Cambridge O Level Mathematics: Histograms and Frequency Density with Unequal Class Widths (4024)

This guide covers section 9.7, Histograms, in the Statistics topic of the Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027. The syllabus is not tiered, so every candidate meets this content. It teaches both outcomes: drawing and interpreting histograms, and calculating with frequency density. Afterwards, check what you have learned with the revision notes and the practice questions. Other topics are listed on the course hub, and you can tick off what you have covered on the printable checklist.

What this unit covers

Syllabus	What you must be able to do
9.7 (1)	Draw and interpret histograms. The vertical axis is labelled “Frequency density”.
9.7 (2)	Calculate with frequency density, where frequency density = frequency ÷ class width.

In practice that means four skills: turning a frequency table into bar heights, turning bar heights back into frequencies, finding a frequency or a height when the vertical scale is missing, and reading the finished diagram sensibly.

Why a histogram needs frequency density

A bar chart for grouped data is easy when every class has the same width: the tallest bar is the most common class. Things go wrong when the widths differ. A class that is twice as wide will usually catch about twice as many values, so a bar drawn to frequency would look far too important.

A histogram fixes this by making the area of each bar, not its height, show the frequency. Then a wide class and a narrow class are compared fairly. The height has to be adjusted to make that work, and the adjusted height is the frequency density.

The key relationships

frequency density = frequency ÷ class width
frequency = frequency density × class width
frequency = area of the bar

All three say the same thing. If you know any two of frequency, class width and frequency density, you can find the third.

Frequency density tells you how many values there are per one unit of the horizontal axis. A density of 3 in a class measured in minutes means 3 values per minute of the class.

Class boundaries and class width

Histograms are used for continuous data such as time, mass or length. The classes are written as inequalities, for example 10 ≤ t < 20. The class width is the upper boundary minus the lower boundary, so this class has width 20 − 10 = 10.

Check each width separately. Do not assume that a gap of 5 between boundaries repeats down the table. Classes such as 0 ≤ t < 2, 2 ≤ t < 5 and 5 ≤ t < 10 have widths 2, 3 and 5.

Skill 1: from a table to bar heights

Method: work out each class width, divide each frequency by its width, then plot.

Worked example 1

The lengths of 71 phone calls, in minutes, are summarised below. Find the frequency density of each class.

Length t (minutes)	Frequency	Class width	Frequency density
0 ≤ t < 2	6	2	6 ÷ 2 = 3
2 ≤ t < 5	15	3	15 ÷ 3 = 5
5 ≤ t < 10	20	5	20 ÷ 5 = 4
10 ≤ t < 20	18	10	18 ÷ 10 = 1.8
20 ≤ t < 40	12	20	12 ÷ 20 = 0.6

The tallest bar is the class 2 ≤ t < 5, even though 20 calls is the largest frequency and sits in the next class. That is the point of the method: the bar height now describes how crowded each minute of the class is.

Drawing the histogram

1. Draw the horizontal axis as a continuous scale for the variable (here t in minutes), marking the class boundaries 0, 2, 5, 10, 20, 40.
2. Label the vertical axis Frequency density. This is the label the syllabus asks for.
3. Choose a vertical scale that fits the largest density (5 here), using equal steps.
4. Draw each bar from its lower boundary to its upper boundary, at the height you calculated.
5. Leave no gaps between bars, because the classes join up.

For worked example 1 the bars are:

Bar from	to	height
0	2	3
2	5	5
5	10	4
10	20	1.8
20	40	0.6

Each bar’s area is its frequency: the last bar is 20 wide and 0.6 tall, and 20 × 0.6 = 12.

Skill 2: from bar heights back to a table

Multiply the height of each bar by its width.

Worked example 2

A clinic records how long patients wait, in minutes. The histogram has these bars. Find the frequency of each class and the total.

Waiting time w	Width	Frequency density	Frequency
0 ≤ w < 10	10	0.8	10 × 0.8 = 8
10 ≤ w < 15	5	2.4	5 × 2.4 = 12
15 ≤ w < 20	5	3.2	5 × 3.2 = 16
20 ≤ w < 30	10	1.5	10 × 1.5 = 15
30 ≤ w < 50	20	0.35	20 × 0.35 = 7

Total frequency = 8 + 12 + 16 + 15 + 7 = 58, so 58 patients were recorded.

Frequencies of whole people must come out as whole numbers. If you get 7.4 patients, recheck the width or the height you read from the scale.

Skill 3: reading the diagram

Once you can move in both directions, interpretation is a matter of choosing the right calculation.

• How many in a class? Height × width.
• How many in several classes? Work out each frequency, then add.
• Which class is the most crowded? The tallest bar, because height is density.
• Which class has the most values? The largest area, which may not be the tallest bar.

Worked example 3

Using the clinic data in worked example 2:

(a) How many patients waited at least 15 minutes?

Classes from 15 upwards: 16 + 15 + 7 = 38

38 patients waited at least 15 minutes.

(b) Estimate how many waited between 20 and 25 minutes.

The class 20 ≤ w < 30 holds 15 patients over a width of 10. The part from 20 to 25 is 5 wide.

5 × 1.5 = 7.5

About 7 or 8 patients. This only works if you assume the patients are spread evenly across the class, which the histogram cannot confirm. Say so when you give an estimate. A fraction of a person is fine for an estimate, so leave 7.5 or round sensibly and state the assumption.

Skill 4: histograms with no vertical scale

Sometimes the vertical axis has been removed, or only one frequency is given. Because frequency is proportional to area, you can find a scale from one known bar and use it for the rest. Measure in centimetres on the diagram, or count squares.

Worked example 4

A histogram of parcel masses has a horizontal scale of 2 cm for every 10 kg. The bar for 10 ≤ m < 20 is 3 cm tall and represents 30 parcels.

(a) Find the number of parcels in the bar for 20 ≤ m < 35, which is 2.4 cm tall.

Bar 10 ≤ m < 20: width 2 cm, height 3 cm, area 6 cm²
6 cm² represents 30 parcels, so 1 cm² represents 30 ÷ 6 = 5 parcels

Bar 20 ≤ m < 35: width (15 ÷ 10) × 2 = 3 cm, height 2.4 cm
Area = 3 × 2.4 = 7.2 cm²
Parcels = 7.2 × 5 = 36

(b) The bar for 35 ≤ m < 40 represents 18 parcels. How tall is it?

Width = (5 ÷ 10) × 2 = 1 cm
Area needed = 18 ÷ 5 = 3.6 cm²
Height = 3.6 ÷ 1 = 3.6 cm

The bar is 3.6 cm tall. This is the same idea as frequency density with a different unit: here the “density” is 5 parcels for every square centimetre.

Skill 5: completing tables and histograms with missing values

Use the information you are given to fill each gap in a sensible order.

Worked example 5

The ages of 60 members of a club are in the table. The frequency of the last class is not given, but the histogram shows its bar has height 0.7.

Age a (years)	Frequency	Width	Frequency density
0 ≤ a < 10	9	10	0.9
10 ≤ a < 18	16	8	2
18 ≤ a < 30	x	12	?
30 ≤ a < 50	y	20	0.7

Step 1: find y from the bar. Frequency = 0.7 × 20 = 14, so y = 14.

Step 2: find x from the total. 9 + 16 + x + 14 = 60, so x = 60 − 39 = 21.

Step 3: find the missing height. Frequency density = 21 ÷ 12 = 1.75.

Always finish by checking that the frequencies add to the stated total.

Unknown heights from a ratio

Sometimes the information is a comparison. Suppose the bar for 10 ≤ x < 20 is twice as tall as the bar for 20 ≤ x < 40. Call the lower height k, so the other is 2k. The frequencies are 2k × 10 = 20k and k × 20 = 20k. Because the widths differ, equal-looking heights do not mean equal frequencies, and writing both in terms of k keeps that clear.

Common errors

• Plotting frequency instead of frequency density. This is the standard slip. With unequal widths it gives bars of the wrong size.
• Using the wrong width. Subtract the boundaries for each class. Do not reuse the previous class’s width.
• Reading the tallest bar as the biggest class. Compare areas, or calculate the frequencies.
• Leaving gaps between bars. Classes touch on a continuous scale.
• Forgetting the axis label. Write “Frequency density” on the vertical axis, with the variable and units on the horizontal.
• Dividing when you should multiply. To go from a bar to a frequency, multiply by the width. Check by asking whether a wide bar should hold more values than a narrow one of the same height.
• Ignoring the assumption in an estimate. Counting part of a class needs the even-spread assumption.

Next steps

Work through the practice questions, then use the revision notes for quick recall. If you are not sure where your gaps are, try the free 10-minute diagnostics.

Official syllabus

Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027, version 2, published by Cambridge University Press & Assessment (Cambridge Assessment International Education). The relevant section is 9.7 Histograms.