QCE Mathematical Methods - Unit 4 - Continuous random variables and the normal distribution
Normal Distributions | QCE Mathematical Methods
Learn QCE Mathematical Methods normal distributions, z-scores, probabilities, quantiles and standardisation.
Updated 2026-05-18 - 4 min read
QCAA official coverage - Mathematical Methods 2025 v1.3
Exact syllabus points covered
- Identify contexts, e.g. naturally occurring variations, that are suitable for modelling by normal random variables.
- Recognise features of the graph of the probability density function of the normal distribution with mean $\mu$ and standard deviation $\sigma$ and the use of the standard normal distribution.
- Recognise and use the link between the normal distribution and the notation $X \sim N(\mu,\sigma^2)$.
- Calculate probabilities and quantiles associated with a given normal distribution, using technology.
- Model and solve problems that involve normal distributions, with and without technology.
The normal distribution is the familiar bell-shaped distribution used for many naturally varying measurements: heights, errors, measurement noise and standardised scores. It is not appropriate for every dataset, but it is one of the most important continuous models in Methods.
Original Sylligence diagram for normal distribution standardisation.
Notation
If $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$, write:
$ X\sim N(\mu,\sigma^2) $
The mean $\mu$ locates the centre. The standard deviation $\sigma$ controls the spread. A larger $\sigma$ makes the bell curve wider and lower because the total area must still be $1$.
The probability density function is:
$ f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2} $
You do not usually integrate this by hand in Methods, but the formula explains the key graph features: the curve is symmetric about $\mu$, always positive, bell-shaped and controlled by $\sigma$.
The standard normal distribution is:
$ Z\sim N(0,1) $
It is the reference normal distribution after standardising.
Standardising with z-scores
A $z$-score tells you how many standard deviations a value is from the mean:
$ z=\frac{x-\mu}{\sigma} $
If $z=2$, the value is two standard deviations above the mean. If $z=-1.5$, the value is one and a half standard deviations below the mean.
Standardising lets you compare values from different normal distributions.
You can also reverse the formula:
$ x=\mu+z\sigma $
This is useful for percentile and quantile questions. Once technology gives a $z$-value, convert it back to the original scale.
Reading the graph
Changing $\mu$ shifts the curve left or right. Changing $\sigma$ changes the spread:
| Change | Effect on graph | What stays true | | --- | --- | --- | | larger $\mu$ | centre moves right | total area is still $1$ | | smaller $\mu$ | centre moves left | shape stays symmetric | | larger $\sigma$ | curve becomes wider and lower | centre stays at $\mu$ | | smaller $\sigma$ | curve becomes narrower and taller | total area is still $1$ |
If the data are strongly skewed, bounded in a way the normal model cannot respect, or contain several clusters, a normal model may be a poor choice even if a calculator can still produce a number.
Probabilities and quantiles
For normal distributions, QCE questions usually expect technology for probabilities and quantiles. Distribution tables are not required in the current syllabus.
- Probability question: find an area, such as $P(X<75)$.
- Quantile question: find the value of $x$ that cuts off a given area, such as the 90th percentile.
For a middle interval:
$ P(a<X<b)=P(X<b)-P(X<a) $
For a right-tail probability:
$ P(X>a)=1-P(X\le a) $
For a quantile, start with the area. If a question asks for the top $10\%$, the cutoff has left-tail area $0.90$, not $0.10$.
Worked example
Worked probability example
Worked quantile example
Comparing scores
Suppose one student scores $86$ on a test with mean $80$ and standard deviation $4$. Another scores $72$ on a test with mean $60$ and standard deviation $10$.
First student:
$ z=\frac{86-80}{4}=1.5 $
Second student:
$ z=\frac{72-60}{10}=1.2 $
Relative to their cohorts, the first score is stronger because it is further above the mean in standard-deviation units.