The normal distribution — modelling continuous data

By the end you can explain when a normal distribution is a sensible model, write it as $X \sim \mathrm{N}(\mu,\sigma^2)$ , sketch the curve, and use the standard normal table to find simple probabilities.

Symbol	Meaning
$\mu$	"mu": the mean (the centre of the distribution)
$\sigma$	"sigma": the standard deviation (a typical distance from the mean)
$\sigma^2$	the variance, which is the standard deviation squared
$\sim$	"has the distribution", so $X \sim \mathrm{N}(\mu,\sigma^2)$ reads " $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$ "
$Z$	the standard normal variable (defined below)
$\Phi$	"phi": the area to the left of a value under the standard normal curve

Measure the heights of 1,000 adults to the nearest centimetre and draw a histogram. Most people cluster near the middle, fewer are very short or very tall, and the outline is roughly symmetric.

Now imagine measuring to the nearest millimetre, then to the nearest micrometre. The bars get narrower and more numerous, and their outline becomes a smooth curve.

A continuous random variable is one that can take any value in an interval (a whole range of numbers, such as anything between 150 cm and 200 cm). Height, time and mass are continuous. A discrete variable, like the score on a die, takes only separate values such as 0, 1, 2.

For many continuous variables, that limiting curve is the normal curve. It is a model: a convenient, accurate description of how the values are spread.

A continuous random variable $X$ has a normal distribution with mean $\mu$ and variance $\sigma^2$ if its probability curve is the symmetric bell-shaped curve centred at $\mu$ with spread controlled by $\sigma$ . We write $X \sim \mathrm{N}(\mu,\sigma^2).$

The probability that $X$ lies between two values is the area under the curve between them. The total area under the curve is 1.

μ slides the curve. Change the mean and the whole bell moves left or right without changing shape.

σ stretches the curve. A small $\sigma$ gives a tall, narrow bell: values hug the mean. A large $\sigma$ gives a short, wide bell: values are more spread out.

The curve is symmetric about $\mu$ , so it has equal areas on each side. It follows that $\mathrm{P}(X<\mu)=\mathrm{P}(X>\mu)=0.5.$

The mean, median and mode are all $\mu$ : the single peak sits above the mean, and symmetry makes the median the same value.

Because the curve is continuous, there is no bar standing over a single value. The area above one exact point has zero width, so $\mathrm{P}(X=x)=0 \quad \text{for every } x.$

This has a practical consequence: for a continuous normal variable, $\mathrm{P}(X \le x)=\mathrm{P}(X<x).$ Including or excluding the endpoint makes no difference to the area.

So do not expect the height of the curve to be a probability. The height is a probability density; only areas give probabilities. A tall narrow curve can still have a peak height greater than 1, which is impossible for a probability.

Every normal curve is the same bell shape, just moved and stretched. In the next lesson you will learn to convert any normal variable into one standard one (this is called standardising), so a single table serves them all. The standard normal variable is $Z \sim \mathrm{N}(0,1),$ with mean 0 and variance 1 (so standard deviation 1). Values of $Z$ are called $z$ -values. The area to the left of $z$ (the probability of being below $z$ ) is written $\Phi(z)=\mathrm{P}(Z<z).$

The normal table in your formulae booklet lists $\Phi(z)$ for $z \ge 0$ .

How to read it. Find $z$ to one decimal place down the left-hand column (for $z = 1.53$ , find the row $1.5$ ). Then move across to the column for the second decimal place ( $3$ ). The number there is $\Phi(1.53) = 0.9370$ .

Some values you will use again and again:

$z$	0.50	1.00	1.50	1.96	2.00
$\Phi(z)$	0.6915	0.8413	0.9332	0.9750	0.9772

For negative $z$ , use symmetry: the area below $-z$ equals the area above $z$ . $\Phi(-z)=1-\Phi(z)$

For example, $\Phi(-1)=1-0.8413=0.1587$ .

For the area to the right of $z$ , take the left area away from the total of 1: $\mathrm{P}(Z>z)=1-\Phi(z)$

For example, $\mathrm{P}(Z>1)=1-0.8413=0.1587$ .

Let $Z \sim \mathrm{N}(0,1)$ . Find (a) $\mathrm{P}(Z<1.5)$ , (b) $\mathrm{P}(Z>-0.5)$ and (c) $\mathrm{P}(Z>2)$ .

The random variable $X$ has the distribution $\mathrm{N}(20,9)$ . What is the standard deviation of $X$ ?

For $Z \sim \mathrm{N}(0,1)$ , find $\mathrm{P}(Z<1.24)$ , using the normal table. Give your answer to 4 decimal places.

For $Z \sim \mathrm{N}(0,1)$ , find $\mathrm{P}(Z>-1)$ . Give your answer to 4 decimal places.

The normal distribution — modelling continuous data

From a histogram to a smooth curve

A probability at one exact value is zero

The standard normal and its table