Statistics · The Normal Distribution · Year 2

The Standard Normal Distribution

Every normal variable can be rescaled onto the same universal curve, Z ~ N(0, 1²).

8 min read 2 worked examples

Standardising a normal variable

It is often useful to standardise normally distributed random variables. You do this by coding the data so that it can be modelled by the standard normal distribution.

Key fact

The standard normal distribution has mean $0$ and standard deviation $1$.

Notation

The standard normal variable is written as $Z\sim \mathrm{N}(0,1^2)$.

If $X\sim \mathrm{N}(\mu,\sigma^2)$ is a normal random variable with mean $\mu$ and standard deviation $\sigma$, then you can code $X$ using the formula:

$Z = \dfrac{X-\mu}{\sigma}$

The resulting $z$-values will be normally distributed with mean $0$ and standard deviation $1$.

Links

If $X=x$ then the corresponding value of $Z$ will be $z=\dfrac{x-\mu}{\sigma}$. The mean of the coded data will be $\dfrac{\mu-\mu}{\sigma}=0$ and the standard deviation will be $\dfrac{\sigma}{\sigma}=1$.

For the standard normal curve $Z\sim \mathrm{N}(0,1^2)$, the probability $\mathrm{P}(Z<a)$ is sometimes written as $\Phi(a)$. You can find it by entering $\mu=0$ and $\sigma=1$ into the normal cumulative distribution function on your calculator.

$\Phi(a)=\mathrm{P}(Z<a)$ — the area under the standard normal curve to the left of $a$.

Worked example 1

Writing probabilities in terms of Φ

The random variable $X\sim \mathrm{N}(70, 5^2)$. Write in terms of $\Phi(z)$ for some value $z$: a $\mathrm{P}(X<77)$ b $\mathrm{P}(X\ge 65)$

a

$z=\dfrac{77-70}{5}=1.4$, so $\mathrm{P}(X<77)=\Phi(1.4)$.

$z=\dfrac{77-70}{5}=1.4$.

Code the data so that it is modelled by the standard normal distribution, using $Z=\dfrac{X-\mu}{\sigma}$.

$\mathrm{P}(X<77) = \mathrm{P}(Z<1.4) = \Phi(1.4)$.

▪ solved

b

$z=\dfrac{65-70}{5}=-1$, so $\mathrm{P}(X\ge 65)=1-\Phi(-1)$.

$\mathrm{P}(X\ge 65) = 1-\mathrm{P}(X<65)$.

The distribution is continuous, so you can use $<$ and $\le$ interchangeably.

$z=\dfrac{65-70}{5}=-1$.

$\mathrm{P}(X\ge 65) = 1-\mathrm{P}(Z<-1) = 1-\Phi(-1)$.

▪ solved

You sometimes need to find $z$-values that correspond to given probabilities. You can find these for some standard values of $p$ using the percentage points of the normal distribution table, which is given in the formulae booklet in your exam. It gives values of $z$ and $p$ such that $\mathrm{P}(Z>z)=p$.

$p$	$z$	$p$	$z$
0.5000	0.0000	0.0500	1.6449
0.4000	0.2533	0.0250	1.9600
0.3000	0.5244	0.0100	2.3263
0.2000	0.8416	0.0050	2.5758
0.1500	1.0364	0.0010	3.0902
0.1000	1.2816	0.0005	3.2905

So $\mathrm{P}(Z>1.96)=0.025$. You can use the symmetry of the distribution to find corresponding negative $z$-values: $\mathrm{P}(Z<-1.96)=0.025$, so $\mathrm{P}(Z>-1.96)=0.975$.

Worked example 2

Using the percentage points table

The time, $T$ minutes, that visitors spend at a museum exhibit is modelled as $T\sim \mathrm{N}(42, 6^2)$. The museum wants to identify visitors in the top 10% of visit durations, for a feedback survey. Find the minimum visit duration for a visitor to be included.

Using the percentage points table with $p=0.10$: $\mathrm{P}(Z>1.2816)=0.10$.

Using the percentage points table: $\mathrm{P}(Z>1.2816)=0.10$.

Use the table with $p=0.10$.

$\dfrac{t-42}{6}=1.2816 \Rightarrow t = 49.69$ (2 d.p.).

Convert the value for $Z$ back into a value for $T$. Remember the denominator is $\sigma$, not $\sigma^2$.

The museum should survey visitors with a visit duration greater than $49.69$ minutes.

▪ solved

Check yourself

The resistance, $R$ ohms, of a batch of resistors is modelled as $R\sim \mathrm{N}(100, 4^2)$. A supplier rejects any resistor with a resistance in the top 5% of the distribution. Using the percentage points table, find the minimum resistance of a rejected resistor.

$\mathrm{P}(Z>1.6449)=0.05$.

$\dfrac{r-100}{4}=1.6449 \Rightarrow r = 106.58$ (2 d.p.).

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Inverse Normal Distribution Function

Finding μ and σ