What is 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.

What is sign error on the z-score?

A value below the mean has a negative z; keep track using a sketch.

SCSA WACE 2022-style practice: Calculator-assumed. The lifetime of a battery is normally distributed with mean μ=40 hours and standard deviation σ=5 hours. (a) Find P(X>48). (b) Find P(35 X 45). (c) Find the lifetime exceeded by 90\% of batteries.

A standardisation question with an inverse part. (a) z=48-405=1.6. P(X>48)=P(Z>1.6)≈ 0.0548. (b) z_1=35-405=-1, z_2=45-405=1. P(-1 Z 1)≈ 0.6827. (c) Exceeded by 90\% means P(X>x)=0.90, so P(Z<z)=0.10, giving z≈ -1.2816. Then x=μ+zσ=40-1.2816(5)≈ 33.6 hours. Markers reward standardising each value, the calculator probabilities, and unstandardising in (c) with a negative z.

§-Syllabus dot point

WAMath MethodsSyllabus dot point

How do we standardise a normal variable and calculate normal probabilities, including inverse problems?

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

WACE Year 12 Mathematics Methods Unit 4 standardisation: converting to a z-score, computing normal probabilities, symmetry of the standard normal, and inverse problems finding a value from a probability, with worked SCSA-style examples.

Generated by Claude Opus 4.88 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
Standardising
Using symmetry
Inverse problems
Comparing values from different distributions
Choosing the direction

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.

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

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.

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.

Exam-style practice questions

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

WACE 20226 marksCalculator-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 20244 marksCalculator-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 $2.5\%$ .