Standardisation and normal probability calculations: WACE Year 12 Mathematics Methods Unit 4

Syllabus dot point

“Standardise a normal variable to a z-score and calculate normal probabilities, including finding a value from a given probability”

8 min readUpdated 2 Jun 2026

Quick answer

Standardise with $z=\dfrac{x-\mu}{\sigma}$ to convert any normal variable to the standard normal $Z\sim N(0,1)$ . Then find probabilities from the standard normal, and for inverse problems reverse the formula as $x=\mu+z\sigma$ to recover a value from a given probability.

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What this dot point is asking
Standardising
Using symmetry
Inverse problems
Comparing values from different distributions
Choosing the direction
Exam-style questions

What this dot point is asking

SCSA Unit 4 uses standardisation to compute probabilities for any normal distribution. This dot point asks you to convert to a $z$ -score, find probabilities (often with the calculator), and solve inverse problems where a value is found from a given probability. It is examined in both sections.

Standardising

A $z$ -score measures how many standard deviations a value lies from the mean.

Standardisation

z = \frac{x-\mu}{\sigma}, \qquad x = \mu + z\sigma.

A positive

z

is above the mean, a negative

z

below. The standardised variable follows

Z\sim N(0,1)

Standardising lets one standard normal handle every normal distribution, and is the basis for both table-based and calculator-based probability work.

A normal probability via standardisation

Probability between two values

Test marks are normal with $\mu=60$ and $\sigma=12$ . Find $P(48\le X\le 72)$ .

Standardise both boundaries.

z_1 = \frac{48-60}{12} = -1, \qquad z_2 = \frac{72-60}{12} = 1.

So $P(48\le X\le 72)=P(-1\le Z\le 1)$ , the central region within one standard deviation.

By the empirical rule this is about $0.68$ (the calculator gives $0.6827$ ). Markers reward standardising both ends and recognising the symmetric central region.

Using symmetry

The standard normal is symmetric about $0$ , so $P(Z<-a)=P(Z>a)$ and $P(Z>a)=1-P(Z<a)$ . These identities convert any tail or interval probability into a form the calculator or the empirical rule handles, and they are essential in the calculator-free section.

Inverse problems

An inverse problem gives a probability and asks for the corresponding value of $X$ . Find the $z$ -score matching that cumulative probability, then unstandardise with $x=\mu+z\sigma$ .

Finding a value from a probability

For marks with $\mu=60$ and $\sigma=12$ , find the mark exceeded by the top $10\%$ .

Step 1: Convert the tail probability to a $z$ -score

We are looking for the $z$ such that $10\%$ of the distribution lies above it, which means $90\%$ lies below. Reading the standard normal table (or calculator) at cumulative probability $0.90$ gives the required $z$ .

The top $10\%$ means $P(Z>z)=0.10$ , so $P(Z<z)=0.90$ . The matching $z$ -score is about $z=1.2816$ .

Step 2: Unstandardise to recover the original scale

The inverse of the standardisation formula is $x = \mu + z\sigma$ . We scale the $z$ -score back to the original units using the given mean and standard deviation.

x = \mu + z\sigma = 60 + 1.2816\times 12 \approx 60 + 15.38 = 75.38

Final answer: a mark of about $75.4$ is the cutoff for the top $10\%$ .

Comparing values from different distributions

A $z$ -score is a universal measure of relative standing because it strips away the original units. A student who scores $z=1.5$ in English and $z=0.8$ in Mathematics performed more strongly in English relative to that cohort, even if the raw marks were higher in Mathematics. SCSA sometimes frames a question this way, asking which of two results is more impressive; the answer is whichever has the larger $z$ -score, since both are then measured on the same standard scale.

An inverse problem with a central interval

Symmetric limits about the mean

Reaction times are normal with $\mu=0.30$ s and $\sigma=0.04$ s. Find the symmetric interval about the mean containing the central $95\%$ of times.

The central $95\%$ corresponds to $z=\pm 1.96$ (the standard normal value cutting off $2.5\%$ in each tail). Unstandardise both ends:

x=\mu\pm 1.96\sigma=0.30\pm 1.96(0.04)=0.30\pm 0.0784.

So the central

95\%

of reaction times lie between about

0.222

s and

0.378

s. Markers reward the

z=\pm 1.96

values and unstandardising symmetrically about the mean.

Choosing the direction

The most common slip is solving the wrong inequality. Sketch the bell, shade the region the probability describes, and check whether the required $z$ is positive (above the mean) or negative (below) before unstandardising.

Common errors

Dividing by $\sigma^{2}$: Standardisation divides by the standard deviation $\sigma$ , not the variance $\sigma^{2}$ .
Wrong tail in an inverse problem: The top $10\%$ corresponds to a cumulative probability of $0.90$ , giving a positive $z$ ; the bottom $10\%$ gives a negative $z$ .
Sign error on the $z$ -score: A value below the mean has a negative $z$ ; keep track using a sketch.
Forgetting to unstandardise: Inverse problems ask for $x$ , so finish with $x=\mu+z\sigma$ , not the $z$ -score itself.

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 2022-style6 marks

Calculator-assumed. The lifetime of a battery is normally distributed with mean

\mu=40

hours and standard deviation

\sigma=5

hours. (a) Find

P(X>48)

. (b) Find

P(35\le X\le 45)

. (c) Find the lifetime exceeded by

90\%

of batteries.

Show worked answer →

A standardisation question with an inverse part.

(a) $z=\dfrac{48-40}{5}=1.6$ . $P(X>48)=P(Z>1.6)\approx 0.0548$ .

(b) $z_1=\dfrac{35-40}{5}=-1$ , $z_2=\dfrac{45-40}{5}=1$ . $P(-1\le Z\le 1)\approx 0.6827$ .

(c) Exceeded by $90\%$ means $P(X>x)=0.90$ , so $P(Z<z)=0.10$ , giving $z\approx -1.2816$ . Then $x=\mu+z\sigma=40-1.2816(5)\approx 33.6$ hours.

Markers reward standardising each value, the calculator probabilities, and unstandardising in (c) with a negative $z$ .

WACE 2024-style4 marks

Calculator-free. For test scores with

\mu=70

and

\sigma=8

, a student scores

86

. (a) Find the student's

z

-score. (b) Using the empirical rule, estimate the proportion of students who scored higher.

Show worked answer →

A calculator-free standardisation question.

(a) $z=\dfrac{86-70}{8}=\dfrac{16}{8}=2$ .

(b) $86$ is $2\sigma$ above the mean. Beyond $2\sigma$ is $5\%$ in two tails, so the upper tail is about $2.5\%$ . About $0.025$ of students scored higher.

Markers reward the $z$ -score of $2$ and using the empirical-rule tail of $2.5\%$ .

Standardisation and normal probability calculations: WACE Year 12 Mathematics Methods Unit 4

What this dot point is asking

Standardising

Probability between two values

Using symmetry

Inverse problems

Step 1: Convert the tail probability to a $z$ -score

Step 2: Unstandardise to recover the original scale

Comparing values from different distributions

Symmetric limits about the mean

Choosing the direction

Exam-style questions

Practise this

Sources & how we know this

What this dot point is asking

Standardising

Probability between two values

Using symmetry

Inverse problems

Step 1: Convert the tail probability to a zzz-score

Step 2: Unstandardise to recover the original scale

Comparing values from different distributions

Symmetric limits about the mean

Choosing the direction

Exam-style questions

Practise this

Sources & how we know this

Step 1: Convert the tail probability to a $z$ -score