When the endpoints map cleanly to μ n σ for n = 1, 2, 3, use the empirical rule.

What is quality control?

Lengths or weights of manufactured items modelled as normal.

What are test scores?

Standardised test scores have a bell-curve distribution.

What are biological measurements?

Heights, blood pressure, gestation periods.

What is empirical rule misapplied?

The rule is for μ n σ specifically. For other endpoints, standardise and use a table or calculator.

What is inverse for the wrong tail?

"Top 10 percent" means P(X > c) = 0.10, so P(X ≤ c) = 0.90. Read carefully.

What is calculator without set-up?

Paper 2 expects the standardisation set-up shown explicitly with the calculator value at the end.

QCAA QCAA 2023-style practice: Paper 1 (technique). The random variable X N(50, 4^2). Use the empirical 68-95-99.7 rule to estimate P(42 ≤ X ≤ 58).

42 = μ - 2 σ and 58 = μ + 2 σ (since 50 - 8 = 42 and 50 + 8 = 58). By the empirical rule, P(μ - 2 σ ≤ X ≤ μ + 2 σ) ≈ 0.95. So P(42 ≤ X ≤ 58) ≈ 0.95. Markers reward identifying the interval as [μ - 2σ, μ + 2σ] and citing the empirical rule.

§-Syllabus dot point

QLDMath MethodsSyllabus dot point

Topic 3: Continuous random variables, the normal distribution, and statistical inference

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

A focused answer to the QCE Maths Methods Unit 4 dot point on the normal distribution. Standardisation, the empirical rule, normal probability and inverse-normal calculations, and worked PSMT and EA examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

QCAA 20244 marksPaper 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.

QCAA 20232 marksPaper 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.