The normal distribution and standardisation: QCE Maths Methods Unit 4

Syllabus dot point

“Apply the normal distribution $N(\mu, \sigma^2)$ and the standardisation $Z = (X - \mu)/\sigma$ to compute normal probabilities and inverse probabilities, including the empirical 68-95-99.7 rule”

9 min readUpdated 7 Oct 2026

Quick answer

The normal distribution $N(\mu, \sigma^2)$ is the bell-shaped distribution with mean $\mu$ and standard deviation $\sigma$ ; standardisation via $Z = (X - \mu)/\sigma$ converts any normal probability question to a standard-normal question, the empirical 68-95-99.7 rule handles Paper 1 exact-value endpoints at $\mu \pm n \sigma$ , and the inverse-normal function handles "find $c$ such that $P(X \leq c) = p$ " questions on Paper 2.

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What this dot point is asking
The normal distribution
The standard normal $Z$
The empirical 68-95-99.7 rule
Computing normal probabilities
Inverse normal
Applications
Exam-style questions

What this dot point is asking

QCAA wants you to recognise the normal distribution, apply the standardisation transformation, use the empirical 68-95-99.7 rule for Paper 1 exact-value questions, and compute general normal probabilities and inverse probabilities using technology in Paper 2.

The normal distribution

A continuous random variable $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$ :

X \sim N(\mu, \sigma^2)

The pdf is:

f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-(x - \mu)^2 / (2 \sigma^2)}

Properties:

Symmetric about $\mu$ .
Bell-shaped, peak at $x = \mu$ .
Mean = median = mode = $\mu$ .
Standard deviation $\sigma$ controls the spread.

The standard normal $Z$

The standard normal is $Z \sim N(0, 1)$ . Any normal can be converted to a standard normal by:

Z = \frac{X - \mu}{\sigma}

This is standardisation. The transformation preserves probabilities:

P(a \leq X \leq b) = P\left( \frac{a - \mu}{\sigma} \leq Z \leq \frac{b - \mu}{\sigma} \right)

The empirical 68-95-99.7 rule

For any normal distribution:

$P(\mu - \sigma \leq X \leq \mu + \sigma) \approx 0.68$
$P(\mu - 2 \sigma \leq X \leq \mu + 2 \sigma) \approx 0.95$
$P(\mu - 3 \sigma \leq X \leq \mu + 3 \sigma) \approx 0.997$

Single-tail derivatives:

$P(X > \mu + \sigma) \approx 0.16$
$P(X > \mu + 2\sigma) \approx 0.025$
$P(X > \mu + 3\sigma) \approx 0.0015$

The rule is the workhorse for Paper 1 exact-value questions when the $x$ -endpoints fall at $\mu \pm n \sigma$ .

Computing normal probabilities

Paper 1 (exact-value). When the endpoints map cleanly to $\mu \pm n \sigma$ for $n = 1, 2, 3$ , use the empirical rule.

Paper 2 (calculator-active).

State the distribution: $X \sim N(\mu, \sigma^2)$ .
Standardise the endpoints: $z_1 = (a - \mu)/\sigma$ , $z_2 = (b - \mu)/\sigma$ .
Compute $P$ using calculator's normCdf: $P = \text{normCdf}(a, b, \mu, \sigma)$ .

Inverse normal

Given a probability $p$ , find $c$ such that $P(X \leq c) = p$ :

Find $z = \text{invNorm}(p)$ such that $P(Z \leq z) = p$ .
Convert back: $c = \mu + z \sigma$ .

Common $z$ values:

$p$	$z$
0.90	1.2816
0.95	1.6449
0.975	1.9600
0.99	2.3263

For an "upper tail" question, $P(X > c) = 1 - p$ , so use $z$ for the complementary $p$ .

Applications

Quality control: Lengths or weights of manufactured items modelled as normal.
Test scores: Standardised test scores have a bell-curve distribution.
Biological measurements: Heights, blood pressure, gestation periods.
Modelling errors: Random measurement errors typically follow a normal distribution.

Battery lifetimes

Battery lifetimes follow $X \sim N(200, 20^2)$ , with mean $\mu = 200$ hours and standard deviation $\sigma = 20$ hours. Three probability questions are solved in sequence.

Step 1: Use the empirical rule to find $P(160 \leq X \leq 240)$

Before reaching for a calculator, check whether the endpoints fall at $\mu \pm n\sigma$ for an integer $n$ .

160 = 200 - 2 \times 20 = \mu - 2\sigma, \qquad 240 = 200 + 2 \times 20 = \mu + 2\sigma

The interval is exactly $[\mu - 2\sigma, \mu + 2\sigma]$ , so the empirical 68-95-99.7 rule applies immediately.

Answer: $P(160 \leq X \leq 240) \approx 0.95$ .

Step 2: Standardise to find $P(X > 230)$

The endpoint $230$ does not fall at a whole-number multiple of $\sigma$ from $\mu$ , so we convert to a $z$ -score. Subtracting the mean and dividing by $\sigma$ maps $X$ to the standard normal $Z \sim N(0, 1)$ .

z = \frac{230 - 200}{20} = \frac{30}{20} = 1.5

Using a standard normal table or calculator:

P(X > 230) = P(Z > 1.5) \approx 0.0668

Step 3: Reverse the process with an inverse-normal calculation

Find $c$ such that the longest-lasting $10\%$ of batteries last more than $c$ hours. The condition $P(X > c) = 0.10$ is equivalent to $P(X \leq c) = 0.90$ .

Looking up the $z$ -score that corresponds to a cumulative probability of $0.90$ :

z_{0.90} \approx 1.2816

Convert back from the standard scale to the original scale using $c = \mu + z\sigma$ .

c = 200 + 1.2816 \times 20 \approx 225.6 \text{ hours}

Final answer: the top $10\%$ of batteries last more than approximately $225.6$ hours.

Common errors

Wrong sign on standardisation: $z = (x - \mu)/\sigma$ , not $(\mu - x)/\sigma$ .
Empirical rule misapplied: The rule is for $\mu \pm n \sigma$ specifically. For other endpoints, standardise and use a table or calculator.
Inverse for the wrong tail: "Top 10 percent" means $P(X > c) = 0.10$ , so $P(X \leq c) = 0.90$ . Read carefully.
Using $\sigma^2$ where $\sigma$ is asked: $z = (x - \mu)/\sigma$ uses the standard deviation, not the variance.
Calculator without set-up: Paper 2 expects the standardisation set-up shown explicitly with the calculator value at the end.

Exam-style questions

Questions in the style of QCAA 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.

Practice4 marks

Paper 2 (complex familiar). Heights are normally distributed with mean

170

cm and standard deviation

8

cm. (a) Determine

P(160 \leq X \leq 180)

. (b) Determine the height

c

such that

P(X > c) = 0.05

Show worked answer →

(a) Probability. $X \sim N(170, 8^2)$ .

Standardise: $z_1 = (160 - 170)/8 = -1.25$ , $z_2 = (180 - 170)/8 = 1.25$ .

$P(160 \leq X \leq 180) = P(-1.25 \leq Z \leq 1.25) \approx 0.7887$ .

So approximately $0.789$ .

(b) Inverse probability. $P(X > c) = 0.05 \implies P(X \leq c) = 0.95$ .

From inverse normal, $z_{0.95} \approx 1.6449$ .

$c = \mu + z \sigma = 170 + 1.6449 \times 8 \approx 183.2$ cm.

Markers reward the standardisation, the use of normCdf or inverse-normal, and converting back from $z$ to $x$ -scale.

Practice2 marks

Paper 1 (technique). The random variable

X \sim N(50, 4^2)

. Use the empirical 68-95-99.7 rule to estimate

P(42 \leq X \leq 58)

Show worked answer →

$42 = \mu - 2 \sigma$ and $58 = \mu + 2 \sigma$ (since $50 - 8 = 42$ and $50 + 8 = 58$ ).

By the empirical rule, $P(\mu - 2 \sigma \leq X \leq \mu + 2 \sigma) \approx 0.95$ .

So $P(42 \leq X \leq 58) \approx 0.95$ .

Markers reward identifying the interval as $[\mu - 2\sigma, \mu + 2\sigma]$ and citing the empirical rule.

Practise this

Sources & how we know this

Mathematical Methods General senior syllabus 2025 — QCAA