Cambridge O Level Statistics: Cumulative frequency tables, curves and polygons (4040) – Practice Questions

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 topic 4 of the Cambridge O Level Statistics 4040 syllabus for examination in 2025, 2026 and 2027: section 4.1 (cumulative frequency tables) and section 4.2 (cumulative frequency curves and polygons for continuous data). Learn the method from the study guide or the revision notes. The wider course is on the 4040 hub and the printable checklist. For graph readings, use the straight-line polygon through the points.

Questions

1. A taxi firm records the number of passengers in each of 60 taxis.

Passengers	1	2	3	4	5
Frequency	7	15	22	11	5

(a) Write down the cumulative frequency for each number of passengers. [2] (b) How many taxis carried more than 3 passengers? [1]

2. The lengths of leaves are measured to the nearest millimetre and grouped in classes 30–39, 40–49 and 50–59.

(a) State the lower and upper class boundaries of the class 40–49. [1] (b) At which length is the cumulative frequency for this class plotted? [1]

3. The waiting times, in minutes, of 24 callers at a help desk are:

2.4, 7.1, 11.8, 4.6, 9.3, 14.2, 6.0, 3.5, 12.7, 8.8, 1.9, 5.5, 10.4, 16.3, 7.7, 4.0, 13.1, 9.9, 6.8, 18.2, 3.1, 12.0, 8.2, 5.0

(a) Complete a frequency table using the classes 0 ≤ t < 5, 5 ≤ t < 10, 10 ≤ t < 15 and 15 ≤ t < 20. [2] (b) Form the cumulative frequency table, labelled with the upper class boundaries. [2]

4. The cumulative frequency table shows the time, t minutes, that 100 pupils spent on homework one evening.

t ≤	30	60	90	120	150	180
Cumulative frequency	8	27	61	84	95	100

(a) Find the frequency of the class 60 < t ≤ 90. [1] (b) How many pupils spent more than 120 minutes? [1] (c) What percentage of the pupils spent 60 minutes or less? [1]

5. The masses of 100 apples are recorded.

Mass, m grams	0 < m ≤ 50	50 < m ≤ 100	100 < m ≤ 150	150 < m ≤ 250	250 < m ≤ 400
Frequency	4	16	30	40	10

(a) Complete the cumulative frequency row. [2] (b) List the coordinates you would plot for a cumulative frequency polygon. [2]

6. A battery test gives the lifetimes, t hours, of 50 batteries. The cumulative frequencies are 3, 12, 29, 43 and 50 for the classes 0 < t ≤ 10, 10 < t ≤ 20, 20 < t ≤ 30, 30 < t ≤ 40 and 40 < t ≤ 50. A pupil plots (5, 3), (15, 12), (25, 29), (35, 43) and (45, 50) and joins them with a smooth curve.

(a) Identify two errors. [2] (b) Write down the correct coordinates. [2]

7. The table shows the waiting times, w minutes, of 140 patients at a clinic.

w ≤	0	10	20	30	40	60
Cumulative frequency	0	11	39	83	119	140

Use a polygon, or straight-line working, to estimate:

(a) the number of patients who waited 25 minutes or less, [2] (b) the number who waited more than 35 minutes, [2] (c) the time that 70 patients did not exceed. [1]

8. Class A (40 pupils) and Class B (50 pupils) sat one test marked out of 60.

Marks, m	0 < m ≤ 10	10 < m ≤ 20	20 < m ≤ 30	30 < m ≤ 40	40 < m ≤ 50	50 < m ≤ 60
Class A frequency	2	5	9	12	8	4
Class B frequency	3	8	14	13	9	3

(a) Form the cumulative frequency table for each class. [2] (b) Explain why percentages, not counts, should be compared. [1] (c) Find the percentage of each class that scored 30 or less. [2] (d) Which class did better on this evidence? Give a reason. [1]

9. A cumulative frequency table for the distances, d km, travelled by 60 shoppers is incomplete.

d ≤	5	10	15	20	25
Cumulative frequency	6	19	x	51	60

The class 10 < d ≤ 15 has frequency 18. Find x, and find the frequency of the class 20 < d ≤ 25. [3]

10. Explain, in terms of a cumulative frequency graph: (a) how a polygon differs from a curve, [1] (b) why the cumulative frequency of a class is plotted at its upper boundary, [1] (c) what a nearly flat section of the graph shows. [1]

11. The heights of 60 sunflowers are measured to the nearest centimetre.

Height (cm)	100–119	120–139	140–159	160–179	180–199	200–229
Frequency	5	11	19	15	7	3

(a) State the boundaries of the first and last classes. [2] (b) Form the cumulative frequency table, with the upper boundaries. [2] (c) Estimate the number of sunflowers shorter than 150 cm. [2] (d) Estimate the height that only 15 sunflowers exceed. [2]

12. A gym records the length of 140 sessions, t minutes.

Length, t minutes	0 < t ≤ 15	15 < t ≤ 30	30 < t ≤ 45	45 < t ≤ 60	60 < t ≤ 90	90 < t ≤ 120
Frequency	8	22	35	40	25	10

(a) Form the cumulative frequency table and list the points to plot, including the starting point. [3] (b) Estimate how many sessions lasted 78 minutes or less. [2] (c) Estimate the time within which 115 sessions finished. [2]

Answers

1. (a) Adding the running total: 7, then 7 + 15 = 22, then 22 + 22 = 44, 44 + 11 = 55, 55 + 5 = 60 [1]. Cumulative frequencies: 7, 22, 44, 55, 60, with the last equal to 60 [1]. [2] (b) 60 − 44 = 16 taxis [1]. [1] Examiner insight: Showing that the last total equals the number of taxis is a quick check on the table.

2. (a) Rounded to the nearest mm, 40 covers 39.5 up to 49.5, so the boundaries are 39.5 and 49.5 [1]. [1] (b) At the upper boundary, 49.5 [1]. [1] Examiner insight: Giving 40 and 49 as the boundaries is a common slip, since they are the class limits.

3. (a) Sorting the values: 0 ≤ t < 5 holds 2.4, 4.6, 3.5, 1.9, 4.0, 3.1 [1]; the classes are 6, 10, 6, 2, with a total of 24 [1]. [2] (b) Running totals: 6, 16, 22, 24 [1]. Table: t < 5: 6, t < 10: 16, t < 15: 22, t < 20: 24 [1]. [2] Examiner insight: Tick values off as you sort them. Here 5.0 belongs in the second class.

4. (a) 61 − 27 = 34 [1]. [1] (b) 100 − 84 = 16 pupils [1]. [1] (c) 27 out of 100 is 27% [1]. [1] Examiner insight: Subtract the cumulative frequency before it, and not the one after, to get a class frequency.

5. (a) Running total: 4, 4 + 16 = 20, 20 + 30 = 50, 50 + 40 = 90, 90 + 10 = 100 [1]. 4, 20, 50, 90, 100 [1]. [2] (b) Upper boundaries 50, 100, 150, 250, 400 with these totals [1], and the start (0, 0). Coordinates: (0, 0), (50, 4), (100, 20), (150, 50), (250, 90), (400, 100) [1]. [2] Examiner insight: Unequal class widths do not change the cumulative frequencies. The coordinates are still upper boundary against running total.

6. (a) The points are plotted at the class midpoints and not the upper boundaries [1], and the graph has no starting point: the first lower boundary, with cumulative frequency 0, is missing [1]. [2] (b) Upper boundaries with totals [1]: (0, 0), (10, 3), (20, 12), (30, 29), (40, 43), (50, 50) [1]. [2] Examiner insight: Name each error exactly, such as “plotted at midpoints”, and avoid a vague comment like “wrong points”.

7. (a) 25 is halfway between 20 and 30, so the value is halfway between 39 and 83 [1]: 39 + 22 = 61 patients [1]. [2] (b) At 35, halfway between 83 and 119: 83 + 18 = 101 [1]. More than 35: 140 − 101 = 39 patients [1]. [2] (c) Between (20, 39) and (30, 83): 20 + (31/44) × 10 ≈ 27 minutes [1]. [1] Examiner insight: For “more than”, subtract your reading from the total. Quoting the reading itself answers the opposite question.

8. (a) Class A: 2, 7, 16, 28, 36, 40 [1]. Class B: 3, 11, 25, 38, 47, 50 [1]. [2] (b) The classes have different sizes (40 and 50 pupils), so counts are not a fair comparison [1]. [1] (c) Class A: 16/40 × 100 = 40% [1]. Class B: 25/50 × 100 = 50% [1]. [2] (d) Class A, because a smaller percentage of Class A (40% against 50%) scored 30 or less [1]. [1] Examiner insight: Use figures from both groups in a comparison, such as “40% against 50%”.

9. Add the class frequency: x = 19 + 18 [1] = 37 [1]. The last class frequency is 60 − 51 = 9 [1]. [3] Examiner insight: Show the addition you used, so your method is visible.

10. (a) A polygon joins the points with straight lines, and a curve joins them with one smooth line [1]. [1] (b) Every item in the class has been counted only once the top of the class is reached [1]. [1] (c) Few items fall in that range, so the cumulative frequency barely rises [1]. [1] Examiner insight: Reason from what the running total counts, and avoid saying only “the graph goes up”.

11. (a) The first class has lower boundary 99.5 [1] and the last class has upper boundary 229.5 [1]. [2] (b) Running total: 5, 16, 35, 50, 57, 60 [1]. Table with the upper boundaries 119.5, 139.5, 159.5, 179.5, 199.5, 229.5 [1]. [2] (c) 150 lies between 139.5 (total 16) and 159.5 (total 35), at a fraction 10.5/20 of the way [1]: 16 + (10.5/20) × 19 = 25.975, so about 26 sunflowers [1]. [2] (d) Read at 60 − 15 = 45 [1]. Between (159.5, 35) and (179.5, 50): 159.5 + (10/15) × 20 = 172.8, so about 173 cm [1]. [2] Examiner insight: “Only 15 exceed” means 45 are at or below, so you read at 45. Reading at 15 answers a different question.

12. (a) Running total: 8, 30, 65, 105, 130, 140 [1]. Table: t ≤ 15: 8, t ≤ 30: 30, t ≤ 45: 65, t ≤ 60: 105, t ≤ 90: 130, t ≤ 120: 140 [1]. Points: (0, 0), (15, 8), (30, 30), (45, 65), (60, 105), (90, 130), (120, 140) [1]. [3] (b) 78 lies between 60 and 90, at a fraction 18/30 of the way [1]: 105 + (18/30) × 25 = 120 sessions [1]. [2] (c) 115 lies between totals 105 and 130, at a fraction 10/25 of the way [1]: 60 + (10/25) × 30 = 72 minutes [1]. [2] Examiner insight: With unequal classes, find the class holding your target first, then take the fraction of that class’s width.

Where marks are usually lost

• Plotting at the midpoint of a class and not at its upper boundary.
• Leaving out the starting point, which has cumulative frequency 0 at the first lower boundary.
• Using class limits (24) instead of boundaries (24.5) for rounded data.
• Errors in the running total, with no check against the total frequency.
• Reading at the wrong height for “more than” or “highest” questions.
• Comparing counts for groups of different sizes and not percentages.
• Quoting graph readings as exact, or without units.
• Leaving out working, although the syllabus says essential working must be shown.

Next steps

• Revise with the revision notes and the study guide.
• Tick it off on the checklist and see the 4040 hub.
• Try all the free 10-minute diagnostics to find your weakest topics.
• Want a teacher? Book a free trial class.

Official syllabus

Cambridge O Level Statistics 4040, syllabus for examination in 2025, 2026 and 2027, Cambridge University Press & Assessment, topic 4, sections 4.1 and 4.2.