Practice Questions
Cambridge O Level Statistics: Frequency polygons and histograms (4040) – Practice Questions
Original practice questions with worked answers on frequency polygons, histograms and frequency density for Cambridge O Level Statistics.
- Subject
- Statistics
- Level
- O LEVELS
- Topic
- Frequency polygons and histograms
- Author
- Marlbridge Academic Team
- Updated
Aligned to Cambridge O Level Statistics (4040), For examination in 2025-2027. Official specification .
Syllabus page (what it covers and how it is assessed): Cambridge O Level Statistics.
Syllabus points this page covers
4040
- 3.2 Representation in frequency polygons and histograms
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These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.
These questions cover section 3.2, representation in frequency polygons and histograms, of the Cambridge O Level Statistics (4040) syllabus for examination in 2025, 2026 and 2027. The syllabus is not tiered. Learn the methods first in the study guide or the revision notes. Course pages: the Cambridge O Level Statistics hub and the printable checklist. For a wider check, use the free 10-minute diagnostics.
Questions
1. Weights are recorded to the nearest kilogram and grouped in a class written 60–64. State the class boundaries and the class midpoint. [2]
2. The masses of 40 parcels are summarised below.
| Mass m (kg) | 2 ≤ m < 4 | 4 ≤ m < 6 | 6 ≤ m < 8 | 8 ≤ m < 10 | 10 ≤ m < 12 |
|---|---|---|---|---|---|
| Frequency | 3 | 9 | 15 | 8 | 5 |
(a) Write down the coordinates of all the points you would plot to draw a frequency polygon, including the end points. [3] (b) Explain why a polygon is extended to the horizontal axis at both ends. [1]
3. The marks (whole numbers) of 50 pupils in a test are grouped as follows.
| Mark | 21–30 | 31–40 | 41–50 | 51–60 | 61–70 |
|---|---|---|---|---|---|
| Frequency | 5 | 12 | 19 | 10 | 4 |
(a) For a histogram, state the boundaries of each bar and the height of each bar. [3] (b) Explain why the bars of the histogram touch. [1]
4. Waiting times at a service desk were recorded for 110 customers.
| Time t (min) | 0 ≤ t < 5 | 5 ≤ t < 10 | 10 ≤ t < 20 | 20 ≤ t < 40 | 40 ≤ t < 60 |
|---|---|---|---|---|---|
| Frequency | 15 | 30 | 35 | 20 | 10 |
(a) Calculate the frequency density for each class. [3] (b) State the modal class and explain your choice. [1]
5. In a histogram, the bar for the class 12 ≤ x < 20 has frequency density 2.5. (a) Find the frequency of this class. [1] (b) Estimate how many items lie between 14 and 20. [1] (c) State the assumption you made in part (b). [1]
6. A histogram of 100 items has four bars.
| Class | 0 ≤ x < 10 | 10 ≤ x < 25 | 25 ≤ x < 30 | 30 ≤ x < 50 |
|---|---|---|---|---|
| Frequency density | 1.2 | 2.0 | not shown | 0.9 |
(a) Find the frequency of each class whose density is shown. [3] (b) Find the frequency and the frequency density of the class 25 ≤ x < 30. [2]
7. On a histogram, the bar for the class 0 ≤ x < 10 is 3 cm tall and represents 24 students. The bar for the class 10 ≤ x < 25 represents 54 students. Find the height of this second bar. [3]
8. Two clubs recorded the ages of their members. Club A has 40 members and club B has 120.
| Age a (years) | 10 ≤ a < 20 | 20 ≤ a < 30 | 30 ≤ a < 40 | 40 ≤ a < 50 |
|---|---|---|---|---|
| Club A frequency | 4 | 16 | 14 | 6 |
| Club B frequency | 12 | 30 | 54 | 24 |
(a) Write down the coordinates of the points for club A’s frequency polygon, including end points. [3] (b) A student draws both polygons on one graph and says club B has the older members because its line is higher. Comment, using proportions of each club to support your answer. [2]
9. The delivery times of 150 parcels are shown below.
| Time t (min) | 10 ≤ t < 20 | 20 ≤ t < 30 | 30 ≤ t < 40 | 40 ≤ t < 60 | 60 ≤ t < 100 |
|---|---|---|---|---|---|
| Frequency | 18 | 42 | 36 | 30 | 24 |
(a) Calculate the frequency density of each class. [3] (b) A histogram is drawn with 2 cm on the vertical axis representing 1 unit of frequency density. Find the height of each bar. [2] (c) Estimate the number of parcels with delivery times between 35 and 50 minutes. [3] (d) State the modal class. [1]
10. A histogram shows the ages of members of a society. The bars are given below.
| Age (years) | 10 ≤ a < 14 | 14 ≤ a < 18 | 18 ≤ a < 26 | 26 ≤ a < 50 |
|---|---|---|---|---|
| Frequency density | 2.5 | 6 | 3 | 0.5 |
(a) Find the frequency of each class. [2] (b) Find the total number of members. [1] (c) Estimate how many members are younger than 16. [2] (d) Later, 18 new members join, all aged 50 ≤ a < 80. Find the frequency density of this class. [1] (e) After they join, find the percentage of all members who are aged 18 or over, giving your answer to 3 significant figures. [2]
11. A student draws a histogram of the data below, using the frequency as the height of each bar.
| Time t (min) | 0 ≤ t < 10 | 10 ≤ t < 20 | 20 ≤ t < 40 | 40 ≤ t < 80 |
|---|---|---|---|---|
| Frequency | 30 | 36 | 48 | 40 |
(a) Explain what is wrong with the student’s method. [1] (b) Calculate the correct heights of the four bars. [2] (c) Which class looks like the modal class on the student’s diagram, and which is the modal class on a correct histogram? [2]
Answers
1. The boundaries are 59.5 and 64.5 [1]. The midpoint is (59.5 + 64.5) ÷ 2 = 62 [1]. [2] Examiner insight: Giving 60 and 64 as the boundaries is a common slip; for whole-number classes, extend each limit by half a unit.
2. (a) The midpoints are 3, 5, 7, 9, 11 [1]. The points are (3, 3), (5, 9), (7, 15), (9, 8), (11, 5) [1]. The end points use the empty classes either side: (1, 0) and (13, 0) [1]. [3] (b) It makes the polygon start and finish on the axis, so the diagram is closed [1]. [1] Examiner insight: Plot polygon points at midpoints and add the closing points; marking the midpoints on your axis makes your intention clear.
3. (a) The boundaries are 20.5–30.5, 30.5–40.5, 40.5–50.5, 50.5–60.5, 60.5–70.5 [1]. Every class width is 10, so the heights are the frequencies [1]: 5, 12, 19, 10, 4 [1]. [3] (b) The data are continuous on the scale, so each class ends where the next begins and there are no gaps [1]. [1] Examiner insight: Labelling the axis with printed limits (21 and 30) leaves gaps; use boundaries and give a reason for “no gaps”.
4. (a) Widths are 5, 5, 10, 20, 20 [1]. Densities are 15 ÷ 5 = 3, 30 ÷ 5 = 6, 35 ÷ 10 = 3.5, 20 ÷ 20 = 1, 10 ÷ 20 = 0.5 [2]. [3] (b) The modal class is 5 ≤ t < 10 because it has the greatest frequency density (6) [1]. [1] Examiner insight: The class 10 ≤ t < 20 has the largest frequency (35) but not the largest density; naming it without referring to density is a common way to lose the mark.
5. (a) Width = 8, so frequency = 2.5 × 8 = 20 [1]. [1] (b) The width from 14 to 20 is 6, so 6 × 2.5 = 15 [1]. [1] (c) The items are assumed to be evenly spread across the class [1]. [1] Examiner insight: A part-class answer is an estimate; writing “estimate” and naming the even-spread assumption shows you know it is not exact.
6. (a) 0 ≤ x < 10: 1.2 × 10 = 12 [1]. 10 ≤ x < 25: 2.0 × 15 = 30 [1]. 30 ≤ x < 50: 0.9 × 20 = 18 [1]. [3] (b) Frequency = 100 − (12 + 30 + 18) = 40 [1]. Density = 40 ÷ 5 = 8.0 [1]. [2] Examiner insight: Giving 40 as the bar’s height is a common slip; the bar height is frequency density, so divide by the width of 5.
7. Density of the first bar = 24 ÷ 10 = 2.4, which is 3 cm, so 1 cm = 0.8 of density [1]. Density of the second bar = 54 ÷ 15 = 3.6 [1]. Height = 3.6 ÷ 0.8 = 4.5 cm [1]. [3] Examiner insight: Comparing the two frequencies directly (54 against 24) ignores the different widths and leads to the wrong height; work through density.
8. (a) Midpoints are 15, 25, 35, 45 and the end classes have midpoints 5 and 55 [1]. Points: (5, 0), (15, 4), (25, 16), (35, 14), (45, 6), (55, 0) [2]. [3] (b) Club B has three times as many members, so every class is higher, which does not show age [1]. As proportions, club A has 14 ÷ 40 = 35% and 6 ÷ 40 = 15% in the top two classes (50% in total), whereas club B has 54 ÷ 120 = 45% and 24 ÷ 120 = 20% (65% in total), so club B does have the older members, but the height of its line is not the evidence [1]. [2] Examiner insight: A convincing comparison uses proportions from each club and states a conclusion.
9. (a) Widths are 10, 10, 10, 20, 40 [1]. Densities: 18 ÷ 10 = 1.8, 42 ÷ 10 = 4.2, 36 ÷ 10 = 3.6 [1]; 30 ÷ 20 = 1.5, 24 ÷ 40 = 0.6 [1]. [3] (b) Height = 2 × density [1], giving 3.6 cm, 8.4 cm, 7.2 cm, 3.0 cm, 1.2 cm [1]. [2] (c) 35 to 40 minutes: 5 × 3.6 = 18 [1]. 40 to 50 minutes: 10 × 1.5 = 15 [1]. Total estimate = 18 + 15 = 33 parcels [1]. [3] (d) The modal class is 20 ≤ t < 30, density 4.2 [1]. [1] Examiner insight: Show each part-class as width × density before adding, so your method is visible even if the total slips.
10. (a) Widths are 4, 4, 8, 24 [1]. Frequencies: 2.5 × 4 = 10, 6 × 4 = 24, 3 × 8 = 24, 0.5 × 24 = 12 [1]. [2] (b) 10 + 24 + 24 + 12 = 70 [1]. [1] (c) Under 16 is the whole of the first bar plus 2 years of the second: 10 + 2 × 6 [1] = 22 [1]. [2] (d) Width = 30, so density = 18 ÷ 30 = 0.6 [1]. [1] (e) New total = 70 + 18 = 88 [1]. Aged 18 or over = 24 + 12 + 18 = 54, so 54 ÷ 88 × 100 = 61.4% [1]. [2] Examiner insight: Remember the total changes when new members join, and keep working unrounded until the final 3 significant figure answer.
11. (a) The class widths are unequal, so the height must be frequency density, not frequency [1]. [1] (b) Widths are 10, 10, 20, 40 [1]. Heights: 30 ÷ 10 = 3, 36 ÷ 10 = 3.6, 48 ÷ 20 = 2.4, 40 ÷ 40 = 1 [1]. [2] (c) The student’s diagram suggests 20 ≤ t < 40, the tallest bar at 48 [1]. The correct modal class is 10 ≤ t < 20, with the greatest density of 3.6 [1]. [2] Examiner insight: For an explanation mark, name the specific error (unequal widths need frequency density) rather than saying only that the heights are wrong.
Where marks are usually lost
- Using printed class limits rather than boundaries for whole-number data, so widths and midpoints are off.
- Leaving out the closing points on a frequency polygon.
- Plotting polygon points at the upper boundary or limit instead of the midpoint.
- Drawing gaps between histogram bars of grouped continuous data.
- Using frequency as bar height when widths are unequal.
- Giving a bar’s height as the frequency, or its frequency as the height.
- Choosing the modal class from frequency instead of frequency density.
- Presenting an estimate for part of a class as exact, or not stating the even-spread assumption.
- Comparing two groups of different sizes using raw counts only.
Next steps
- Revisit the method in the study guide.
- Use the revision notes as a quick reference.
- Go to the Cambridge O Level Statistics hub and tick off the printable checklist.
- Try all free 10-minute diagnostics.
- Book a free trial class.
Official syllabus
Cambridge O Level Statistics 4040, syllabus for examination in 2025, 2026 and 2027, published by Cambridge University Press & Assessment. These questions follow section 3.2.
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