Z-scores and normal probabilities (SACE Stage 2 Mathematical Methods)

Syllabus dot point

“A z-score standardises a normal value to the standard normal distribution, enabling exact probabilities to be found by technology or tables.”

8 min readUpdated 2 Jun 2026

Quick answer

A z-score $z=\dfrac{x-\mu}{\sigma}$ measures how many standard deviations a value $x$ is from the mean. It converts any normal distribution to the standard normal $Z\sim N(0,1)$ , so probabilities can be found exactly from technology or tables. A positive $z$ is above the mean; a negative $z$ is below.

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What this dot point is asking
The z-score
Finding probabilities
The inverse problem
Finding an unknown mean or standard deviation
Reading the right area
Common errors
Why it matters
Exam-style questions

What this dot point is asking

The empirical rule only handles whole-number multiples of $\sigma$ . For any value, you standardise it to a z-score and read the probability from the standard normal distribution $N(0,1)$ .

The z-score

Standardising a value

z=\frac{x-\mu}{\sigma}.

The z-score is the number of standard deviations

x

lies above (

z>0

) or below (

z<0

) the mean.

Standardising maps $X\sim N(\mu,\sigma^2)$ onto $Z\sim N(0,1)$ , which has mean $0$ and standard deviation $1$ . Equal z-scores represent equally extreme positions in different distributions, so z-scores let you compare values from different normal distributions.

Comparing two test results

Anna scored $78$ on a test with mean $70$ and $\sigma=5$ , while Ben scored $84$ on a different test with mean $75$ and $\sigma=6$ . This example uses z-scores to determine who performed relatively better.

Step 1: standardise Anna's score

A z-score measures how many standard deviations a result sits above or below its test's mean, placing results from different tests on a common scale. For Anna:

z_{\text{Anna}}=\frac{78-70}{5}=\frac{8}{5}=1.6.

Step 2: standardise Ben's score

Applying the same formula to Ben's result on his test:

z_{\text{Ben}}=\frac{84-75}{6}=\frac{9}{6}=1.5.

Step 3: compare the z-scores

Both z-scores are positive, meaning both students scored above their respective test means. Anna's $z=1.6$ exceeds Ben's $z=1.5$ , so Anna is further above her test's mean in standard-deviation terms and performed relatively better.

Final answer: Anna performed relatively better, with $z=1.6$ compared to Ben's $z=1.5$ .

Finding probabilities

In Stage 2 you use a calculator's normal CDF (or the standard normal table) to find the area. The standardisation tells you which area to look up.

Probability below a value

A machine fills bottles with volume $X\sim N(500,\,4^2)$ mL. This example finds the probability that a bottle contains less than $494$ mL.

Step 1: standardise 494 mL to a z-score

To use the standard normal tables or calculator, we convert $494$ mL to a z-score by subtracting the mean and dividing by the standard deviation. A negative result means the value lies below the mean:

z=\frac{494-500}{4}=\frac{-6}{4}=-1.5.

Step 2: look up the standard normal probability

The event $P(X<494)$ is equivalent to $P(Z<-1.5)$ under the standard normal distribution. Reading this left-tail area from tables or a calculator:

P(Z<-1.5)\approx 0.0668.

Final answer: approximately $6.7\%$ of bottles contain less than $494$ mL.

Probability between two values

For the same bottles $N(500,4^2)$ , this example finds $P(497\le X\le 506)$ , the probability a bottle volume falls between two values.

Step 1: standardise both endpoints

A between-probability requires converting each boundary to a z-score. Subtracting the mean and dividing by $\sigma=4$ for each endpoint:

z_1=\frac{497-500}{4}=-0.75, \qquad z_2=\frac{506-500}{4}=1.5.

Step 2: find each cumulative probability from the standard normal

The calculator or tables give left-tail (cumulative) areas. Looking up both z-scores:

P(Z<1.5)\approx 0.9332 \quad \text{and} \quad P(Z<-0.75)\approx 0.2266.

Step 3: subtract to get the between-probability

The area between the two z-scores is the difference of the two cumulative areas:

P(497\le X\le 506)=P(-0.75\le Z\le 1.5)=0.9332-0.2266=0.7066.

Final answer: approximately $70.7\%$ of bottles have a volume between $497$ mL and $506$ mL.

The inverse problem

Sometimes you know the probability and want the value - the inverse normal. Find the z-score for that area, then unstandardise with $x=\mu+z\sigma$ .

Inverse normal: finding a cut-off

Exam marks are $N(60,12^2)$ . The top $10\%$ receive a distinction. This example finds the minimum mark required for a distinction.

Step 1: convert the probability statement to a cumulative area

The condition "top $10\%$ " means $P(X>x)=0.10$ , which is a right-tail area. Because tables and calculators give left-tail cumulative areas, we rewrite this as a left-tail condition: $P(X\le x)=0.90$ .

Step 2: find the z-score for the 90th percentile

Using the inverse standard normal function with cumulative area $0.90$ , the corresponding z-score is:

z\approx 1.2816.

Step 3: unstandardise to get the mark

The z-score formula $z=\tfrac{x-\mu}{\sigma}$ rearranges to $x=\mu+z\sigma$ . Substituting the known values:

x=60+1.2816\times 12\approx 60+15.4=75.4.

Final answer: a mark of approximately $75.4$ (so $76$ in whole marks) is needed for a distinction.

Finding an unknown mean or standard deviation

The standardisation formula can be rearranged to find $\mu$ or $\sigma$ when a probability is given. If you know that $P(X<x)=p$ for a particular $x$ , find the matching $z$ from the inverse normal, then solve $z=\dfrac{x-\mu}{\sigma}$ for the unknown parameter. For example, if a normal variable has $\sigma=5$ and $P(X<30)=0.84$ , then $z\approx 1.0$ , so $1.0=\dfrac{30-\mu}{5}$ , giving $\mu=25$ . Two such conditions give two equations and let you solve for both $\mu$ and $\sigma$ simultaneously, a standard higher-tariff SACE question.

Reading the right area

Every normal probability question reduces to one of three area types: a left area $P(Z<z)$ , a right area $P(Z>z)=1-P(Z<z)$ , or a between area $P(z_1<Z<z_2)=P(Z<z_2)-P(Z<z_1)$ . The calculator returns left areas by default, so right and between probabilities are built from them by subtraction. Deciding which type you need is the single most important step, which is why a quick sketch of the shaded region is worth the few seconds it takes.

Common errors

Sign errors in z and forgetting to unstandardise

A value below the mean has a negative z-score - dropping the minus sign flips the probability to the wrong tail. In inverse problems, students often find $z$ but forget the final step $x=\mu+z\sigma$ , reporting the z-score as the answer. Always convert back to the original units.

Sketch and shade the required region

Draw the bell curve, mark the mean and the value(s), and shade the region the probability refers to. This decides whether you need the left area, the right area ( $1-$ left), or a difference of two areas - the most common source of full-versus-zero marks in normal questions.

Why it matters

Z-scores are the exact-probability engine of Topic 5 and a guaranteed exam skill, both forward (value to probability) and inverse (probability to value). Standardisation is also the mechanism behind Topic 6's confidence intervals, where critical z-values such as $1.96$ come straight from inverse-normal reasoning.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2024-style2 marks

Calculator-assumed. Let

X

be normally distributed with mean

54

and standard deviation

6

. Given that

\Pr(X > k) = 0.10

, determine the value of

k

Show worked answer →

This is an inverse-normal problem. $\Pr(X > k) = 0.10$ means $\Pr(X \le k) = 0.90$ .

The $z$ -score with $0.90$ of the area below it is $z = 1.2816$ (the 90th percentile of the standard normal).

Unstandardise with $x = \mu + z\sigma$ :

k = 54 + 1.2816 \times 6 = 54 + 7.69 \approx 61.7.

So $k \approx 61.7$ . Sense check: only $10\%$ exceed $k$ , so $k$ sits well above the mean $54$ . Marks: one for $z = 1.2816$ , one for unstandardising to $61.7$ .

SACE 2023-style2 marks

Calculator-assumed. The daily milk production of a cow is normally distributed with mean

21.9

litres and standard deviation

3.26

litres. Given that

15\%

of cows produce

c

litres or more, determine

c

Show worked answer →

" $15\%$ produce $c$ litres or more" means $\Pr(X \ge c) = 0.15$ , so $\Pr(X < c) = 0.85$ .

The $z$ -score for the 85th percentile is $z = 1.0364$ .

Unstandardise with $c = \mu + z\sigma$ :

c = 21.9 + 1.0364 \times 3.26 = 21.9 + 3.38 \approx 25.3 \text{ litres}.

So $c \approx 25.3$ litres. Marks: one for $z = 1.0364$ , one for unstandardising. Sense check: only the top $15\%$ exceed $c$ , so it lies above the mean.

SACE 2022-style2 marks

Calculator-free. Two students sit different tests. Mai scores

82

on a test with mean

74

and standard deviation

8

. Tom scores

66

on a test with mean

60

and standard deviation

4

. Use

z

-scores to determine who performed relatively better.

Show worked answer →

Standardise each: $z_{\text{Mai}} = \dfrac{82 - 74}{8} = 1.0$ and $z_{\text{Tom}} = \dfrac{66 - 60}{4} = 1.5$ .

Tom's $z = 1.5$ exceeds Mai's $z = 1.0$ , so Tom performed relatively better - he is further above his test's mean in standard-deviation terms.

Marks: one for both $z$ -scores, one for the correct conclusion with justification.

Practise this

Sources & how we know this

Stage 2 Mathematical Methods subject outline (2025 onwards) — SACE Board
Stage 2 Mathematical Methods: external assessment — SACE Board