---|---|---|---|---|
| frequency | 40 | 35 | 45 | 25 |
| height of bar (cm) | 4 |
Stanley wants to draw a histogram. Complete the table.
Show solution
- Frequency density = frequency ÷ class width.
- First class: width = 90 − 70 = 20, so density = 40 ÷ 20 = 2. It is drawn 4 cm tall, so the height in cm is 2 × frequency density.
- 90 < L ≤ 100: width 10, density 35 ÷ 10 = 3.5, height 2 × 3.5 = 7
- 100 < L ≤ 115: width 15, density 45 ÷ 15 = 3, height 2 × 3 = 6
- 115 < L ≤ 120: width 5, density 25 ÷ 5 = 5, height 2 × 5 = 10
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q19
The time spent in a shop by each of 40 customers is recorded.
| time (t minutes) | 0 < t ≤ 5 | 5 < t ≤ 10 | 10 < t ≤ 20 | 20 < t ≤ 40 |
|---|---|---|---|---|
| frequency | 3 | 18 | 13 | 6 |
(a) Calculate an estimate of the mean time. (b) Complete the histogram to show the information in the frequency table.
Show solution
- Use the midpoint of each interval as an estimate: 2.5, 7.5, 15, 30.
- Multiply each midpoint by its frequency: 3 × 2.5 = 7.5, 18 × 7.5 = 135, 13 × 15 = 195, 6 × 30 = 180.
- Total = 7.5 + 135 + 195 + 180 = 517.5.
- Mean = 517.5 ÷ 40 = 12.9375 = 12.9 minutes
- In a histogram the height of each bar is the frequency density = frequency ÷ class width.
- 0 < t ≤ 5: 3 ÷ 5 = 0.6
- 5 < t ≤ 10: 18 ÷ 5 = 3.6
- 10 < t ≤ 20: 13 ÷ 10 = 1.3
- 20 < t ≤ 40: 6 ÷ 20 = 0.3
- Draw each bar across its whole interval at this height.
Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q18(a,b)
The table shows information about the time that each of 90 trains is delayed.
| time (m minutes) | 0 < m ≤ 10 | 10 < m ≤ 30 | 30 < m ≤ 45 | 45 < m ≤ 75 |
|---|---|---|---|---|
| frequency | 13 | 35 | 36 | 6 |
(a) Calculate an estimate of the mean.
(b) A histogram is drawn to show the information. The height of the bar for the interval 10 < m ≤ 30 is 14 cm. Calculate the height of the bar for each of the other intervals.
Show solution
- Use the midpoint of each interval as an estimate: 5, 20, 37.5, 60.
- Multiply each midpoint by its frequency: 5 × 13 = 65, 20 × 35 = 700, 37.5 × 36 = 1350, 60 × 6 = 360.
- Total = 65 + 700 + 1350 + 360 = 2475.
- Mean = 2475 ÷ 90 = 27.5 minutes
- In a histogram, frequency density = frequency ÷ class width.
- 0–10: 13 ÷ 10 = 1.3. 10–30: 35 ÷ 20 = 1.75. 30–45: 36 ÷ 15 = 2.4. 45–75: 6 ÷ 30 = 0.2.
- 1.75 is drawn as 14 cm, so 1 unit of frequency density = 14 ÷ 1.75 = 8 cm.
- 0 < m ≤ 10: 1.3 × 8 = 10.4 cm
- 30 < m ≤ 45: 2.4 × 8 = 19.2 cm
- 45 < m ≤ 75: 0.2 × 8 = 1.6 cm
Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q25
Kai sorts parcels into light (0 < m ≤ 2 kg) and heavy (2 < m ≤ 5 kg). The histogram shows the light parcels with frequency density 20.
(a) Find the number of light parcels. (b) There are 102 heavy parcels. Complete the histogram.
Show solution
- In a histogram, frequency = frequency density × class width.
- Light parcels: class width = 2 − 0 = 2 kg and frequency density = 20.
- Frequency = 20 × 2 = 40
- Heavy parcels: class width = 5 − 2 = 3 kg.
- Frequency density = frequency ÷ class width = 102 ÷ 3 = 34.
- Draw a bar from 2 to 5 with height 34.
Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q16
The histogram shows the masses of some coconuts in four categories: A (0.9 to 1.1 kg, frequency density 30), B (1.1 to 1.25 kg, frequency density 100), C (1.25 to 1.35 kg, frequency density 170) and D (1.35 to 1.6 kg, frequency density 40).
(a) Show that there are 10 coconuts in category D.
(b) Two of the coconuts from those in category C and category D are chosen at random. Find the probability that both are from category D.
(c) Calculate an estimate of the mean mass of the coconuts.
Show solution
- In a histogram, frequency = frequency density × class width.
- Category D: class width = 1.6 − 1.35 = 0.25 kg.
- Frequency = 40 × 0.25 = 10 ✓
- Category C: 170 × (1.35 − 1.25) = 170 × 0.1 = 17 coconuts. C and D together: 17 + 10 = 27.
- Two are chosen without replacement, so after the first D there are 9 D left out of 26.
- P(both D) = 10⁄27 × 9⁄26 = 90⁄702.
- Divide the top and bottom by 18: 5⁄39
- Find each frequency: A = 30 × 0.2 = 6, B = 100 × 0.15 = 15, C = 17, D = 10. Total = 48.
- Midpoints: A 1, B 1.175, C 1.3, D 1.475.
- Multiply and add: 6 × 1 + 15 × 1.175 + 17 × 1.3 + 10 × 1.475 = 6 + 17.625 + 22.1 + 14.75 = 60.475.
- Mean = 60.475 ÷ 48 = 1.259…
- 1.26 kg
Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q23
The table shows the mass of each of 200 oranges.
| mass (m grams) | 180 < m ≤ 200 | 200 < m ≤ 210 | 210 < m ≤ 215 | 215 < m ≤ 230 |
|---|---|---|---|---|
| frequency | 32 | 64 | 74 | 30 |
(a) Calculate an estimate of the mean mass of an orange.
(b) Sarah draws a histogram. The height of the bar for 210 < m ≤ 215 is 7.4 cm. Complete the table of bar heights.
Show solution
- Use the midpoint of each class: 190, 205, 212.5, 222.5.
- Multiply each midpoint by its frequency: 32 × 190 = 6080, 64 × 205 = 13 120, 74 × 212.5 = 15 725, 30 × 222.5 = 6675.
- Total = 6080 + 13 120 + 15 725 + 6675 = 41 600.
- Mean = 41 600 ÷ 200 = 208 g
- Bar heights are proportional to frequency density = frequency ÷ class width.
- 180–200: 32 ÷ 20 = 1.6. 200–210: 64 ÷ 10 = 6.4. 210–215: 74 ÷ 5 = 14.8. 215–230: 30 ÷ 15 = 2.
- 14.8 is drawn as 7.4 cm, so 1 unit of frequency density = 7.4 ÷ 14.8 = 0.5 cm.
- 180 < m ≤ 200: 1.6 × 0.5 = 0.8 cm
- 200 < m ≤ 210: 6.4 × 0.5 = 3.2 cm
- 215 < m ≤ 230: 2 × 0.5 = 1 cm
May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q13
(a) 100 students solve a puzzle. The table shows information about the time taken by each student to solve the puzzle.
| Time (t seconds) | 20 < t ≤ 40 | 40 < t ≤ 60 | 60 < t ≤ 100 |
| Frequency | 30 | 40 | 30 |
(i) Work out an estimate of the mean.
(ii) Complete the histogram to show the information in the table.
(b) 80 adults solve the same puzzle as the students. The cumulative frequency table shows information about the time taken by each adult to solve the puzzle.
| Time (t seconds) | t ≤ 20 | t ≤ 40 | t ≤ 60 | t ≤ 80 | t ≤ 100 | t ≤ 120 |
| Cumulative frequency | 0 | 12 | 36 | 60 | 74 | 80 |
(i) On the grid, draw a cumulative frequency diagram.
(ii) Use your cumulative frequency diagram to find an estimate for (a) the median (b) the lower quartile.
Show solution
- The exact times are not known, so use the midpoint of each class: 30, 50 and 80.
- Multiply each midpoint by its frequency: 30 × 30 = 900, 40 × 50 = 2000, 30 × 80 = 2400.
- Add: 900 + 2000 + 2400 = 5300.
- Divide by the total frequency: 5300 ÷ 100 = 53 seconds
- In a histogram, the height of each bar is the frequency density = frequency ÷ class width.
- 40 < t ≤ 60: width 20, density 40 ÷ 20 = 2.
- 60 < t ≤ 100: width 40, density 30 ÷ 40 = 0.75.
- Draw a bar from 40 to 60 of height 2 and a bar from 60 to 100 of height 0.75.
- Plot each cumulative frequency at the upper end of its class: (20, 0), (40, 12), (60, 36), (80, 60), (100, 74), (120, 80).
- Join the points with a smooth curve that keeps going up.
- There are 80 adults, so the median is at a cumulative frequency of 80 ÷ 2 = 40.
- Read across from 40 to the curve and down: median ≈ 63 seconds
- The lower quartile is at 80 ÷ 4 = 20.
- Read across from 20 to the curve and down: lower quartile ≈ 47 seconds
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