Probability and statistics (WACE Year 12 Mathematics Applications)

Syllabus dot point

“Use the normal distribution and the 68-95-99.7 rule, standardise to z-scores, and construct and interpret sample proportions and confidence intervals.”

9 min readUpdated 2 Jun 2026

Quick answer

The normal distribution is the symmetric bell curve described by its mean $\mu$ and standard deviation $\sigma$ . The 68-95-99.7 rule gives the percentage of data within $1$ , $2$ and $3$ standard deviations of the mean. A z-score, $z = \dfrac{x - \mu}{\sigma}$ , says how many standard deviations a value lies from the mean. A sample proportion $\hat{p}$ estimates a population proportion, and a confidence interval gives a range likely to contain the true value.

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What this dot point is asking
The normal distribution and the empirical rule
z-scores
Sample proportions and confidence intervals
Margin of error and sample size
Working backwards through a normal distribution
Exam-style questions

What this dot point is asking

You must apply the normal distribution and empirical rule, compute z-scores, and build and interpret confidence intervals for a proportion.

The normal distribution and the empirical rule

The normal distribution is a symmetric bell-shaped curve centred at its mean $\mu$ , with spread set by its standard deviation $\sigma$ .

The 68-95-99.7 rule

For normally distributed data:

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

Because the curve is symmetric, each interval splits evenly about the mean, so for example $34\%$ lies between the mean and $+1\sigma$ . Sketching the bell curve and marking the standard-deviation lines is the safest way to read off percentages.

z-scores

A z-score standardises a value by stating how many standard deviations it is from the mean.

z-score

$z = \dfrac{x - \mu}{\sigma}$ , where $x$ is the value, $\mu$ the mean and $\sigma$ the standard deviation.

A positive z-score is above the mean, negative below. z-scores let you compare values from different normal distributions: a score with $z = 1.5$ in one test is more impressive than $z = 0.8$ in another, regardless of the raw marks.

Worked example

Using the empirical rule and z-scores

Exam marks are normally distributed with mean $\mu = 64$ and standard deviation $\sigma = 8$ . Find the percentage of students scoring above $80$ , and the z-score for a mark of $54$ .

Step 1: Express 80 as a number of standard deviations from the mean.

Before applying the empirical rule we must identify where $80$ sits relative to the mean. We have $80 = 64 + 16 = \mu + 2\sigma$ , so $80$ lies exactly two standard deviations above the mean. Locating the value on the standard-deviation scale is the key first step for any empirical-rule problem.

Step 2: Use the empirical rule to find the percentage above 80.

The 68-95-99.7 rule tells us that $95\%$ of marks lie within $\mu \pm 2\sigma$ , meaning $5\%$ lies outside this interval altogether. Because the normal distribution is symmetric, that $5\%$ splits equally between the two tails: $2.5\%$ below $\mu - 2\sigma$ and $2.5\%$ above $\mu + 2\sigma$ . Therefore about $2.5\%$ of students score above $80$ .

Step 3: Calculate the z-score for a mark of 54.

A z-score converts a raw value to a number of standard deviations from the mean using $z = \dfrac{x - \mu}{\sigma}$ . Substituting the values: $z = \dfrac{54 - 64}{8} = \dfrac{-10}{8} = -1.25$ . The negative sign shows the mark is below the mean: the mark of $54$ is $1.25$ standard deviations below the mean.

Sample proportions and confidence intervals

A sample proportion $\hat{p} = \dfrac{\text{number with the trait}}{n}$ estimates the unknown population proportion $p$ . Because samples vary, we report a range.

Confidence interval for a proportion

A confidence interval is $\hat{p} \pm z\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}}$ , where $z = 1.96$ for $95\%$ confidence (and $z = 2.58$ for $99\%$ ). The quantity $\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ is the standard error, and $z$ times it is the margin of error.

The interpretation is about the population: "we are $95\%$ confident the true population proportion lies between the lower and upper limits". A larger sample $n$ shrinks the margin of error, giving a narrower, more precise interval.

Margin of error and sample size

The margin of error is $E = z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ . Two levers change it: the confidence level (a higher level uses a larger $z$ , widening the interval) and the sample size (a larger $n$ shrinks the margin). Because $n$ sits under a square root, quartering the margin requires multiplying the sample size by about $16$ , so improving precision gets expensive quickly. SCSA questions often ask you to state the effect of doubling the sample size: the margin shrinks by a factor of $\sqrt{2} \approx 1.41$ , not by half.

Working backwards through a normal distribution

As well as finding a percentage from a value, you may be asked for the value at a given percentile. Use the empirical rule in reverse: the value below which $97.5\%$ of a normal population lies is $\mu + 2\sigma$ (since $2.5\%$ lies above it), and the value below which $16\%$ lies is $\mu - \sigma$ (since $34\%$ lies between $\mu - \sigma$ and the mean, leaving $16\%$ below). For percentiles that do not land on whole standard deviations, you rearrange the z-score formula: $x = \mu + z\sigma$ , reading $z$ from the required tail area. This connects the normal model to the confidence-interval $z$ values of $1.96$ and $2.58$ .

Common errors

Forgetting to halve the tail: $95\%$ within $\pm 2\sigma$ leaves $5\%$ split as $2.5\%$ in each tail; the upper tail alone is $2.5\%$ , not $5\%$ .
Sign error in the z-score: A value below the mean has a negative z-score; keep the sign.
Using $n$ instead of the standard error: The margin uses $\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ , not $\hat{p}/n$ or $1/n$ .
Interpreting the interval as about the sample: A confidence interval estimates the population proportion, not the sample proportion (which is already known exactly).

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-style6 marks

The heights of a population of plants are normally distributed with mean

\mu = 80

cm and standard deviation

\sigma = 6

cm. (a) Use the 68-95-99.7 rule to find the percentage of plants between

74

cm and

92

cm. (b) Calculate the z-score for a plant of height

89

cm and interpret it.

Show worked answer →

Locate each value as a number of standard deviations from the mean.

(a) $74$ cm is $1$ standard deviation below the mean ( $80 - 6$ ), and $92$ cm is $2$ standard deviations above ( $80 + 12$ ). The percentage from $-1\sigma$ to $+2\sigma$ is half of the $68\%$ (that is $34\%$ ) plus half of the $95\%$ (that is $47.5\%$ ), giving $34 + 47.5 = 81.5\%$ . (4 marks)

(b) $z = \dfrac{89 - 80}{6} = \dfrac{9}{6} = 1.5$ . The plant is $1.5$ standard deviations above the mean. (2 marks)

Markers reward splitting the empirical rule by half-intervals and a correct z-score with interpretation.

WACE 2023-style6 marks

In a sample of

400

voters,

240

support a policy. (a) Find the sample proportion. (b) Construct an approximate

95\%

confidence interval for the population proportion (use

z = 1.96

). (c) Interpret the interval.

Show worked answer →

Use the sample proportion and its standard error.

(a) $\hat{p} = \dfrac{240}{400} = 0.60$ . (1 mark)

(b) Standard error $= \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\dfrac{0.6 \times 0.4}{400}} = \sqrt{0.0006} = 0.0245$ . Margin $= 1.96 \times 0.0245 = 0.048$ . Interval $= 0.60 \pm 0.048 = (0.552, 0.648)$ . (4 marks)

(c) We are $95\%$ confident the true proportion of all voters who support the policy lies between $55.2\%$ and $64.8\%$ . (1 mark)

Markers reward the sample proportion, the standard-error and margin calculation, and a confidence-interval interpretation about the population.