Continuous random variables and the normal distribution: WACE Year 12 Mathematics Methods Unit 4

Syllabus dot point

“Use probability density functions to find probabilities, mean and variance for continuous random variables, and apply the normal distribution with standardisation”

9 min readUpdated 2 Jun 2026

Quick answer

A continuous random variable is described by a probability density function whose total area is 1; probabilities are areas (integrals). The normal distribution is the key model, and standardising to $z = \dfrac{x - \mu}{\sigma}$ lets you find any normal probability.

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What this dot point is asking
Probability density functions
The normal distribution
Why standardisation works
Standardisation and z-scores
Exam-style questions

What this dot point is asking

SCSA Unit 4 moves from the discrete distributions of Unit 3 to continuous random variables, where outcomes form an interval rather than a list. Probabilities become areas under a probability density function (pdf), and the normal distribution is the central continuous model examined in the WACE written examination.

Probability density functions

Probability density function

A function $f(x)$ is a pdf for a continuous random variable $X$ if $f(x) \ge 0$ for all $x$ and the total area is 1:

\int_{-\infty}^{\infty} f(x)\,dx = 1.

For a continuous variable, the probability of any single value is 0, so $P(X = a) = 0$ . Probabilities are areas:

P(a \le X \le b) = \int_a^b f(x)\,dx.

Because single points have probability 0,

P(a \le X \le b) = P(a < X < b)

The mean (expected value) and variance are found by integration:

\mu = E(X) = \int_{-\infty}^{\infty} x\,f(x)\,dx, \qquad \mathrm{Var}(X) = \int_{-\infty}^{\infty} (x - \mu)^2 f(x)\,dx = E(X^2) - \mu^2.

Working with a pdf

The continuous random variable $X$ has pdf $f(x) = kx$ for $0 \le x \le 2$ and $f(x) = 0$ otherwise. Find $k$ , then $P(X \le 1)$ .

Step 1: Use the total-area condition to find $k$

Every valid pdf must integrate to 1 over its support. Setting this integral equal to 1 is the equation we solve for the unknown constant $k$ .

\int_0^2 kx\,dx = k\left[\frac{x^2}{2}\right]_0^2 = k \cdot \frac{4}{2} = 2k = 1

So $k = \dfrac{1}{2}$ .

Step 2: Compute the probability by integration

A probability for a continuous variable equals the area under the pdf over the required interval. We integrate the now-known pdf from $0$ to $1$ .

P(X \le 1) = \int_0^1 \frac{1}{2}x\,dx = \frac{1}{2}\left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}

Final answer: $k = \dfrac{1}{2}$ and $P(X \le 1) = 0.25$ .

The normal distribution

The normal distribution $X \sim N(\mu, \sigma^2)$ has a symmetric bell-shaped pdf centred at the mean $\mu$ with standard deviation $\sigma$ . Its key features:

It is symmetric about $x = \mu$ , so $P(X < \mu) = P(X > \mu) = 0.5$ .
The total area under the curve is 1.
Larger $\sigma$ gives a wider, flatter curve.

The 68-95-99.7 rule

For a normal distribution, approximately:

$68\%$ of values lie within $1$ standard deviation of the mean ( $\mu \pm \sigma$ ),
$95\%$ within $2$ standard deviations ( $\mu \pm 2\sigma$ ),
$99.7\%$ within $3$ standard deviations ( $\mu \pm 3\sigma$ ).

Why standardisation works

Every normal distribution has the same shape; only its centre and width differ. Subtracting the mean shifts the centre to $0$ , and dividing by the standard deviation rescales the width to $1$ , producing the single standard normal $Z\sim N(0,1)$ . Because this transformation preserves areas, a probability for any normal variable equals the matching probability for $Z$ , which is why one set of standard-normal values (or one calculator function) serves every normal model. This is the practical payoff of the symmetry and two-parameter structure described above.

Standardisation and z-scores

To compare values from different normal distributions, or to use tables and the calculator, convert to the standard normal $Z \sim N(0, 1)$ :

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

The

z

-score is the number of standard deviations a value lies above (positive) or below (negative) the mean.

A normal probability with standardisation

Test scores are normally distributed with $\mu = 70$ and $\sigma = 8$ . Find the proportion of students scoring above $86$ .

Step 1: Standardise the boundary value

Converting $x = 86$ to a $z$ -score tells us how many standard deviations above the mean it sits, so we can use the standard normal distribution.

z = \frac{86 - 70}{8} = \frac{16}{8} = 2

Step 2: Apply the empirical rule to find the tail probability

Since $z = 2$ , we need $P(Z > 2)$ . By the 68-95-99.7 rule, $95\%$ of values lie within $2$ standard deviations, leaving $5\%$ in the two tails combined. Symmetry splits this equally, so each tail holds $2.5\%$ .

P(X > 86) = P(Z > 2) \approx 0.025

Final answer: about $2.5\%$ of students score above $86$ . (In the calculator section you would obtain $P(Z > 2) \approx 0.0228$ directly.)

Common errors

Treating $P(X = a)$ as nonzero: For a continuous variable the probability of an exact value is 0; probabilities are areas over intervals.
Forgetting to make total area 1: When a pdf has an unknown constant, you must set $\int f(x)\,dx = 1$ over the support to find it before computing probabilities.
Mixing up $\sigma$ and $\sigma^2$: The notation $N(\mu, \sigma^2)$ uses the variance, but standardisation $z = \frac{x-\mu}{\sigma}$ uses the standard deviation. Take the square root before dividing.
Sign of the z-score: A value below the mean gives a negative $z$ . Using symmetry, $P(Z < -a) = P(Z > a)$ , so keep track of which tail you want.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style7 marks

Calculator-assumed. The volume of soft drink in a can is normally distributed with mean

\mu=375

mL and standard deviation

\sigma=4

mL. (a) Find

P(X<370)

. (b) Find

P(372\le X\le 378)

. (c) Cans below a certain volume are rejected; the bottom

2\%

are rejected. Find the rejection volume.

Show worked answer →

A normal-distribution question with an inverse part.

(a) $z=\dfrac{370-375}{4}=-1.25$ . $P(X<370)=P(Z<-1.25)\approx 0.106$ .

(b) $z_1=\dfrac{372-375}{4}=-0.75$ , $z_2=\dfrac{378-375}{4}=0.75$ . $P(-0.75\le Z\le 0.75)\approx 0.547$ .

Markers reward standardising, the calculator probabilities, and unstandardising with a negative $z$ in (c).

WACE 2023-style6 marks

Calculator-assumed. A continuous random variable has pdf

f(x)=\dfrac{3}{4}(1-x^{2})

for

-1\le x\le 1

. (a) Verify the total area is

1

. (b) Find

P(0\le X\le 0.5)

. (c) State the mean, justifying by symmetry.

Show worked answer →

A pdf question combining integration and symmetry.

(a) $\displaystyle\int_{-1}^{1}\dfrac{3}{4}(1-x^{2})\,dx=\dfrac{3}{4}\left[x-\dfrac{x^{3}}{3}\right]_{-1}^{1}=\dfrac{3}{4}\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{3}{4}\cdot\dfrac{4}{3}=1$ .

(b) $\displaystyle\int_{0}^{0.5}\dfrac{3}{4}(1-x^{2})\,dx=\dfrac{3}{4}\left[x-\dfrac{x^{3}}{3}\right]_{0}^{0.5}=\dfrac{3}{4}\left(0.5-\dfrac{0.125}{3}\right)\approx 0.344$ .

Markers reward the area check, the definite integral, and a symmetry justification for the mean.

Continuous random variables and the normal distribution: WACE Year 12 Mathematics Methods Unit 4

What this dot point is asking

Probability density functions

Step 1: Use the total-area condition to find $k$

Step 2: Compute the probability by integration

The normal distribution

Why standardisation works

Standardisation and z-scores

Step 1: Standardise the boundary value

Step 2: Apply the empirical rule to find the tail probability

Exam-style questions

Practise this

Sources & how we know this

What this dot point is asking

Probability density functions

Step 1: Use the total-area condition to find kkk

Step 2: Compute the probability by integration

The normal distribution

Why standardisation works

Standardisation and z-scores

Step 1: Standardise the boundary value

Step 2: Apply the empirical rule to find the tail probability

Exam-style questions

Practise this

Sources & how we know this

Step 1: Use the total-area condition to find $k$