WACE Mathematics Methods probability and statistics deep dive: the 2026 exam

WACEMath MethodsStudy guide17 min readUpdated 23 Sep 2026

Revision deep dive for the probability and statistics in WACE Mathematics Methods Units 3 and 4: discrete random variables, Bernoulli and binomial distributions, probability density functions, the normal distribution, the sample proportion, confidence intervals and sample size, with worked exam-style examples and links to every statistics dot point.

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How statistics is examined
1. Discrete random variables
2. Bernoulli and binomial distributions
3. Continuous random variables
4. The normal distribution
5. The sample proportion
6. Confidence intervals and margin of error
Common mistakes
Check your knowledge

How statistics is examined

Probability and statistics make up a large share of the WACE Methods exam, and they reward precise communication more than any other part of the course. Markers want to see the random variable defined, the distribution named with its parameters, the probability statement written, and the answer interpreted in context. This deep dive links every statistics dot point on the site in the order you would revise them.

For the calculus half of the course, see the calculus deep dive. For exam format and timing, see the exam strategy guide.

1. Discrete random variables

Dot points: discrete probability distributions, expected value of a discrete random variable, variance and standard deviation of a discrete random variable, discrete random variables and the binomial distribution.

Discrete random variables

\sum P(X = x) = 1 \qquad E(X) = \sum x\,P(X = x) \qquad \text{Var}(X) = E(X^2) - [E(X)]^2

E(aX + b) = aE(X) + b \qquad \text{Var}(aX + b) = a^2\,\text{Var}(X)

Unknown probability, mean and variance

The random variable $X$ has this distribution:

$x$	0	1	2	3
$P(X = x)$	0.1	$k$	$2k$	0.3

Find $k$: $0.1 + k + 2k + 0.3 = 1$ , so $3k = 0.6$ and $k = 0.2$ .
Mean: $E(X) = 0(0.1) + 1(0.2) + 2(0.4) + 3(0.3) = 1.9$ .
Variance: $E(X^2) = 0 + 1(0.2) + 4(0.4) + 9(0.3) = 4.5$ , so $\text{Var}(X) = 4.5 - 1.9^2 = 0.89$ .
Transform: If $Y = 3X - 2$ , then $E(Y) = 3(1.9) - 2 = 3.7$ and $\text{Var}(Y) = 9(0.89) = 8.01$ . The $-2$ shifts the mean but does not change the spread.

Fair games. A game is fair when the expected gain is zero. If a ticket costs $5 and wins $20 with probability 0.2, the expected gain is $20(0.2) - 5 = -\$1$ per game, so the game favours the operator.

2. Bernoulli and binomial distributions

Dot points: the Bernoulli distribution, binomial probabilities and cumulative probabilities, mean and variance of the binomial distribution.

A Bernoulli variable is a single trial: $P(X = 1) = p$ , $P(X = 0) = 1 - p$ , with mean $p$ and variance $p(1 - p)$ . A binomial variable counts successes in $n$ independent Bernoulli trials with the same $p$ .

Binomial distribution

X \sim \text{Bin}(n, p)

P(X = x) = \binom{n}{x}p^x(1 - p)^{n - x} \qquad E(X) = np \qquad \text{Var}(X) = np(1 - p)

At least one, at most two

Each of 10 independent customers buys a coffee with probability 0.3. Let $X \sim \text{Bin}(10, 0.3)$ .

At least one: use the complement. $P(X \ge 1) = 1 - P(X = 0) = 1 - 0.7^{10} \approx 0.9718$ .
At most two: $P(X \le 2) = P(X=0) + P(X=1) + P(X=2) \approx 0.3828$ (cumulative binomial on the calculator).
Mean and standard deviation: $E(X) = 3$ and $\sigma = \sqrt{10(0.3)(0.7)} = \sqrt{2.1} \approx 1.449$ .

Before using the binomial in context, check the conditions. Sampling without replacement from a small population breaks independence, and a probability that changes from trial to trial breaks the "same $p$ " condition. A two-mark "explain why the binomial may not be appropriate" question wants one of these, tied to the context.

3. Continuous random variables

Dot points: probability density functions, continuous random variables and the normal distribution.

For a continuous random variable, probability is area under the density curve, and $P(X = a) = 0$ for any single value.

Probability density functions

f(x) \ge 0, \quad \int f(x)\,dx = 1 \qquad P(a \le X \le b) = \int_a^b f(x)\,dx

E(X) = \int x f(x)\,dx \qquad \text{Var}(X) = \int x^2 f(x)\,dx - [E(X)]^2

A quadratic density

$f(x) = kx^2$ for $0 \le x \le 3$ , and 0 elsewhere.

Find $k$: $\displaystyle\int_0^3 kx^2\,dx = k\left[\frac{x^3}{3}\right]_0^3 = 9k = 1$ , so $k = \dfrac{1}{9}$ .
Probability: $P(X \le 2) = \displaystyle\int_0^2 \frac{x^2}{9}\,dx = \frac{8}{27} \approx 0.296$ .
Mean: $E(X) = \displaystyle\int_0^3 \frac{x^3}{9}\,dx = \frac{1}{9}\cdot\frac{81}{4} = 2.25$ .
Variance: $E(X^2) = \displaystyle\int_0^3 \frac{x^4}{9}\,dx = \frac{1}{9}\cdot\frac{243}{5} = 5.4$ , so $\text{Var}(X) = 5.4 - 2.25^2 = 0.3375$ .

4. The normal distribution

Dot points: the normal distribution and its properties, standardisation and normal probability calculations.

$X \sim N(\mu, \sigma^2)$ is symmetric about $\mu$ . About 68, 95 and 99.7 percent of values lie within 1, 2 and 3 standard deviations of the mean. The standardised score $z = \dfrac{x - \mu}{\sigma}$ converts any normal to $Z \sim N(0, 1)$ .

Forward and inverse normal

Adult heights in a population are modelled by $X \sim N(170, 8^2)$ cm.

Forward: $P(X > 180) = P(Z > 1.25) \approx 0.1056$ .
Rule of thumb: $P(154 < X < 186)$ is within two standard deviations, so it is about 0.95.
Inverse: The height exceeded by only 10 percent: find $k$ with $P(X < k) = 0.9$ . Then $k = 170 + 1.2816 \times 8 \approx 180.3$ cm.

Notice the mean and standard deviation are stated as $N(170, 8^2)$ : the second parameter is the variance. Enter the standard deviation, 8, into the calculator.

5. The sample proportion

Dot point: the sample proportion and its distribution.

For random samples of size $n$ from a population with proportion $p$ , the sample proportion $\hat{p}$ varies from sample to sample. Its distribution has

E(\hat{p}) = p \qquad \text{SD}(\hat{p}) = \sqrt{\frac{p(1 - p)}{n}}

and for large $n$ it is approximately normal.

Probability about a sample proportion

Thirty percent of a town's residents cycle to work. For random samples of 100 residents, $\text{SD}(\hat{p}) = \sqrt{\dfrac{0.3 \times 0.7}{100}} \approx 0.0458$ .

Using the normal approximation, $P(\hat{p} > 0.35) \approx P\left(Z > \dfrac{0.05}{0.0458}\right) = P(Z > 1.091) \approx 0.138$ .

6. Confidence intervals and margin of error

Dot points: confidence intervals for proportions, margin of error and sample size.

Approximate confidence interval for

p

\hat{p} \pm z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \qquad z = 1.645 \ (90\%), \quad 1.96 \ (95\%), \quad 2.576 \ (99\%)

The margin of error is

E = z\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}}

Choosing a sample size

A council wants a 95 percent confidence interval with margin of error at most 0.03. With no prior estimate, use $\hat{p} = 0.5$ (the value that makes the margin largest).

1.96\sqrt{\frac{0.25}{n}} \le 0.03 \implies n \ge \frac{1.96^2 \times 0.25}{0.03^2} \approx 1067.1.

Round up: the council needs at least 1068 people. Rounding down would give a margin slightly larger than 0.03.

Three facts examiners test repeatedly:

Higher confidence means a wider interval (bigger $z$ ).
Quadrupling $n$ halves the margin of error.
Different random samples give different intervals; roughly 5 percent of 95 percent intervals miss the true $p$ .

Common mistakes

Where statistics marks go missing

Using the standard deviation where the variance is needed (or the reverse). $\text{Var}(aX + b) = a^2\text{Var}(X)$ , not $a\,\text{Var}(X)$ .
Getting "at least" wrong: $P(X \ge 3) = 1 - P(X \le 2)$ , not $1 - P(X \le 3)$ .
Calling a confidence interval a range that contains 95 percent of sample proportions.
Not stating the distribution, so a correct number has no method mark to fall back on.
Rounding a required sample size down.

Check your knowledge

$X \sim \text{Bin}(20, 0.25)$ . Find $E(X)$ and $\text{Var}(X)$ . (Answer: 5 and 3.75.)
A 95 percent interval for $p$ is $(0.42, 0.50)$ . Find $\hat{p}$ and the margin of error. (Answer: 0.46 and 0.04.)
For $X \sim N(50, 4^2)$ , find $P(X < 46)$ using the 68-95-99.7 rule. (Answer: about 0.16.)

Then try the statistics practice quiz.

Sources & how we know this

Mathematics Methods ATAR course Year 12 syllabus (for teaching from 2026) — School Curriculum and Standards Authority (SCSA)
Mathematics Methods past ATAR course examinations — School Curriculum and Standards Authority (SCSA)
Mathematics Methods ATAR course: syllabus and support materials — SCSA

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