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Cambridge O Level Mathematics: Statistics: Classifying Data, Averages, Spread and Statistical Charts (4024) – Practice Questions

Original questions with worked answers on tables, averages, grouped estimates, pie charts, pictograms and comparing data for Cambridge O Level Mathematics.

Subject
Mathematics
Level
O LEVELS
Topic
Statistics: Classifying Data, Averages, Spread and Statistical Charts
Updated

Aligned to Cambridge O Level Mathematics (4024), 2025-2027. Official specification .

Syllabus page (what it covers and how it is assessed): Cambridge O Level Mathematics.

Syllabus points this page covers

4024

  • 9.1 Classifying statistical data
  • 9.2 Interpreting statistical data
  • 9.3 Averages and measures of spread
  • 9.4 Statistical charts and diagrams

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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 sections 9.1 to 9.4 of the Cambridge O Level Mathematics (Syllabus D) syllabus for exams in 2025, 2026 and 2027. Use the study guide and revision notes if a method is unfamiliar. See also the course hub, the checklist and the free diagnostics. Show all working. Give answers to 3 significant figures unless told otherwise.

Questions

1. State whether each of the following is categorical data, discrete numerical data or continuous numerical data. (a) The shoe sizes of 30 pupils. [1] (b) The time taken by each pupil to run 100 m. [1] (c) The favourite colour of each pupil. [1]

2. Twenty households were asked how many pets they own. The results were:

1 0 2 1 3 1 0 2 2 1 4 1 0 1 2 3 1 2 1 0

(a) Complete a tally and frequency table for the data. [2] (b) Write down the mode. [1] (c) Write down the fraction of households that own more than two pets. [1]

3. The table shows the clubs joined by 60 pupils in Years 10 and 11. Some entries are missing.

Chess Drama Art Total
Year 10 12 14 35
Year 11 25
Total 20 24 16 60

(a) Copy and complete the table. [3] (b) Find the percentage of the Drama club who are in Year 11. Give your answer to 1 decimal place. [2]

4. The number of text messages sent by a pupil on eight days was 12, 7, 15, 9, 7, 20, 11, 7.

(a) Find the mean. [2] (b) Find the median. [2] (c) Find the mode and the range. [2]

5. The mean of six numbers is 8.5. A seventh number is added and the mean of the seven numbers is 9.

(a) Find the seventh number. [3] (b) The six original numbers are 4, 6, 8, 9, 11 and 13. Find the median of all seven numbers. [2]

6. The table shows the number of passengers in each of 30 cars.

Passengers 1 2 3 4 5
Cars 9 12 5 3 1

(a) Write down the mode. [1] (b) Find the median. [2] (c) Calculate the mean. [3]

7. The weekly savings, £s, of 40 pupils are summarised below.

Savings s (£) 0 < s ≤ 5 5 < s ≤ 10 10 < s ≤ 15 15 < s ≤ 20 20 < s ≤ 25
Frequency 6 13 11 7 3

(a) Write down the modal class. [1] (b) Calculate an estimate of the mean weekly savings. [3] (c) Explain why your answer to part (b) is only an estimate. [1]

8. A survey of 90 households recorded their main type of heating: gas 40, electricity 25, oil 15, wood 10.

(a) Calculate the angle of each sector of a pie chart for this data. [3] (b) In a pie chart for a different survey, a sector of 135° represents 27 people. Find the total number of people in that survey. [2]

9. A pictogram shows the number of bicycles hired on four days. The key is: one full symbol represents 6 bicycles. Monday has 4 full symbols. Tuesday has 3 full symbols and a half symbol. Wednesday has 2 full symbols and a half symbol. Thursday has 5 full symbols.

(a) How many bicycles were hired on Wednesday? [1] (b) Find the total number hired over the four days. [2] (c) On Friday, 33 bicycles were hired. How many symbols are needed? [1]

10. The table shows the hot and cold drinks sold at a kiosk.

Month Jan Feb Mar Apr
Hot drinks 30 25 20 15
Cold drinks 10 15 25 35

A composite bar chart is drawn with hot drinks at the bottom of each bar.

(a) Find the height of each of the four bars. [2] (b) Between which two heights is the cold section of the March bar drawn? [1] (c) Describe two changes shown by the data from January to April. [2]

11. Eight plants were grown with fertiliser A and eight with fertiliser B. Their heights in cm after four weeks were:

A: 12 15 16 16 18 19 20 24
B: 15 16 17 17 18 18 19 20

(a) Find the mean height for each fertiliser. [2] (b) Find the median height for each fertiliser. [2] (c) Find the range for each fertiliser. [1] (d) Compare the two sets of plants, using your answers. [2] (e) Give one reason why you should be careful about drawing a firm conclusion from this data. [1]

12. A survey asked 48 pupils how many hours t they spent online last week.

Hours t 0 < t ≤ 4 4 < t ≤ 8 8 < t ≤ 12 12 < t ≤ 16 16 < t ≤ 20
Frequency 8 x 15 9 4

In a pie chart of all 48 pupils, the sector for 4 < t ≤ 8 has an angle of 90°.

(a) Find the value of x. [2] (b) Write down the modal class. [1] (c) Calculate an estimate of the mean number of hours. [3] (d) Find the angle of the sector for 12 < t ≤ 16. [2]

Answers

1. (a) Shoe sizes can only take separate values, so discrete numerical [1]. (b) Time can take any value in a range, so continuous numerical [1]. (c) A colour is described in words, so categorical [1]. [3] Examiner insight: A short reason, such as “separate values only”, shows you know the difference between discrete and continuous.

2. (a) Tally strokes grouped in fives, then counted [1]. Frequencies: 0 → 4, 1 → 8, 2 → 5, 3 → 2, 4 → 1, total 20 [1]. (b) Mode = 1 [1]. (c) Households with 3 or 4 pets: 2 + 1 = 3, so the fraction is 3/20 [1]. [4] Examiner insight: Check that the frequencies add up to 20. Writing the mode as 8 (its frequency) gives the count, not the value.

3. (a) Year 10 Art = 35 − 12 − 14 = 9 [1]. Year 11 Chess = 20 − 12 = 8 and Year 11 Drama = 24 − 14 = 10 [1]. Year 11 Art = 25 − 8 − 10 = 7, which agrees with 16 − 9 = 7 [1]. (b) Fraction = 10/24 [1]. Percentage = 10 ÷ 24 × 100 = 41.7% [1]. [5] Examiner insight: Use the Drama column total (24) as the denominator, not 60. Checking a missing entry two ways guards against slips.

4. (a) Sum = 12 + 7 + 15 + 9 + 7 + 20 + 11 + 7 = 88 [1]. Mean = 88 ÷ 8 = 11 [1]. (b) Ordered: 7, 7, 7, 9, 11, 12, 15, 20 [1]. Median = (9 + 11) ÷ 2 = 10 [1]. (c) Mode = 7 [1]. Range = 20 − 7 = 13 [1]. [6] Examiner insight: Show the addition. Taking the median from the unordered list is a common slip.

5. (a) Total of six numbers = 6 × 8.5 = 51 [1]. Total of seven numbers = 7 × 9 = 63 [1]. Seventh number = 63 − 51 = 12 [1]. (b) Ordered: 4, 6, 8, 9, 11, 12, 13 [1]. Median = the fourth value = 9 [1]. [5] Examiner insight: Work with totals, not means. Averaging the two means does not use the totals.

6. (a) The highest frequency is 12, for 2 passengers, so the mode is 2 [1]. (b) There are 30 cars, so the median lies between the 15th and 16th values [1]. Cumulative frequencies are 9, then 21, so both values are 2 and the median is 2 [1]. (c) Σf = 30 [1]. Σfx = 9 + 24 + 15 + 12 + 5 = 65 [1]. Mean = 65 ÷ 30 = 2.17 (3 s.f.) [1]. [6] Examiner insight: Write each product f × x down so your method is visible. Dividing by 5 columns instead of 30 cars is a frequent slip.

7. (a) The highest frequency is 13, so the modal class is 5 < s ≤ 10 [1]. (b) Midpoints: 2.5, 7.5, 12.5, 17.5, 22.5 [1]. Σ(f × midpoint) = 15 + 97.5 + 137.5 + 122.5 + 67.5 = 440 [1]. Estimated mean = 440 ÷ 40 = £11 [1]. (c) The exact savings are not known, so each value is assumed to be at the midpoint of its class [1]. [5] Examiner insight: Use midpoints, not class limits. Saying that exact values are unknown is stronger than just writing “approximate”.

8. (a) One household is 360° ÷ 90 = 4° [1]. Gas = 40 × 4 = 160° and electricity = 25 × 4 = 100° [1]. Oil = 15 × 4 = 60° and wood = 10 × 4 = 40°; the total is 360° [1]. (b) 135° ÷ 27 = 5° per person [1]. Total = 360° ÷ 5° = 72 people [1]. [5] Examiner insight: Add your angles to check they make 360°, and label each sector.

9. (a) 2.5 × 6 = 15 bicycles [1]. (b) Monday 24, Tuesday 21, Wednesday 15, Thursday 30 [1]. Total = 90 bicycles [1]. (c) 33 ÷ 6 = 5.5 symbols (five full and one half) [1]. [4] Examiner insight: Write the key value in your working. Giving the number of symbols instead of bicycles is a common misreading.

10. (a) Totals: January 30 + 10 = 40, February 25 + 15 = 40 [1]. March 20 + 25 = 45, April 15 + 35 = 50 [1]. (b) The cold section runs from 20 to 45 [1]. (c) Hot drinks fell from 30 to 15 [1]. Cold drinks rose from 10 to 35, and the total rose from 40 to 50 [1]. [5] Examiner insight: Quote values from the table. “Hot drinks went down” with no numbers is a weaker answer.

11. (a) Each set has a total of 140, so each mean is 140 ÷ 8 = 17.5 cm [1] [1]. (b) A: (16 + 18) ÷ 2 = 17 cm [1]. B: the middle two values are 17 and 18, so the median is (17 + 18) ÷ 2 = 17.5 cm [1]. (c) Range A = 24 − 12 = 12 and range B = 20 − 15 = 5 [1]. (d) The means are equal (17.5 cm), so the two fertilisers give the same average height [1]. The range for B is much smaller, so B gives more consistent heights, while A gives more variation [1]. (e) Only eight plants were used for each fertiliser, so the sample is small [1]. [8] Examiner insight: A good comparison has one statement on an average and one on spread, with figures and context. “B is better” alone gives no evidence.

12. (a) 90° ÷ 360° = 1/4 of the pupils, so x = 48 ÷ 4 = 12 [1]. Check: 8 + 12 + 15 + 9 + 4 = 48 [1]. (b) The highest frequency is 15, so the modal class is 8 < t ≤ 12 [1]. (c) Midpoints: 2, 6, 10, 14, 18 [1]. Σ(f × midpoint) = 16 + 72 + 150 + 126 + 72 = 436 [1]. Estimated mean = 436 ÷ 48 = 9.08 hours (3 s.f.) [1]. (d) One pupil is 360° ÷ 48 = 7.5° [1]. Angle = 9 × 7.5° = 67.5° [1]. [8] Examiner insight: Part (a) feeds later parts, so check x against the total of 48 before moving on.

Where marks are usually lost

  • Taking the median from a list that has not been put in order.
  • Giving the frequency of the mode instead of the mode.
  • Dividing the sum of f × x by the number of rows instead of by the total frequency.
  • Using class limits rather than midpoints when estimating a mean.
  • Writing a grouped mean as an exact value, with no word such as “estimate”.
  • Pie chart angles that do not add to 360°.
  • Comparing two data sets with an average only and giving no comment on spread.

Next steps

Official syllabus

Cambridge O Level Mathematics (Syllabus D) syllabus for examination in 2025, 2026 and 2027, Cambridge University Press & Assessment, section 9: Statistics, parts 9.1 to 9.4.

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