Cambridge O Level Statistics: Quartiles, percentiles and combining data sets (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 sections 5.3 and 5.4 of the Cambridge O Level Statistics 4040 syllabus for examination in 2025, 2026 and 2027, which is not tiered. Learn the methods in the study guide and the revision notes. Check your place on the 4040 course hub and the printable checklist. Unsure where to start? Try the free 10-minute diagnostics.

Use n for a total frequency and take the target positions as n/4, n/2, 3n/4 and (P/100) × n. Give estimates to 3 significant figures unless told otherwise.

Questions

1. A sample has 200 values in a cumulative frequency table. State the cumulative frequency you would read to estimate the lower quartile, the median, the upper quartile and the 70th percentile. [2]

2. A cycle courier records the length, d km, of 90 journeys.

Distance, d km	0 < d ≤ 5	5 < d ≤ 10	10 < d ≤ 20	20 < d ≤ 40	40 < d ≤ 60
Frequency	14	29	27	14	6

(a) Complete a cumulative frequency table for these data. [1] (b) Estimate the median distance by linear interpolation. [3] (c) Estimate the interquartile range. [3]

3. Use the table in question 2.

(a) Estimate the 90th percentile of the distances. [3] (b) Estimate how many journeys were longer than 30 km. [2]

4. The times, to the nearest second, taken by 80 people to complete a puzzle are grouped as follows.

Time (seconds)	10–19	20–29	30–39	40–49	50–59
Frequency	6	18	30	20	6

(a) Write down the boundaries of the first class and the cumulative frequency at each upper boundary. [2] (b) Estimate the median time. [2] (c) Estimate the interquartile range. [3]

5. A cumulative frequency polygon for 60 values of a variable x passes through these points.

x	0	20	40	60	80	100
Cumulative frequency	0	6	21	45	57	60

(a) Estimate the median. [2] (b) Estimate the lower quartile and the upper quartile. [3] (c) Estimate the 80th percentile. [2]

6. In a reading test, school A has median 54 and interquartile range 12. School B has median 58 and interquartile range 24. The 90th percentile of school B is 71. (a) Explain what the 90th percentile of 71 means. [1] (b) Compare the results of the two schools. [3]

7. A class of 36 pupils has mean mark 58. A second class of 24 pupils has mean mark 67. Find the mean mark of all 60 pupils. [3]

8. Group A has 12 values with mean 40 and standard deviation 4. Group B has 8 values with mean 45 and standard deviation 3. (a) Find the mean and standard deviation of the 20 values combined. [5] (b) A student says the combined standard deviation should be 3.5, the mean of 4 and 3 weighted by group size. Explain why this is wrong. [1]

9. Set A is 12, 15, 9, 18, 12, 12. Set B is 20, 14, 17, 13. Find the mean and standard deviation of the ten values together. [6]

10. Twenty-four values have mean 52. They are combined with a second set of n values with mean 60, and the mean of the combined set is 57. Find n. [4]

11. Ten readings have mean 36 and standard deviation 4. (a) An eleventh reading, 36, is added. Find the new mean and standard deviation. [5] (b) Explain why the standard deviation has fallen. [2] (c) The eleven readings are combined with nine more readings of mean 41 and standard deviation 6. Find the mean and standard deviation of all twenty. [3]

Answers

1. Lower quartile: 200/4 = 50. Median: 100. Upper quartile: 150. [1] 70th percentile: 0.7 × 200 = 140. [1] [2] Examiner insight: Write the target next to each measure before reading the table, since a wrong target position is the commonest way to lose the whole question.

2. (a) Cumulative frequencies: 14, 43, 70, 84, 90 at 5, 10, 20, 40, 60. [1] [1] (b) Target 90/2 = 45, in the class 10–20 (43 before it, 27 in it). [1] Median = 10 + (45 − 43)/27 × 10 [1] = 10.7 km. [1] [3] (c) Q1: target 22.5, in the class 5–10: 5 + (22.5 − 14)/29 × 5 = 6.47. [1] Q3: target 67.5, in the class 10–20: 10 + (67.5 − 43)/27 × 10 = 19.07. [1] Interquartile range = 19.07 − 6.47 = 12.6 km. [1] [3] Examiner insight: Show the class chosen, the cumulative frequency before it and the class width, because a slip in one of them then still leaves visible, correct method.

3. (a) Target 0.9 × 90 = 81, in the class 20–40 (70 before it, 14 in it, width 20). [1] 20 + (81 − 70)/14 × 20 [1] = 35.7 km. [1] [3] (b) At 30 km the cumulative frequency is 70 + (10/20) × 14 = 77. [1] Journeys longer than 30 km: 90 − 77 = 13. [1] [2] Examiner insight: “Longer than” needs a subtraction from the total, so say that step out loud in your working.

4. (a) The first class runs from 9.5 to 19.5. [1] Cumulative frequencies at 19.5, 29.5, 39.5, 49.5, 59.5 are 6, 24, 54, 74, 80. [1] [2] (b) Target 40, in the class 29.5 to 39.5: 29.5 + (40 − 24)/30 × 10 [1] = 34.8 seconds. [1] [2] (c) Q1: target 20: 19.5 + (20 − 6)/18 × 10 = 27.3. [1] Q3: target 60: 39.5 + (60 − 54)/20 × 10 = 42.5. [1] Interquartile range = 42.5 − 27.3 = 15.2 seconds. [1] [3] Examiner insight: Using the limits 10, 20, 30 instead of the boundaries gives answers that are all off by half a second, so state the boundaries first.

5. (a) Target 30, between (40, 21) and (60, 45). [1] 40 + (30 − 21)/24 × 20 = 47.5. [1] [2] (b) Lower quartile: target 15, between (20, 6) and (40, 21): 20 + (15 − 6)/15 × 20 [1] = 32. [1] Upper quartile: target 45 is a plotted point, so 60. [1] [3] (c) Target 0.8 × 60 = 48, between (60, 45) and (80, 57): 60 + (48 − 45)/12 × 20 [1] = 65. [1] [2] Examiner insight: If a target is exactly one of the plotted cumulative frequencies, say so and read the value directly instead of interpolating.

6. (a) 90% of pupils in school B scored 71 or less (and 10% scored more). [1] [1] (b) School B has the higher median (58 against 54), so its typical pupil scored more. [1] School B has the larger interquartile range (24 against 12), so its middle half is more spread out and the results are less consistent. [1] In context, school A’s pupils are more alike in reading ability, while school B’s pupils score higher typically but vary more. [1] [3] Examiner insight: A comparison earns credit when each statement uses a figure and names the property (average or spread) and the context, not just “B is bigger”.

7. Total marks of the first class = 36 × 58 = 2088. [1] Total of the second = 24 × 67 = 1608. [1] Mean = (2088 + 1608)/60 = 3696/60 = 61.6. [1] [3] Examiner insight: Averaging 58 and 67 to get 62.5 fails to weight by class size, and a bare wrong answer then has no method to credit.

8. (a) Σx: 12 × 40 = 480 and 8 × 45 = 360, total 840, so the mean is 840/20 = 42. [1] Σx² for A = 12 × (16 + 1600) = 19 392. [1] Σx² for B = 8 × (9 + 2025) = 16 272. [1] Total Σx² = 35 664, so variance = 35 664/20 − 42² = 1783.2 − 1764 = 19.2. [1] Standard deviation = 4.38 (3 s.f.). [1] [5] (b) Standard deviation depends on squared distances from the combined mean, and the two group means (40 and 45) differ, which adds spread that a simple average of the two standard deviations cannot include. [1] [1] Examiner insight: Keep the combined mean unrounded and subtract its square last, as an early rounding slip changes the variance noticeably.

9. Σx = 78 + 64 = 142, so the mean is 14.2. [1] Σx² for A = 144 + 225 + 81 + 324 + 144 + 144 = 1062. [1] Σx² for B = 400 + 196 + 289 + 169 = 1054. [1] Total Σx² = 2116. [1] Variance = 2116/10 − 14.2² = 211.6 − 201.64 = 9.96. [1] Standard deviation = 3.16 (3 s.f.). [1] [6] Examiner insight: Checking each set’s total against a quick estimate catches an arithmetic slip early, which matters because every later step depends on the totals.

10. Total of the first set = 24 × 52 = 1248. [1] Total of the second = 60n. [1] Combined: 1248 + 60n = 57 × (24 + n) = 1368 + 57n. [1] 3n = 120, so n = 40. [1] [4] Examiner insight: Form an equation from totals, not from means, and check by substituting n back into the combined mean.

11. (a) New mean = (10 × 36 + 36)/11 = 36. [1] Σx² for the ten = 10 × (16 + 1296) = 13 120. [1] Adding 36² = 1296 gives 14 416. [1] Variance = 14 416/11 − 36² = 14.545. [1] Standard deviation = 3.81 (3 s.f.). [1] [5] (b) The new reading equals the mean, so it adds no squared deviation to the total. [1] The number of readings rises, so the average squared deviation falls. [1] [2] (c) Σx = 396 + 9 × 41 = 396 + 369 = 765, so the mean is 765/20 = 38.25. [1] Σx² for the nine = 9 × (36 + 1681) = 15 453, so total Σx² = 14 416 + 15 453 = 29 869. [1] Variance = 29 869/20 − 38.25² = 1493.45 − 1463.0625 = 30.3875, so standard deviation = 5.51 (3 s.f.). [1] [3] Examiner insight: Use the exact sum of squares from part (a) in part (c), not the rounded 3.81, to avoid a small inaccuracy that carries through.

Where marks are usually lost

• Reading the wrong target position, such as n/2 for a quartile.
• Interpolating with the cumulative frequency of the class itself instead of the one before it.
• Using the class limits instead of the true boundaries for rounded data.
• Forgetting the subtraction from n when a question asks for “more than” or “the highest”.
• Combining means without allowing for group sizes, or taking the mean of two standard deviations.
• Writing Σx² = n × σ² and leaving out the mean squared.
• Rounding the combined mean or variance before the final square root.
• Comparing two data sets with no figures, or comparing without naming both an average and a spread.
• Quoting a graph or interpolation estimate to far more figures than the data supports.

Next steps

• Re-read the revision notes for the formulas and distinctions.
• Go back to the study guide for any method you missed.
• Visit the 4040 course hub and the printable checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class: /trial/.

Official syllabus

Cambridge O Level Statistics 4040, syllabus for examination in 2025, 2026 and 2027, Cambridge University Press & Assessment, topic 5, sections 5.3 and 5.4.