What is displaying data?

A histogram displays grouped continuous data, where the area of each bar (not its height) is proportional to the frequency, so the vertical axis is frequency density = frequencyclass width. A box plot shows the minimum, lower quartile, median, upper quartile and maximum, giving a five-number summary that makes comparing two data sets easy.

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How do we summarise, display and interpret a set of data?

Measures of central tendency and spread, calculating the mean, median, mode, range, interquartile range, variance and standard deviation, displaying data with histograms and box plots, and identifying outliers.

A CCEA A-Level Mathematics answer on measures of central tendency and spread, calculating the mean, median, mode, range, interquartile range, variance and standard deviation, presenting data with histograms and box plots, and identifying outliers.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

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

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What this dot point is asking
The answer
Examples in context
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What this dot point is asking

CCEA wants you to summarise data using measures of central tendency and spread, calculate the mean, median, mode, range, interquartile range, variance and standard deviation, present data in histograms and box plots, and identify outliers using a stated rule. These descriptive tools turn raw data into interpretable summaries and lead into the probability distributions later in the unit.

The answer

Measures of central tendency

Measures of spread

The interquartile range, like the median, resists outliers, whereas the range and standard deviation are pulled by extreme values.

Displaying data

A histogram displays grouped continuous data, where the area of each bar (not its height) is proportional to the frequency, so the vertical axis is frequency density $= \dfrac{\text{frequency}}{\text{class width}}$ . A box plot shows the minimum, lower quartile, median, upper quartile and maximum, giving a five-number summary that makes comparing two data sets easy.

Outliers

Worked example: frequency density in a histogram

Find a frequency from a histogram bar

In a histogram, a class of width $5$ is drawn with a frequency density of $8$ . Find the frequency in that class, and the bar height if another class of width $10$ has the same frequency.

step Use the frequency-density definition

\text{frequency} = \text{frequency density} \times \text{class width} = 8 \times 5 = 40.

step Find the second bar's density

The second class has the same frequency $40$ but width $10$ , so

\text{frequency density} = \frac{40}{10} = 4.

step Interpret

The wider class is half as tall even though it holds the same number of items, because in a histogram area, not height, represents frequency. This is why unequal class widths must use frequency density.

Examples in context

Example 1. Comparing two classes. Two classes with the same mean test score can differ in standard deviation: a small one means consistent results, a large one means a wide spread. Reporting both an average and a measure of spread is essential to describe data fairly.

Example 2. Skewed incomes. Income data has a few very high values, so the mean is dragged up while the median better reflects a typical person. Choosing the median over the mean for skewed data is the standard interpretive judgement.

Try this

Q1. Find the mean of $4, 7, 9, 12, 18$ . [1 mark]

Cue. $\frac{50}{5} = 10$ .

Q2. A data set has $Q_1 = 12$ and $Q_3 = 20$ . Find the interquartile range. [1 mark]

Cue. $20 - 12 = 8$ .

Q3. A histogram bar has class width $4$ and frequency density $6$ . Find the frequency. [2 marks]

Cue. $6 \times 4 = 24$ .

Exam-style practice questions

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

CCEA 20206 marksThe times, in seconds, for eight runners are 12.1, 12.4, 12.8, 13.0, 13.1, 13.5, 14.2 and 18.0. Find the median and the interquartile range, and determine whether 18.0 is an outlier using the rule that an outlier lies more than 1.5 times the interquartile range above the upper quartile.

Show worked answer →

With eight ordered values, the median is the mean of the 4th and 5th: $\frac{13.0 + 13.1}{2} = 13.05\,\text{s}$ .

The lower quartile is the mean of the 2nd and 3rd: $\frac{12.4 + 12.8}{2} = 12.6$ . The upper quartile is the mean of the 6th and 7th: $\frac{13.5 + 14.2}{2} = 13.85$ .

The interquartile range is $13.85 - 12.6 = 1.25\,\text{s}$ .

The outlier boundary is $Q_3 + 1.5 \times \text{IQR} = 13.85 + 1.5(1.25) = 13.85 + 1.875 = 15.725$ .

Since $18.0 > 15.725$ , the value $18.0$ is an outlier.

Markers reward the median, both quartiles, the interquartile range, the boundary calculation, and the outlier conclusion.

CCEA 20185 marksFor a sample of values,

\sum x = 240

\sum x^2 = 6100

and

n = 10

. Find the mean and the standard deviation.

Show worked answer →

The mean is $\bar{x} = \frac{\sum x}{n} = \frac{240}{10} = 24$ .

The variance is $\frac{\sum x^2}{n} - \bar{x}^2 = \frac{6100}{10} - 24^2 = 610 - 576 = 34$ .

The standard deviation is $\sqrt{34} = 5.83$ (to three significant figures).

Markers reward the mean, the variance formula, the substitution, and the square root for the standard deviation.

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Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)