Cambridge O Level Statistics: Averages: mean, median, mode and modal class (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 section 5.1 of the Cambridge O Level Statistics (4040) syllabus for examination in 2025, 2026 and 2027: the mean, median, mode and modal class from a set of numbers, ungrouped frequency distributions and grouped frequency distributions, and the choice of average. Work from the study guide or the revision notes if you need to recall a method. See also the Cambridge O Level Statistics hub and the printable checklist. Show your working and give answers to 3 significant figures unless a question says otherwise.

Questions

1. A cafe records the number of cakes sold each day for seven days: 6, 9, 4, 9, 12, 7, 9. Find the mean, the median and the mode. [3]

2. The marks of eight students in a test are 53, 48, 61, 48, 57, 66, 52 and 59.

(a) Find the median mark. [2] (b) A ninth student scores 80. Find the new mean mark. [2] (c) Find the new median. [1]

3. A team’s goals in 40 matches are shown.

Goals	0	1	2	3	4	5	6
Matches	3	7	10	9	6	3	2

(a) Calculate the mean number of goals. [3] (b) Find the median. [2] (c) State the mode. [1]

4. The times, in seconds, taken by 80 runners to complete a course are grouped below.

Time t (seconds)	15 ≤ t < 25	25 ≤ t < 35	35 ≤ t < 45	45 ≤ t < 65	65 ≤ t < 95
Frequency	8	15	22	30	5

(a) Calculate the frequency density of each class. [2] (b) State the modal class and explain your choice. [2] (c) Estimate the mean time. [3] (d) Explain why your answer to (c) is an estimate. [1]

5. Say which average (mean, median or mode) is most suitable in each case.

(a) The favourite sport of each of 30 pupils. [1] (b) The salaries of 12 staff, one of whom is a director paid far more than everyone else. Give a reason. [2]

6. Five numbers have a mean of 12.4. Four of them are 9, 15, 11 and 14.

(a) Find the fifth number. [2] (b) Find the median of the five numbers. [2]

7. The table shows the number of rooms, x, in some flats. The frequency of 3 rooms is k and the mean number of rooms is 3.1.

Rooms (x)	1	2	3	4	5
Flats	2	7	k	6	4

(a) Find the value of k. [4] (b) State the mode. [1]

8. The number of texts sent in a day by 60 people is grouped below.

Texts	0-4	5-9	10-14	15-19	20-24
People	6	14	21	12	7

(a) State the modal class. [1] (b) Estimate the mean number of texts. [3] (c) Estimate the median, using the class boundaries. [2]

9. The delivery times, in minutes, of two couriers are recorded for ten deliveries each.

Courier A: 22, 24, 25, 25, 26, 27, 28, 29, 30, 64 Courier B: 23, 25, 26, 27, 27, 27, 28, 30, 31, 33

(a) Find the mean of each courier. [2] (b) Find the median of each courier. [2] (c) Find the mode of each courier. [1] (d) A customer wants the courier who is usually quicker. Choose an average, justify it and state which courier you recommend. [3]

10. Twelve scores are 3, 7, 8, 12, 13, 14, 16, 17, 21, 22, 24 and 28.

(a) Find the exact mean. [1] (b) The scores are grouped as 0-9 (3 scores), 10-19 (5 scores) and 20-29 (4 scores). Estimate the mean from the groups. [2] (c) Explain why the two means differ. [1]

Answers

1. Sum = 6 + 9 + 4 + 9 + 12 + 7 + 9 = 56, so mean = 56 ÷ 7 = 8 [1]. Ordered: 4, 6, 7, 9, 9, 9, 12, so the 4th value gives median 9 [1]. Mode 9 [1]. [3] Examiner insight: Writing the ordered list is the safest way to show where the median came from, because it lets a reader see where the middle value came from.

2. (a) Ordered: 48, 48, 52, 53, 57, 59, 61, 66 [1]. Median = (53 + 57) ÷ 2 = 55 [1]. (b) Sum = 444, new sum = 524 [1]. New mean = 524 ÷ 9 = 58.2 [1]. (c) Nine values; the 5th is 57 [1]. [5] Examiner insight: With an even number of values the median lies between two values, so quote the halfway value rather than either middle one.

3. (a) Σfx = 0 + 7 + 20 + 27 + 24 + 15 + 12 = 105 [1], Σf = 40 [1], mean = 105 ÷ 40 = 2.625 = 2.63 [1]. (b) Cumulative: 3, 10, 20, 29, 35, 38, 40. The 20th value is 2 and the 21st is 3 [1], so the median is 2.5 [1]. (c) Mode = 2 goals [1]. [6] Examiner insight: Here the middle falls exactly between two groups, so say which two values you averaged; a median of 2 or 3 alone misses the point.

4. (a) Widths 10, 10, 10, 20, 30 [1]. Densities: 0.8, 1.5, 2.2, 1.5, 0.167 [1]. (b) Modal class 35 ≤ t < 45 [1], because it has the highest frequency density (2.2), and the widths are unequal [1]. (c) Midpoints 20, 30, 40, 55, 80 [1]. Σfx = 160 + 450 + 880 + 1650 + 400 = 3540 [1]. Mean ≈ 3540 ÷ 80 = 44.25 s (44.3 s) [1]. (d) The exact times are not known, so every runner is assumed to be at the class midpoint [1]. [8] Examiner insight: The class 45 ≤ t < 65 has the largest frequency (30), but comparing frequency alone ignores its greater width; showing the densities lets a reader see why you chose otherwise.

5. (a) The mode, because the data is categories [1]. (b) The median [1], because the director’s salary is an extreme value that would inflate the mean [1]. [3] Examiner insight: Naming the average earns little without a reason tied to the data given in the question.

6. (a) Total = 5 × 12.4 = 62 [1]. Fifth number = 62 − 49 = 13 [1]. (b) Ordered: 9, 11, 13, 14, 15 [1]. Median = 13 [1]. [4] Examiner insight: Show the total of 62 before subtracting, so a slip in the subtraction is still visible to a reader.

7. (a) Σf = 19 + k [1]. Σfx = 2 + 14 + 3k + 24 + 20 = 60 + 3k [1]. Equation: 60 + 3k = 3.1(19 + k) [1]. 60 + 3k = 58.9 + 3.1k, so 1.1 = 0.1k and k = 11 [1]. (b) The highest frequency, 11, is at x = 3, so the mode is 3 rooms [1]. [5] Examiner insight: A common slip is to divide the total by 5 (the number of rows); the denominator is Σf, which includes k.

8. (a) 10-14 (frequency 21) [1]. (b) Midpoints 2, 7, 12, 17, 22 [1]. Σfx = 12 + 98 + 252 + 204 + 154 = 720 [1]. Mean ≈ 720 ÷ 60 = 12 texts [1]. (c) The 30th value lies in 10-14, which has boundaries 9.5 and 14.5, with 20 people below [1]. Median ≈ 9.5 + (10 ÷ 21) × 5 = 11.9 [1]. [6] Examiner insight: Starting interpolation from the boundary 9.5, not the stated limit 10, is the choice that fits whole-number classes.

9. (a) A: 300 ÷ 10 = 30.0 [1]; B: 277 ÷ 10 = 27.7 [1]. (b) A: (26 + 27) ÷ 2 = 26.5 [1]; B: (27 + 27) ÷ 2 = 27 [1]. (c) A: 25; B: 27 [1]. (d) Use the median [1], because courier A’s 64 minutes is an extreme value that pushes the mean up to 30.0 [1]. A’s median of 26.5 is lower than B’s 27, so recommend courier A for usually being quicker [1]. [8] Examiner insight: A recommendation is stronger when it names the figures compared and admits that the mean points the other way.

10. (a) 185 ÷ 12 = 15.4 [1]. (b) Midpoints 4.5, 14.5, 24.5: 13.5 + 72.5 + 98 = 184 [1]. 184 ÷ 12 = 15.3 [1]. (c) The grouped estimate assumes every score sits at its class midpoint, but the actual scores are spread differently [1]. [4] Examiner insight: A good comparison states that the estimate is close but not equal, and gives the assumption as the reason.

Where marks are usually lost

• Finding the median without ordering the data.
• Quoting a frequency as the mode.
• Using the number of rows as the divisor instead of Σf.
• Using class limits rather than midpoints for an estimated mean.
• Picking the modal class from the largest frequency when the classes are not equally wide.
• Calling an estimate exact, or not saying why it is an estimate.
• Giving an average without a reason that refers to the data.
• Rounding part-way through, then losing accuracy in the final line.

Next steps

• Review the methods in the revision notes and the study guide.
• Visit the Cambridge O Level Statistics hub and tick off the checklist.
• Take the free 10-minute diagnostics to find weak spots.
• Ask for a free trial class for help with any question you could not finish.

Official syllabus

Cambridge O Level Statistics (4040), syllabus for examination in 2025, 2026 and 2027, Cambridge University Press & Assessment (Cambridge International). Section 5.1: measures of central tendency, mean, median, mode and modal class.