Pearson Edexcel Edexcel 1ST0 2020-style practice: The table shows the times, in minutes, taken by 60 runners. 0 < t 20: frequency 12; 20 < t 30: frequency 18; 30 < t 50: frequency 20; 50 < t 80: frequency 10. Calculate the frequency density for each class so a histogram can be drawn.

Frequency density = frequencyclass width. 0 < t 20: width 20, density (12)/(20) = 0.6. 20 < t 30: width 10, density (18)/(10) = 1.8. 30 < t 50: width 20, density (20)/(20) = 1.0. 50 < t 80: width 30, density (10)/(30) ≈ 0.33. Markers reward dividing each frequency by the correct class width and four correct densities.

Pearson Edexcel Edexcel 1ST0 2022-style practice: A histogram has a bar for the class 40 65.

(a) Frequency = frequency density × class width. 40 65, take the part of the 60 to 70 class above 65: width 5, so frequency = 4 × 5 = 20. Markers reward frequency = density × width for both classes, and using the proportion of the class above 65 to estimate 20 values.

EnglandStatisticsSyllabus dot point

How do histograms display continuous data, and why is area the key idea?

Histograms for continuous data with equal and unequal class widths; frequency density; using area to represent frequency; estimating frequencies within a class; correct use of class boundaries.

A focused answer to Edexcel GCSE Statistics on histograms, covering continuous data and class boundaries, equal and unequal class widths, frequency density, why area represents frequency, and estimating frequencies within a class interval at Higher tier.

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
Continuous data and class boundaries
Why area represents frequency
Frequency density
Histograms versus bar charts
Estimating frequencies within a class

What this dot point is asking

Edexcel code 2a.03 includes histograms, and code 2a.06 stresses correct use of class boundaries. You must represent continuous grouped data with a histogram, calculate and use frequency density for unequal class widths (Higher tier), understand why area represents frequency, and estimate frequencies within part of a class. At Foundation tier histograms with equal class widths are used (and the vertical axis may be labelled frequency); frequency density is Higher tier only.

Continuous data and class boundaries

A histogram is used for continuous grouped data. Because the data is continuous, the bars touch (no gaps), and the horizontal axis is a continuous number line. Getting the class boundaries right is essential: a class written as $20 < t \le 30$ runs from $20$ to $30$ and has width $10$ . For data rounded to whole units, boundaries may sit at the half unit (a class " $10$ to $19$ " of ages has boundaries $9.5$ to $19.5$ ). Errors in boundaries are a common source of graphical misrepresentation.

Why area represents frequency

The central idea of a histogram is that frequency is proportional to area, not to height. This matters when class widths differ: a wide class can hold many values without being tall. If you plotted raw frequency as height with unequal widths, the wide classes would look misleadingly large. Using area keeps the comparison fair.

Frequency density

To draw a histogram you compute the frequency density of each class and plot it as the bar height. To read a histogram you multiply the density by the width to recover the frequency. This pair of operations is the single most examined skill in this topic.

A useful shortcut for questions that give you one known class is to find the frequency per unit of area on the graph. If a class with a known frequency covers a known number of small squares on the grid, then each square represents the same number of items everywhere on the histogram, so you can read off any other class by counting its squares. Examiners often supply one frequency precisely so you can unlock the rest of the diagram this way.

Histograms versus bar charts

Edexcel expects you to know when a histogram is appropriate and to contrast it with a bar chart. A histogram is for continuous grouped data: the horizontal axis is a continuous measurement scale, the bars touch, and area represents frequency. A bar chart is for discrete or categorical data: the bars have gaps, equal widths, and height (not area) shows frequency. Choosing the wrong diagram, or leaving gaps in a histogram, is treated as a misrepresentation. When a question asks you to "justify" your choice of diagram, name the data type (continuous, grouped) as the reason a histogram is correct.

Estimating frequencies within a class

Because area represents frequency, you can estimate how many values fall in part of a class by taking the matching fraction of that class's area. For the class $60 < x \le 70$ with frequency $40$ , the number with $x > 65$ is the part from $65$ to $70$ , which is half the class width, giving an estimate of $\frac{1}{2} \times 40 = 20$ . This assumes values are spread evenly across the class, which is the standard GCSE assumption.

Drawing and reading a histogram

A data set has three classes with unequal widths: $0 < x \le 10$ (frequency $20$ ), $10 < x \le 25$ (frequency $45$ ) and $25 < x \le 30$ (frequency $15$ ). Draw the histogram and estimate how many values fall in the range $x > 20$ .

Step 1: Find each class width from the boundaries

With unequal class widths we cannot use height to represent frequency directly, or a wider class would look misleadingly large. We need to work out how wide each class is.

Class widths: $10 - 0 = 10$ , $25 - 10 = 15$ , $30 - 25 = 5$ .

Step 2: Calculate the frequency density for each class

Frequency density is the height we plot on the vertical axis, so that area (height times width) equals frequency.

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

Densities: $\dfrac{20}{10} = 2$ , $\dfrac{45}{15} = 3$ , $\dfrac{15}{5} = 3$ . Plot these heights as the bars on the histogram.

Step 3: Verify by reading a frequency back from the histogram

As a check, recover the frequency of the $10 < x \le 25$ bar: density $3$ times width $15$ gives $3 \times 15 = 45$ , which matches the table.

Step 4: Estimate the number with $x > 20$ using proportion of the class area

Within the $10 < x \le 25$ class, the part from $20$ to $25$ has width $5$ . Because values are assumed to be spread evenly across the class, we use the proportion of the area.

\text{estimated frequency} = \text{density} \times \text{width of part} = 3 \times 5 = 15

Final answer: The histogram bars have heights (frequency densities) $2$ , $3$ and $3$ . About $15$ values are estimated to have $x > 20$ .

Exam-style practice questions

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

Edexcel 1ST0 20204 marksThe table shows the times, in minutes, taken by

60

runners.

0 < t \le 20

: frequency

12

;

20 < t \le 30

: frequency

18

;

30 < t \le 50

: frequency

20

;

50 < t \le 80

: frequency

10

. Calculate the frequency density for each class so a histogram can be drawn.

Show worked answer →

Frequency density $= \frac{\text{frequency}}{\text{class width}}$ .

$0 < t \le 20$ : width $20$ , density $\frac{12}{20} = 0.6$ .

$20 < t \le 30$ : width $10$ , density $\frac{18}{10} = 1.8$ .

$30 < t \le 50$ : width $20$ , density $\frac{20}{20} = 1.0$ .

$50 < t \le 80$ : width $30$ , density $\frac{10}{30} \approx 0.33$ .

Markers reward dividing each frequency by the correct class width and four correct densities.

Edexcel 1ST0 20223 marksA histogram has a bar for the class

40 < x \le 60

with frequency density

2.5

, and a bar for

60 < x \le 70

with frequency density

4

. (a) Find the frequency of each class. (b) Estimate the number of values with

x > 65

Show worked answer →

(a) Frequency $=$ frequency density $\times$ class width.

$40 < x \le 60$ : $2.5 \times 20 = 50$ . $60 < x \le 70$ : $4 \times 10 = 40$ .

(b) For $x > 65$ , take the part of the $60$ to $70$ class above $65$ : width $5$ , so frequency $= 4 \times 5 = 20$ .

Markers reward frequency $=$ density $\times$ width for both classes, and using the proportion of the class above $65$ to estimate $20$ values.

Related dot points

Sources & how we know this

Pearson Edexcel GCSE (9-1) Statistics (1ST0) specification — Pearson Edexcel (2017)