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How do you sample a population fairly, summarise data, and measure its centre and spread?

Statistical sampling methods, presenting and interpreting data, measures of central tendency (mean, median, mode) and measures of variation (range, interquartile range, variance and standard deviation).

A CCEA AS Further Maths Statistics answer on sampling methods, presenting and interpreting data, the mean, median and mode, and measures of spread including the range, interquartile range, variance and standard deviation.

Generated by Claude Opus 4.812 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
Try this

What this dot point is asking

CCEA wants you to describe and compare sampling methods, present and interpret data, and calculate measures of central tendency (mean, median, mode) and measures of variation (range, interquartile range, variance and standard deviation). Choosing the right summary statistic and interpreting it in context are central.

The answer

Sampling

Measures of central tendency

Measures of variation

The standard deviation measures the typical distance of values from the mean and is in the same units as the data; the interquartile range measures the spread of the middle 50% and resists outliers.

Presenting data

Mean and standard deviation from grouped data

The frequencies of a discrete variable are: value $2$ (freq $3$ ), value $4$ (freq $5$ ), value $6$ (freq $2$ ). Find the mean and standard deviation.

step Find the totals

$n = 3 + 5 + 2 = 10$ . The sum is $\sum fx = 2(3) + 4(5) + 6(2) = 6 + 20 + 12 = 38$ .

step Find the mean

\bar{x} = \frac{\sum fx}{n} = \frac{38}{10} = 3.8.

step Find the variance and standard deviation

$\sum fx^2 = 4(3) + 16(5) + 36(2) = 12 + 80 + 72 = 164$ .

\text{variance} = \frac{\sum fx^2}{n} - \bar{x}^2 = \frac{164}{10} - 3.8^2 = 16.4 - 14.44 = 1.96,

\text{standard deviation} = \sqrt{1.96} = 1.4.

Examples in context

Example 1. A national household survey. Statisticians use stratified sampling by region, age and income so the sample mirrors the country's structure, then report both the average (centre) and the standard deviation (spread) of, say, household spending. The method choice protects against bias.

Example 2. Quality control on a production line. An engineer sampling every 50th item (systematic sampling) tracks the mean and standard deviation of a dimension. A rising standard deviation signals the process is drifting out of control even when the mean still looks fine.

Try this

Q1. Find the mean of $4, 7, 7, 10, 12$ . [1 mark]

Cue. $\frac{40}{5} = 8$ .

Q2. Name a sampling method that guarantees each subgroup is represented in proportion to its size. [1 mark]

Cue. Stratified sampling.

Q3. For data with $\sum x = 50$ , $\sum x^2 = 540$ and $n = 5$ , find the variance. [2 marks]

Cue. $\bar{x} = 10$ ; variance $= \frac{540}{5} - 10^2 = 108 - 100 = 8$ .

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 AS 20216 marksThe times, in seconds, for eight runners are 11, 12, 12, 13, 14, 14, 15, 17. Find the mean and the standard deviation of these times.

Show worked answer →

The mean is the total divided by the number of values:

$\bar{x} = \dfrac{11 + 12 + 12 + 13 + 14 + 14 + 15 + 17}{8} = \dfrac{108}{8} = 13.5\,\text{s}.$

For the standard deviation, find $\sum x^2 = 121 + 144 + 144 + 169 + 196 + 196 + 225 + 289 = 1484$ .

Variance $= \dfrac{\sum x^2}{n} - \bar{x}^2 = \dfrac{1484}{8} - 13.5^2 = 185.5 - 182.25 = 3.25.$

Standard deviation $= \sqrt{3.25} = 1.80\,\text{s}$ (to 3 significant figures).

Markers reward the mean, the use of the $\frac{\sum x^2}{n} - \bar{x}^2$ formula, and the standard deviation.

CCEA AS 20195 marksA researcher wants a sample of 50 students from a school of 1000 students, organised into year groups of different sizes. Name a suitable sampling method, explain how it would be carried out, and give one advantage over simple random sampling.

Show worked answer →

A suitable method is stratified sampling. The population is divided into strata (the year groups), and the sample is taken from each stratum in proportion to its size.

To carry it out, find the proportion each year group is of the school, multiply by 50 to get the number to take from each, then sample randomly within each year group.

One advantage over simple random sampling is that stratified sampling guarantees every year group is represented in proportion to its size, so the sample reflects the structure of the population and reduces sampling bias.

Markers reward naming stratified sampling, the proportional allocation method, and a valid advantage such as guaranteed representative coverage of each stratum.

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Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)