Standardising — from any normal curve to the table

By the end of this lesson

By the end you can standardise any normal variable with $z=\dfrac{x-\mu}{\sigma}$ , sketch the relevant area, and find $\mathrm{P}(X<x)$ , $\mathrm{P}(X>x)$ and $\mathrm{P}(a<X<b)$ using the standard normal table.

Key to the symbols

The main symbols used in this lesson:

Symbol	Meaning
$X \sim \mathrm{N}(\mu,\sigma^2)$	$X$ is normal with mean $\mu$ and variance $\sigma^2$
$\mu$	the mean of $X$
$\sigma$	the standard deviation of $X$ (the square root of the variance)
$x$	one particular value of $X$ , such as $38$
$Z \sim \mathrm{N}(0,1)$	the standard normal variable: mean 0, standard deviation 1
$z$	a value of $Z$ (a " $z$ -value")
$\Phi(z)$	$\mathrm{P}(Z<z)$ , the area to the left of $z$ , read from the table
$z_a$ , $z_b$ (or $z_1$ , $z_2$ )	the $z$ -values for two boundaries $a$ and $b$

One table, infinitely many normal curves

Heights of adults, waiting times and measurement errors all follow normal curves, but with different means and spreads. Printing a table for every possible pair $(\mu,\sigma)$ is impossible.

The trick is that all normal curves are the same shape once you measure horizontally in units of $\sigma$ from the mean. That single standardised curve is the standard normal $Z \sim \mathrm{N}(0,1)$ .

Standardising does not change areas. It slides the curve so its centre is 0 and rescales the horizontal axis so one standard deviation is 1. Every shaded region keeps the same share of the total area.

Definition

Standardised normal variable

If $X \sim \mathrm{N}(\mu,\sigma^2)$ , the standardised variable is $Z=\frac{X-\mu}{\sigma}.$ Then $Z \sim \mathrm{N}(0,1)$ , and for any value $x$ , $\mathrm{P}(X<x)=\mathrm{P}\left(Z<\frac{x-\mu}{\sigma}\right)=\Phi\left(\frac{x-\mu}{\sigma}\right).$

The value $z=\dfrac{x-\mu}{\sigma}$ says how many standard deviations $x$ lies from the mean: positive above the mean, negative below.

Why the formula is (x − μ) over σ

Start at the mean $\mu$ . Moving to $x$ is a distance $x-\mu$ along the axis.

Divide by $\sigma$ to express that distance in standard deviations. If $x$ is one standard deviation above the mean, then $z=1$ , whatever the original units.

Subtracting $\mu$ slides the centre to 0; dividing by $\sigma$ stretches or squashes the axis to a standard width. Both operations are horizontal only, so the area between two values is preserved.

Worked example: Finding a lower-tail probability

The mass $X$ grams of a component satisfies $X \sim \mathrm{N}(30,25)$ . Find $\mathrm{P}(X<38)$ .

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