Cambridge O Level Statistics: Averages: mean, median, mode and modal class (4040) – Revision Notes

These notes condense 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, calculated or estimated from a set of numbers, an ungrouped frequency distribution or a grouped frequency distribution, and the choice of the appropriate average. Full explanations and longer worked examples are in the study guide. For exam-style questions, use the practice questions.

Other links: the Cambridge O Level Statistics hub, the printable checklist, formation of frequency distributions for class limits and boundaries, and the free 10-minute diagnostics.

Definitions

Term	Meaning
Mean	Sum of values ÷ number of values
Median	Middle value of the ordered data; halfway between the two middle values if n is even
Mode	Most frequent value; there may be none, one or several
Modal class	Class with the highest frequency (equal widths) or frequency density (unequal widths)
Midpoint	Halfway between a class’s lower and upper boundaries
Frequency density	Frequency ÷ class width

Formulas

Case	Mean	Median position
List of n values	(Σx) / n	(n + 1)/2 th value
Ungrouped table	(Σfx) / (Σf)	(Σf + 1)/2 th value, found from cumulative frequency
Grouped table	(Σfx) / (Σf), x = midpoint, an estimate	n/2 th value, then interpolate
Grouped median	L + ((n/2 − F) / f) × w	L lower boundary of the median class, F cumulative frequency before it, f its frequency, w its width

Method in steps

From a list

1. Mean: add, then divide by how many values there are.
2. Median: put the values in order first. Count to the middle.
3. Mode: count repeats. Say “no mode” if nothing repeats.

From an ungrouped table

1. Add a column f × x and find Σf and Σfx.
2. Mean = Σfx ÷ Σf.
3. Median: build cumulative frequencies, find which value covers the middle position.
4. Mode: the value (the x entry) with the largest f.

From a grouped table

1. Write the midpoint of each class, then f × midpoint.
2. Estimated mean = Σfx ÷ Σf. Label it as an estimate.
3. Modal class: highest frequency, or highest frequency density if widths differ.
4. Median class: the class containing the n/2 th value. Interpolate inside it from its lower boundary.

Small worked reminders

List with a repeated value

Marks: 34, 41, 29, 41, 38, 50, 41, 36.

Ordered: 29 34 36 38 41 41 41 50
Mean   = 310 ÷ 8 = 38.75
Median = (38 + 41) ÷ 2 = 39.5
Mode   = 41

Grouped estimate

Lengths of 20 leaves in centimetres: 0 ≤ l < 10 has 3, 10 ≤ l < 20 has 9, 20 ≤ l < 30 has 6, 30 ≤ l < 40 has 2.

Midpoints 5, 15, 25, 35
Σfx = 3×5 + 9×15 + 6×25 + 2×35 = 15 + 135 + 150 + 70 = 370
Estimated mean = 370 ÷ 20 = 18.5 cm
Modal class: 10 ≤ l < 20
Median: 10th value; 3 below 10 → 10 + (7/9) × 10 = 17.8 cm (to 3 s.f.)

Ungrouped table with an odd total

Shoe sizes 10, 11, 12, 13, 14 with frequencies 4, 9, 13, 8, 3.

Σf = 37,  Σfx = 40 + 99 + 156 + 104 + 42 = 441
Mean = 441 ÷ 37 = 11.9 (to 3 s.f.)
Cumulative frequencies: 4, 13, 26, 34, 37
Median position: (37 + 1)/2 = 19th value, which lies in the 12 group (14th to 26th)
Median = 12      Mode = 12 (frequency 13)

Effect of an extreme value

Take 6, 7, 7, 8, 9: mean 7.4, median 7, mode 7. Add the value 40. The total becomes 77 and the mean 12.83. The median becomes 7.5 and the mode stays 7. One extreme value changed the mean by more than 5 and the median by half a unit. This is the standard reason for preferring the median when data has outliers.

Class midpoints for whole-number classes

For classes written as 1-5 and 6-10, the midpoint of the first is (1 + 5)/2 = 3, and of the second is (6 + 10)/2 = 8. Using the boundaries 0.5 and 5.5 gives the same midpoint, 3. Either route works.

Working backwards from a mean

The mean of 12 numbers is 6.5, so their total is 12 × 6.5 = 78. If one value of 10 is removed, the total is 68 and the mean of the remaining 11 is 68 ÷ 11 = 6.18 (to 3 s.f.).

Reading the question

• “Calculate” means show the working and give a number, usually to 3 significant figures unless told otherwise.
• “Estimate” signals grouped data and midpoints, so a word such as “approximately” belongs in your answer.
• “State” the mode or modal class needs no working, but a quick underline of the highest frequency is safe.
• “Which average” or “justify” needs a named average and a reason, so one sentence is not enough without the reason.
• Units matter: write minutes, kg or years after the answer when the question gives them.

Must-know distinctions

• Mode is a value, not a frequency. In a table the mode is the entry in the x column.
• Mean uses every value, so one extreme value shifts it. Median and mode barely move.
• Grouped answers are estimates because the midpoint stands in for every value in the class.
• Median needs ordered data. A frequency table is already ordered, so use cumulative frequency.
• Modal class is for grouped data. Mode is for a list or an ungrouped table.
• Choose by the data:
  • categories → mode
  • skewed data or outliers → median
  • symmetrical numerical data, or when totals matter → mean

Choosing and justifying an average

A strong answer has three parts: the average, the figure, and the reason tied to the data. For example: “The median is better here, because one very high value raises the mean above most of the values.”

Sense checks before you move on

• The mean must lie between the smallest and largest values. For grouped data it must lie between the lowest and highest class boundaries.
• The estimated median must lie inside the median class. If it does not, the position or the cumulative frequency is wrong.
• Σf must equal the total number of observations given in the question. If it does not, a frequency was misread.
• If the mean is much higher than the median, expect a few large values. Say so when asked to comment.
• A mean of a whole-number count (people, books, goals) can be a decimal. Do not round it to a whole number unless asked.

Quick self-test

1. Find the mean of 4, 8, 9 and 15.
2. Find the median of 12, 7, 15, 9, 20, 7.
3. Find the mode of 3, 5, 5, 6, 3, 7, 5, 3.
4. A table gives x = 1, 2, 3, 4 with frequencies 5, 8, 4, 3. Find the mean and the median.
5. What is the midpoint of the class 20 ≤ x < 30?
6. Classes 0 ≤ x < 10, 10 ≤ x < 20, 20 ≤ x < 30 have frequencies 5, 10 and 5. Estimate the mean.
7. Six numbers have a mean of 7.5. A seventh number, 12, is added. Find the new mean.
8. A shop wants to know which shoe size to stock most of. Which average should it use?
9. A class 5 ≤ x < 10 has frequency 15 and a class 10 ≤ x < 20 has frequency 24. Which is the modal class?
10. Which position holds the median in 51 ordered values?
11. A data set has an extreme high value. Which average does it affect most?
12. Why is a mean from grouped data called an estimate?

Answers

1. 36 ÷ 4 = 9.
2. Ordered: 7, 7, 9, 12, 15, 20. The middle two are 9 and 12, so the median is 10.5.
3. 3 and 5 each occur three times, so the data is bimodal: 3 and 5.
4. Σf = 20, Σfx = 5 + 16 + 12 + 12 = 45, mean = 2.25. Cumulative frequencies are 5, 13, 17, 20. The 10th and 11th values are both 2, so the median is 2.
6. Midpoints 5, 15, 25: (25 + 150 + 125) ÷ 20 = 15.
7. Total 6 × 7.5 = 45, new total 57, new mean 57 ÷ 7 = 8.14 (to 3 s.f.).
8. The mode.
9. Frequency densities are 15 ÷ 5 = 3 and 24 ÷ 10 = 2.4, so the modal class is 5 ≤ x < 10.
10. The 26th value.
11. The mean.
12. Individual values are not known, so the midpoint is assumed for every value in a class.

Where marks are usually lost

• Finding the median from unordered data.
• Quoting the frequency as the mode.
• Dividing Σfx by the number of classes rather than Σf.
• Using the class limit instead of the midpoint.
• Leaving out the f × x column, so a slip cannot be traced.
• Writing “mean” instead of “estimated mean” for grouped data.
• Using the stated limit instead of the boundary when interpolating for the median.
• Choosing the average without a reason linked to the data.
• Rounding too early. Keep full figures until the last line, then give 3 significant figures unless told otherwise.

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.