Rukasha Maths

GCSE maths Beta

Histograms and frequency density · Higher notes

Higher · Statistics · AQA references S3

In a histogram, frequency is represented by the area of a bar, not simply its height. This matters when the class intervals have unequal widths. Dividing each frequency by its class width gives frequency density, which is plotted as the height when the vertical axis is labelled that way. Multiplying density by width recovers frequency. Continuous classes meet at their boundaries, unlike separated categorical bars. Read both axes carefully, and remember that grouped data cannot reveal the individual values inside a class.

Learning objectives

  • Calculate frequency density for unequal class widths.
  • Recover frequency from a histogram bar area.

Key ideas

Use the class boundaries to find each width.

class width=upper boundary−lower boundary\text{class width}=\text{upper boundary}-\text{lower boundary}

Divide frequency by class width to calculate the bar height.

frequency density=frequencyclass width\text{frequency density}=\frac{\text{frequency}}{\text{class width}}

Multiply density by width to recover frequency.

frequency=density×class width\text{frequency}=\text{density}\times\text{class width}

Histogram bar area represents frequency. Adjoining bars can have unequal widths.

Worked examples

Example 1

Times are grouped in intervals from zero to ten, ten to thirty and thirty to fifty, with frequencies 8, 30 and 12. Find their densities.

0≤t<10,10≤t<30,30≤t<500\le t<10,\quad10\le t<30,\quad30\le t<50
class widths:10, 20, 20\text{class widths}:10,\ 20,\ 20
densities:810, 3020, 1220\text{densities}:\frac8{10},\ \frac{30}{20},\ \frac{12}{20}

Answer

0.8,1.5,0.60.8,\quad1.5,\quad0.6

Example 2

A class has width eight and frequency twenty-four. Find its density.

density=248\text{density}=\frac{24}{8}

Answer

33

Question styles to prepare for

Original practice guidance aligned to the specification, not a count of past-paper appearances or a prediction. The AO labels below describe the intended practice focus, not official AQA mark allocations.

AO1 · Methods

Calculate frequency density or complete a histogram with unequal class widths.

AO2 · Reasoning

Explain why frequency is represented by bar area rather than height alone.

AO3 · Problem solving

Estimate a frequency within part of a class, stating the distribution assumption.

When a task asks you to explain, include a mathematical reason. When solving a problem, show how the information becomes a calculation, equation or diagram, then interpret your result.

Avoid this mistake

Bar height is density, not frequency. Equal heights can represent different frequencies when class widths differ.

Try it yourself

Attempt each question before opening its answer.

Check 1: A class has width 8 and frequency 24. Find its density.

Answer: 3

Divide frequency by class width: 24/8.

Check 2: A class has width 15 and density 1.2. Find its frequency.

Answer: 18

Multiply width by density: 15 × 1.2.

Explain it in your own words

  • Explain why unequal-width histogram bars must use density.
  • Write the formula connecting frequency, density and width.
  • Describe what information grouping loses about individual values.

Keep practising

Download the compact topic guide (PDF) · Practise this topic in Blitz · All higher notes

Specification source

AQA Statistics specification · Assessment objectives. Checked 16 October 2026. Topic references identify related content; these notes do not cover every specification statement.