AQA AQA 2020-style practice: A grouped table of times (minutes) has classes 0 t<10 (frequency 20), 10 t<25 (frequency 45) and 25 t<30 (frequency 15). (a) Calculate the frequency density for each class. (b) Explain how the frequency for the middle class is shown on the histogram.

(a) Widths are 10, 15, 5. Frequency densities: (20)/(10) = 2; (45)/(15) = 3; (15)/(5) = 3. (b) The frequency of the middle class is the area of its bar: frequency density 3 times class width 15 gives 3 × 15 = 45, which is the frequency. Markers reward all three frequency densities and the key idea that frequency equals area (frequency density times width), not the bar height.

AQA AQA 2022-style practice: A histogram has a bar over the class 40 x<60 with frequency density 1.5, and another over 60 x<70 with frequency density 4. Calculate the total frequency represented by these two classes.

First class: width 20, frequency = 1.5 × 20 = 30. Second class: width 10, frequency = 4 × 10 = 40. Total = 30 + 40 = 70. Markers reward frequency = frequency density × width for each bar, then adding to give 70.

EnglandStatisticsSyllabus dot point

How do you display continuous data with unequal class widths?

Histograms with equal and unequal class widths, frequency density, frequency polygons and population pyramids.

A focused answer to AQA GCSE Statistics on histograms, covering equal and unequal class widths, frequency density, reading frequencies as areas, frequency polygons and population pyramids.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Histograms and frequency density
Frequency polygons
Population pyramids

What this dot point is asking

AQA wants you to draw and interpret histograms for continuous data with equal and unequal class widths, calculate and use frequency density, read frequencies as areas, and use frequency polygons and population pyramids. The single most-tested idea here is that on a histogram area, not height, represents frequency.

Histograms and frequency density

The reason for frequency density is fairness. If you plotted raw frequency as the bar height, a wide class would look taller simply because it covers more of the axis, exaggerating it. Dividing by the class width turns frequency into a density (frequency per unit), so a wide, sparsely populated class and a narrow, busy class are shown on the same footing. With equal class widths the heights are already comparable, so a plain frequency axis is fine; frequency density is essential only when widths differ.

Frequency density and reading frequency as area

Find each class width

A grouped table has $10 \le t < 30$ with frequency $40$ , and $30 \le t < 35$ with frequency $25$ . The widths are $30 - 10 = 20$ and $35 - 30 = 5$ .

Calculate the frequency densities

First class: $\frac{40}{20} = 2$ . Second class: $\frac{25}{5} = 5$ . So the narrow class is drawn much taller despite having fewer values.

Read a frequency back as an area

For a bar of frequency density $3$ over a class width of $5$ , the frequency is the area: $3 \times 5 = 15$ .

Use the area idea to find a partial frequency

If a question asks how many lie in part of a class, take that fraction of the bar's area: half of the $30 \le t < 35$ bar (so $30 \le t < 32.5$ ) represents $\frac{1}{2} \times 25 = 12.5$ values.

Frequency polygons

A frequency polygon is drawn by plotting the frequency (or frequency density) against the midpoint of each class and joining the points with straight lines. Because it is just a line, two frequency polygons can be drawn on the same axes to compare the shapes of two distributions directly, which is harder to do with overlapping histograms.

Population pyramids

Reading a population pyramid is about shape: comparing the widths of bars at different ages tells you about birth rates, life expectancy and the balance between the sexes, which is why they are used to compare countries or to track one country over time.

A frequent higher-tariff histogram question gives a part-complete histogram and a part-complete frequency table and asks you to fill in both. The method is to use frequency $=$ frequency density $\times$ class width wherever you can read a bar, and frequency density $=$ frequency $\div$ class width wherever the table gives a frequency, working back and forth until everything is filled. Another common task is to find the number of values in a part of a class, which you do by taking the matching fraction of that bar's area, assuming the values are spread evenly across the class. Keeping the "area equals frequency" rule at the front of your mind turns all of these into a single idea applied repeatedly.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20205 marksA grouped table of times (minutes) has classes

0\le t<10

(frequency

20

10\le t<25

(frequency

45

) and

25\le t<30

(frequency

15

). (a) Calculate the frequency density for each class. (b) Explain how the frequency for the middle class is shown on the histogram.

Show worked answer →

(a) Widths are $10$ , $15$ , $5$ . Frequency densities: $\frac{20}{10} = 2$ ; $\frac{45}{15} = 3$ ; $\frac{15}{5} = 3$ .

(b) The frequency of the middle class is the area of its bar: frequency density $3$ times class width $15$ gives $3 \times 15 = 45$ , which is the frequency.

Markers reward all three frequency densities and the key idea that frequency equals area (frequency density times width), not the bar height.

AQA 20223 marksA histogram has a bar over the class

40\le x<60

with frequency density

1.5

, and another over

60\le x<70

with frequency density

4

. Calculate the total frequency represented by these two classes.

Show worked answer →

First class: width $20$ , frequency $= 1.5 \times 20 = 30$ .

Second class: width $10$ , frequency $= 4 \times 10 = 40$ .

Total $= 30 + 40 = 70$ .

Markers reward frequency $=$ frequency density $\times$ width for each bar, then adding to give $70$ .

Related dot points

Sources & how we know this

AQA GCSE Statistics (8382) specification — AQA (2017)