The normal distribution

Statistics · Senior school and first-year statistics · about 10 min

Heights, exam marks and measurement errors often pile up around an average in a symmetric bell shape: the normal distribution. Knowing its shape lets you answer questions like "what fraction of people are taller than 182 cm?" from just the mean and standard deviation.

The key idea

A normal distribution is set by its mean $\mu$ and standard deviation $\sigma$ . To find probabilities, convert a value to a z-score, how many standard deviations it is from the mean:

z = \frac{x - \mu}{\sigma}

then look up $P(Z < z)$ in a table or calculator.

Try it yourself

Change the mean and spread and shade areas to see the probabilities.

The 68-95-99.7 rule

For any normal distribution:

about $68\%$ of values lie within $1\sigma$ of the mean,
about $95\%$ within $2\sigma$ ,
about $99.7\%$ within $3\sigma$ .

It's the quickest sense-check for any answer.

A symmetric bell curve in dark blue over a horizontal axis with ticks at the seven points listed. Shade the band within one standard deviation darkest, the band within two lighter, and the band within three lightest. Above the curve, three horizontal dimension brackets, one for each band, with the widest at the top. — How much of a normal distribution lies within 1, 2 and 3 standard deviations.

Using the table

Standard tables give $P(Z < z)$ , the area to the left of $z$ .

Area to the right: $P(Z > z) = 1 - P(Z < z)$
Between two values: $P(a < Z < b) = P(Z < b) - P(Z < a)$
By symmetry, $P(Z < -z) = P(Z > z)$ .

A symmetric bell curve in dark blue over a horizontal axis. A thin vertical line at a point to the right of centre. The area under the curve to the left of that line is shaded pale blue. The area to the right stays white. — Tables give the area to the left of z.

Worked example: heights of students

Heights are normally distributed with mean $170$ cm and standard deviation $8$ cm. What fraction of students are shorter than $182$ cm? What fraction are between $162$ and $178$ cm?

Show the solution

Shorter than $182$ cm:

z = \frac{182 - 170}{8} = 1.5 \qquad P(Z < 1.5) = 0.9332

So about $93.3\%$ are shorter than $182$ cm.

Between $162$ and $178$ cm: these are $z = -1$ and $z = 1$ .

P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6827

About $68\%$ , exactly as the 68-95-99.7 rule predicts.

Common mistakes

Forgetting to convert to a z-score

Tables are for the standard normal ( $\mu = 0$ , $\sigma = 1$ ). Convert with $z = \frac{x - \mu}{\sigma}$ first.

Reading the table as the area to the right

Most tables give the area to the left. Subtract from $1$ for "greater than".

Using the variance instead of the standard deviation

$\sigma$ is the square root of the variance. If you're given $\sigma^2 = 64$ , then $\sigma = 8$ .

Check your understanding

1. Marks have mean $60$ and standard deviation $10$ . What is the z-score of a mark of $75$ ?

2. About what percentage of a normal distribution lies within two standard deviations of the mean?

3. If $P(Z < 1) = 0.8413$ , what is $P(Z > 1)$ ?

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