EnglandStatisticsSyllabus dot point

How do you measure spread with standard deviation and compare values from different data sets?

Standard deviation for a set of values and for grouped data; using the mean and standard deviation to compare data sets; standardising values with the standardised score to compare across distributions.

A focused answer to Edexcel GCSE Statistics (Higher tier) on standard deviation and standardised scores, covering the standard deviation formulae for a set of values and grouped data, comparing data sets with the mean and standard deviation, and standardising values to compare across distributions.

Generated by Claude Opus 4.810 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
What standard deviation measures
The standard deviation formulae
Comparing data sets with mean and standard deviation
Standardised scores

What this dot point is asking

Edexcel Higher tier codes 2c.01 and 2c.06 require you to calculate the standard deviation for a set of values and for grouped data, to use the mean and standard deviation to compare data sets, and to standardise values so they can be compared across different distributions. The standard deviation formulae are given on the formulae sheet, but the standardised score $\frac{x - \mu}{\sigma}$ is not given, so you must memorise it.

What standard deviation measures

Unlike the range and IQR, the standard deviation uses every value, which makes it the natural partner of the mean. Edexcel pairs mean with standard deviation, and median with IQR; mixing them is marked wrong.

The standard deviation formulae

The formulae sheet gives the formulae for a set of values; you apply them, and adapt them to grouped data using frequencies.

The second form ("mean of the squares minus the square of the mean") is usually quicker in the exam. For grouped data, use the class midpoints exactly as in the grouped mean.

Comparing data sets with mean and standard deviation

To compare two roughly symmetric data sets, compare the means (which is higher on average) and the standard deviations (which is more spread out, hence less consistent). For example, two machines fill bottles to a mean of $500$ ml; the one with the smaller standard deviation is more reliable because its fills vary less. Always state both comparisons in context.

Standardised scores

This lets you compare values that come from different distributions, such as marks in two tests with different means and spreads. A positive score is above the mean, a negative score below; the larger the standardised score, the better the value relative to its own distribution. This is the key tool for "in which did the student do better relative to the class?" questions, and the formula is not on the sheet, so learn it.

Comparing performances with standardised scores

Write the standardised score formula

The standardised score is $z = \frac{x - \mu}{\sigma}$ , the number of standard deviations from the mean.

Standardise the first value

A swimmer's $100$ m time is in a race where $\mu = 60$ s and $\sigma = 4$ s; her time is $54$ s, so $z = \frac{54 - 60}{4} = \frac{-6}{4} = -1.5$ . (A faster time is below the mean, hence negative, which is good here.)

Standardise the second value

In a different race $\mu = 58$ s, $\sigma = 5$ s, and she swims $51$ s, so $z = \frac{51 - 58}{5} = \frac{-7}{5} = -1.4$ .

Compare and conclude

Her first swim is $1.5$ standard deviations below the mean versus $1.4$ in the second, so relative to the field her first race was marginally better.

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 20214 marksFive students score the following marks in a test:

4, 6, 8, 10, 12

. Calculate the mean and the standard deviation of the marks. (The standard deviation formula is given on the formulae sheet.)

Show worked answer →

Mean $\bar{x} = \frac{4 + 6 + 8 + 10 + 12}{5} = \frac{40}{5} = 8$ .

Using $\sigma = \sqrt{\frac{\sum x^2}{n} - \bar{x}^2}$ : $\sum x^2 = 16 + 36 + 64 + 100 + 144 = 360$ .

$\sigma = \sqrt{\frac{360}{5} - 8^2} = \sqrt{72 - 64} = \sqrt{8} \approx 2.83$ .

Markers reward the mean, $\sum x^2$ , substitution into the formula, and $\sigma \approx 2.83$ marks.

Edexcel 1ST0 20224 marksIn a maths test the mean is

60

with standard deviation

12

. In an English test the mean is

50

with standard deviation

8

. Priya scores

72

in maths and

66

in English. Use standardised scores to decide in which subject she did better relative to her class.

Show worked answer →

Standardised score $= \frac{x - \mu}{\sigma}$ .

Maths: $\frac{72 - 60}{12} = \frac{12}{12} = 1$ . English: $\frac{66 - 50}{8} = \frac{16}{8} = 2$ .

Priya's English score is $2$ standard deviations above the mean, compared with $1$ in maths, so she performed better in English relative to her class.

Markers reward both standardised scores and the conclusion that the higher standardised score (English) means the better relative performance.

Related dot points

Sources & how we know this

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