The normal distribution and its properties: WACE Year 12 Mathematics Methods Unit 4

Syllabus dot point

“Describe the normal distribution, its symmetry and parameters, and apply the empirical 68-95-99.7 rule”

8 min readUpdated 2 Jun 2026

Quick answer

The normal distribution $X\sim N(\mu,\sigma^{2})$ is a symmetric bell-shaped density centred at the mean $\mu$ , with spread set by the standard deviation $\sigma$ . The empirical rule gives about $68\%$ , $95\%$ and $99.7\%$ of values within $1$ , $2$ and $3$ standard deviations of the mean.

Jump to a section

What this dot point is asking
Defining properties
The empirical rule
Building probabilities from the rule
Comparing distributions
Exam-style questions

What this dot point is asking

SCSA Unit 4 makes the normal distribution the central continuous model. This dot point asks you to describe its shape and parameters and to use the empirical 68-95-99.7 rule for quick probability estimates, examined in both sections, with the rule especially useful in the calculator-free section.

Defining properties

Properties of the normal distribution

The density is symmetric and bell-shaped about $x=\mu$ .
The total area under the curve is $1$ .
$\mu$ locates the centre; $\sigma$ controls the width.
Because of symmetry, $P(X<\mu)=P(X>\mu)=0.5$ .
A larger $\sigma$ gives a wider, flatter curve; a smaller $\sigma$ gives a taller, narrower one.

The mean, median and mode of a normal distribution all coincide at $\mu$ because of the symmetry. The notation $N(\mu,\sigma^{2})$ uses the variance $\sigma^{2}$ as the second parameter, not the standard deviation.

The empirical rule

For any normal distribution the proportion of values within a fixed number of standard deviations is the same, giving the 68-95-99.7 rule.

The 68-95-99.7 rule

P(\mu-\sigma \le X \le \mu+\sigma) \approx 0.68,

P(\mu-2\sigma \le X \le \mu+2\sigma) \approx 0.95,

P(\mu-3\sigma \le X \le \mu+3\sigma) \approx 0.997.

By symmetry, each tail beyond $2\sigma$ holds about $2.5\%$ , and beyond $3\sigma$ about $0.15\%$ . These let you answer many questions without a calculator.

Using the empirical rule

Seedling heights are normal with $\mu=20$ cm and $\sigma=3$ cm. Estimate the proportion taller than $26$ cm.

Step 1: Express the boundary as a multiple of $\sigma$

The empirical rule is stated in terms of standard deviations from the mean, so the first task is to check how many $\sigma$ the boundary lies from $\mu$ .

$26 = 20 + 2\times 3$ , so $26$ is exactly $2\sigma$ above the mean.

Step 2: Apply the empirical rule and use symmetry

By the 68-95-99.7 rule, $95\%$ of values lie within $2\sigma$ of the mean, leaving $5\%$ in the two tails combined. Because the normal distribution is symmetric, each tail holds the same proportion.

By symmetry each tail holds $2.5\%$ , so $P(X>26) \approx 0.025$ .

Final answer: about $2.5\%$ of seedlings exceed $26$ cm.

Building probabilities from the rule

The empirical rule gives more than the three headline figures. Because the curve is symmetric, you can combine bands to answer many calculator-free questions. The region from the mean to $1\sigma$ holds half of $68\%$ , that is $34\%$ . The region between $1\sigma$ and $2\sigma$ holds half of $(95-68)\%$ , that is $13.5\%$ . Beyond $2\sigma$ each tail holds $2.5\%$ , and beyond $3\sigma$ each holds about $0.15\%$ . Sketching the bell and labelling these slices lets you assemble probabilities such as $P(\mu+\sigma\le X\le\mu+2\sigma)\approx 0.135$ without a calculator.

Combining empirical-rule bands

IQ scores are normal with $\mu=100$ and $\sigma=15$ . Estimate the proportion scoring between $115$ and $130$ .

Step 1: Express both boundaries in standard deviations

To use the empirical rule, convert both limits to multiples of $\sigma$ from the mean.

$115 = 100 + 1\times 15 = \mu + \sigma$ and $130 = 100 + 2\times 15 = \mu + 2\sigma$ .

Step 2: Find the area between the two bands by subtraction

The empirical rule gives total areas within $1\sigma$ and $2\sigma$ . The band between $1\sigma$ and $2\sigma$ on one side is found by subtracting the central proportions and halving for symmetry.

The region within $2\sigma$ holds $95\%$ and within $1\sigma$ holds $68\%$ , so the two-sided gap is $(95-68)\% = 27\%$ . Half of this lies on the positive side (between $\mu+\sigma$ and $\mu+2\sigma$ ).

P(115\le X\le 130)\approx \frac{27\%}{2} = 0.135

Final answer: about $13.5\%$ of scores fall in this band.

Comparing distributions

Because $\sigma$ sets the width, two normal distributions with the same mean but different standard deviations look very different: the smaller $\sigma$ concentrates values near the mean. The empirical rule helps compare them: a value $2\sigma$ above the mean is in the top $2.5\%$ of either distribution, regardless of the actual numbers.

Common errors

Confusing $\sigma$ with $\sigma^{2}$: The notation $N(\mu,\sigma^{2})$ uses the variance; the standard deviation is its square root.
Applying the rule to non-multiples of $\sigma$: The 68-95-99.7 figures apply at exactly $1$ , $2$ and $3$ standard deviations; for other boundaries you standardise and use the calculator.
Forgetting to halve the tail: The $5\%$ outside $2\sigma$ is split between two tails, so each holds $2.5\%$ .
Assuming all bell shapes are normal: A normal curve has the specific symmetric form with the empirical proportions; not every mound-shaped distribution qualifies.

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-style5 marks

Calculator-free. The masses of apples are normally distributed with mean

\mu=150

g and standard deviation

\sigma=10

g. Using the empirical rule, find (a)

P(140\le X\le 160)

, (b)

P(X>170)

and (c) the proportion between

130

g and

150

Show worked answer →

A calculator-free empirical-rule question.

(a) $140=\mu-\sigma$ and $160=\mu+\sigma$ , so this is the central $1\sigma$ band: $P\approx 0.68$ .

(b) $170=\mu+2\sigma$ . Beyond $2\sigma$ is $5\%$ split into two tails, so the upper tail is $P(X>170)\approx 0.025$ .

(c) $130=\mu-2\sigma$ and $150=\mu$ . From $-2\sigma$ to the mean is half of the central $95\%$ band, so $P\approx 0.475$ .

Markers reward expressing each boundary as a multiple of $\sigma$ and using symmetry.

WACE 2023-style4 marks

Calculator-free. A normal distribution has

P(X<\mu)=0.5

and

\sigma=4

. Given

\mu=30

, find (a)

P(X>34)

using the empirical rule and (b) the value

3

standard deviations above the mean.

Show worked answer →

A short symmetry-and-parameters question.

(a) $34=\mu+\sigma=30+4$ . Within $1\sigma$ is $68\%$ , leaving $32\%$ in the two tails, so the upper tail is $P(X>34)\approx 0.16$ .

(b) $\mu+3\sigma=30+3(4)=42$ .

Markers reward identifying $34$ as $1\sigma$ above the mean, halving the remaining $32\%$ , and the $3\sigma$ value.

The normal distribution and its properties: WACE Year 12 Mathematics Methods Unit 4

What this dot point is asking

Defining properties

The empirical rule

Step 1: Express the boundary as a multiple of $\sigma$

Step 2: Apply the empirical rule and use symmetry

Building probabilities from the rule

Step 1: Express both boundaries in standard deviations

Step 2: Find the area between the two bands by subtraction

Comparing distributions

Exam-style questions

Practise this

Sources & how we know this

What this dot point is asking

Defining properties

The empirical rule

Step 1: Express the boundary as a multiple of σ\sigmaσ

Step 2: Apply the empirical rule and use symmetry

Building probabilities from the rule

Step 1: Express both boundaries in standard deviations

Step 2: Find the area between the two bands by subtraction

Comparing distributions

Exam-style questions

Practise this

Sources & how we know this

Step 1: Express the boundary as a multiple of $\sigma$