EnglandStatisticsSyllabus dot point

What are the features of a Normal distribution, and what proportions lie within one, two and three standard deviations?

Characteristics of a Normal distribution; the notation N(mu, sigma squared); the symmetrical bell shape with equal mean, median and mode; the 68, 95 and 99.7 per cent proportions; conditions for a Normal model.

A focused answer to Edexcel GCSE Statistics (Higher tier) on the Normal distribution, covering its symmetrical bell shape, the notation N(mu, sigma squared), equal mean, median and mode, the proportions within one, two and three standard deviations, and the conditions that make a Normal model suitable.

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
Characteristics of a Normal distribution
The 68, 95, 99.7 proportions
Using the proportions
Conditions for a Normal model

What this dot point is asking

Edexcel Higher tier codes 3p.12 and 3p.13 require you to know and interpret the characteristics of a Normal distribution, to use the notation $N(\mu, \sigma^2)$ , to know it is symmetrical and bell-shaped with equal mean, median and mode, and to know the proportions of data within one, two and three standard deviations of the mean. Normal distribution tables are not expected; the work is based on the symmetry and the $68$ , $95$ , $99.7$ percentages.

Characteristics of a Normal distribution

The notation $X \sim N(\mu, \sigma^2)$ records the mean $\mu$ and the variance $\sigma^2$ (the square of the standard deviation $\sigma$ ). Many natural measurements, such as heights, masses and exam marks, are approximately Normal, which is why the distribution is so widely used.

The 68, 95, 99.7 proportions

The symmetry and shape of the Normal distribution give fixed proportions within standard-deviation bands:

Edexcel summarises this as "approximately $95\%$ within two standard deviations and about $68\%$ (just over two thirds) within one", and that values more than three standard deviations from the mean are very unusual. Because the curve is symmetric, the half outside one band splits equally between the two tails, which is how you find one-sided percentages (for example $16\%$ above $\mu + \sigma$ ).

Using the proportions

Most questions give a mean and standard deviation and ask for a range or a percentage. You work in standard-deviation steps from the mean:

A range for "about $95\%$ " is $\mu \pm 2\sigma$ .
"Almost all" (about $99.7\%$ ) is $\mu \pm 3\sigma$ .
For a one-sided percentage, use the symmetry: the proportion above $\mu + \sigma$ is half of the $32\%$ outside one standard deviation, namely about $16\%$ .

No tables or standardising are needed; the marks come from applying the percentages and the symmetry correctly.

Conditions for a Normal model

A Normal model is suitable when the data is continuous, the distribution is roughly symmetrical and bell-shaped, and the mean, median and mode are approximately equal. Skewed data (such as incomes) is not well modelled by a Normal distribution, so checking these conditions before assuming Normality is part of the analysis.

Using the 68, 95, 99.7 rule

The heights of plants follow a Normal distribution $N(\mu = 50, \sigma^2 = 25)$ . Find (a) the range containing about $95\%$ of heights, (b) the approximate percentage taller than $55$ cm, and (c) whether a plant of $65$ cm height is unusual.

Step 1: Read off the mean and standard deviation from the notation

The notation $N(\mu, \sigma^2)$ gives the mean and the variance. The second parameter is the variance, so we must take its square root to get the standard deviation.

Mean $\mu = 50$ cm, variance $\sigma^2 = 25$ , so standard deviation $\sigma = \sqrt{25} = 5$ cm.

Step 2: Apply the 95 per cent rule to find the central range

About $95\%$ of values in a Normal distribution lie within two standard deviations of the mean. We add and subtract $2\sigma$ from $\mu$ .

\mu \pm 2\sigma = 50 \pm 2 \times 5 = 50 \pm 10, \text{ so between } 40\,\text{cm and } 60\,\text{cm}

Step 3: Use symmetry to find the one-sided percentage above 55 cm

About $68\%$ lie within one standard deviation: $\mu \pm \sigma = 45$ to $55$ cm. This means $100 - 68 = 32\%$ lie outside this band. Because the distribution is symmetric, that $32\%$ is split equally between the two tails, so about $16\%$ are taller than $55$ cm.

Step 4: Identify whether 65 cm is a rare value

$50 + 3 \times 5 = 65$ cm, which is exactly three standard deviations above the mean. The $99.7\%$ rule tells us that only about $0.3\%$ of values fall beyond three standard deviations, so a plant this tall is very unusual.

Final answer: About $95\%$ of plants are between $40$ cm and $60$ cm tall; about $16\%$ are taller than $55$ cm; a height of $65$ cm is very unusual.

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 marksThe masses of apples are Normally distributed with mean

160

g and standard deviation

20

g. (a) Between which two masses do about

95\%

of the apples lie? (b) Approximately what percentage of apples have a mass greater than

180

Show worked answer →

(a) About $95\%$ lie within two standard deviations of the mean: $160 \pm 2 \times 20 = 160 \pm 40$ , so between $120$ g and $200$ g.

(b) $180$ g is one standard deviation above the mean. About $68\%$ lie within one standard deviation ( $140$ g to $180$ g), leaving $32\%$ outside, split equally, so about $16\%$ lie above $180$ g.

Markers reward $120$ g to $200$ g using the $95\%$ rule, and about $16\%$ using the $68\%$ rule with the symmetry of the distribution.

Edexcel 1ST0 20223 marksA teacher says a set of test marks follows a Normal distribution. (a) State two features of a Normal distribution. (b) The marks have mean

50

and standard deviation

10

. State the range of marks within which almost all (about

99.7\%

) of the students lie.

Show worked answer →

(a) Any two features: it is symmetrical about the mean; it is bell-shaped; the mean, median and mode are equal; values more than three standard deviations from the mean are very rare.

(b) Almost all the data lies within three standard deviations: $50 \pm 3 \times 10 = 50 \pm 30$ , so between $20$ and $80$ marks.

Markers reward two correct features and the range $20$ to $80$ using three standard deviations.

Related dot points

Sources & how we know this

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