What is sign of the z-score?

A value below the mean gives a negative z. Using symmetry, P(Z a), so keep track of which tail you want.

SCSA WACE 2021-style practice: Calculator-assumed. The volume of soft drink in a can is normally distributed with mean μ=375 mL and standard deviation σ=4 mL. (a) Find P(X<370). (b) Find P(372 X 378). (c) Cans below a certain volume are rejected; the bottom 2\% are rejected. Find the rejection volume.

A normal-distribution question with an inverse part. (a) z=370-3754=-1.25. P(X<370)=P(Z<-1.25)≈ 0.106. (b) z_1=372-3754=-0.75, z_2=378-3754=0.75. P(-0.75 Z 0.75)≈ 0.547. (c) Bottom 2\%: P(Z<z)=0.02, so z≈ -2.054. Then x=μ+zσ=375-2.054(4)≈ 366.8 mL. Markers reward standardising, the calculator probabilities, and unstandardising with a negative z in (c).

SCSA WACE 2023-style practice: Calculator-assumed. A continuous random variable has pdf f(x)=34(1-x^2) for -1 x 1. (a) Verify the total area is 1. (b) Find P(0 X 0.5). (c) State the mean, justifying by symmetry.

A pdf question combining integration and symmetry. (a) _-1^134(1-x^2)\,dx=34[x-x^33]_-1^1=34(23+23)=34·43=1. (b) _0^0.534(1-x^2)\,dx=34[x-x^33]_0^0.5=34(0.5-0.1253)≈ 0.344. (c) The pdf is even (symmetric about x=0), so the mean is E(X)=0. Markers reward the area check, the definite integral, and a symmetry justification for the mean.

§-Syllabus dot point

WAMath MethodsSyllabus dot point

How do we model continuous data using probability density functions and the normal distribution?

Use probability density functions to find probabilities, mean and variance for continuous random variables, and apply the normal distribution with standardisation

WACE Year 12 Mathematics Methods Unit 4 continuous random variables: probability density functions, mean and variance by integration, the normal distribution, standardisation with z-scores and the 68-95-99.7 rule, with worked SCSA-style examples.

Generated by Claude Opus 4.89 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
Probability density functions
The normal distribution
Why standardisation works
Standardisation and z-scores

What this dot point is asking

SCSA Unit 4 moves from the discrete distributions of Unit 3 to continuous random variables, where outcomes form an interval rather than a list. Probabilities become areas under a probability density function (pdf), and the normal distribution is the central continuous model examined in the WACE written examination.

Probability density functions

For a continuous variable, the probability of any single value is 0, so $P(X = a) = 0$ . Probabilities are areas:

P(a \le X \le b) = \int_a^b f(x)\,dx.

Because single points have probability 0,

P(a \le X \le b) = P(a < X < b)

The mean (expected value) and variance are found by integration:

\mu = E(X) = \int_{-\infty}^{\infty} x\,f(x)\,dx, \qquad \mathrm{Var}(X) = \int_{-\infty}^{\infty} (x - \mu)^2 f(x)\,dx = E(X^2) - \mu^2.

Working with a pdf

The continuous random variable $X$ has pdf $f(x) = kx$ for $0 \le x \le 2$ and $f(x) = 0$ otherwise. Find $k$ , then $P(X \le 1)$ .

Step 1: Use the total-area condition to find $k$

Every valid pdf must integrate to 1 over its support. Setting this integral equal to 1 is the equation we solve for the unknown constant $k$ .

\int_0^2 kx\,dx = k\left[\frac{x^2}{2}\right]_0^2 = k \cdot \frac{4}{2} = 2k = 1

So $k = \dfrac{1}{2}$ .

Step 2: Compute the probability by integration

A probability for a continuous variable equals the area under the pdf over the required interval. We integrate the now-known pdf from $0$ to $1$ .

P(X \le 1) = \int_0^1 \frac{1}{2}x\,dx = \frac{1}{2}\left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}

Final answer: $k = \dfrac{1}{2}$ and $P(X \le 1) = 0.25$ .

The normal distribution

The normal distribution $X \sim N(\mu, \sigma^2)$ has a symmetric bell-shaped pdf centred at the mean $\mu$ with standard deviation $\sigma$ . Its key features:

It is symmetric about $x = \mu$ , so $P(X < \mu) = P(X > \mu) = 0.5$ .
The total area under the curve is 1.
Larger $\sigma$ gives a wider, flatter curve.

Why standardisation works

Every normal distribution has the same shape; only its centre and width differ. Subtracting the mean shifts the centre to $0$ , and dividing by the standard deviation rescales the width to $1$ , producing the single standard normal $Z\sim N(0,1)$ . Because this transformation preserves areas, a probability for any normal variable equals the matching probability for $Z$ , which is why one set of standard-normal values (or one calculator function) serves every normal model. This is the practical payoff of the symmetry and two-parameter structure described above.

Standardisation and z-scores

To compare values from different normal distributions, or to use tables and the calculator, convert to the standard normal $Z \sim N(0, 1)$ :

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

The

z

-score is the number of standard deviations a value lies above (positive) or below (negative) the mean.

A normal probability with standardisation

Test scores are normally distributed with $\mu = 70$ and $\sigma = 8$ . Find the proportion of students scoring above $86$ .

Step 1: Standardise the boundary value

Converting $x = 86$ to a $z$ -score tells us how many standard deviations above the mean it sits, so we can use the standard normal distribution.

z = \frac{86 - 70}{8} = \frac{16}{8} = 2

Step 2: Apply the empirical rule to find the tail probability

Since $z = 2$ , we need $P(Z > 2)$ . By the 68-95-99.7 rule, $95\%$ of values lie within $2$ standard deviations, leaving $5\%$ in the two tails combined. Symmetry splits this equally, so each tail holds $2.5\%$ .

P(X > 86) = P(Z > 2) \approx 0.025

Final answer: about $2.5\%$ of students score above $86$ . (In the calculator section you would obtain $P(Z > 2) \approx 0.0228$ directly.)

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 20217 marksCalculator-assumed. The volume of soft drink in a can is normally distributed with mean

\mu=375

mL and standard deviation

\sigma=4

mL. (a) Find

P(X<370)

. (b) Find

P(372\le X\le 378)

. (c) Cans below a certain volume are rejected; the bottom

2\%

are rejected. Find the rejection volume.

Show worked answer →

A normal-distribution question with an inverse part.

(a) $z=\dfrac{370-375}{4}=-1.25$ . $P(X<370)=P(Z<-1.25)\approx 0.106$ .

(b) $z_1=\dfrac{372-375}{4}=-0.75$ , $z_2=\dfrac{378-375}{4}=0.75$ . $P(-0.75\le Z\le 0.75)\approx 0.547$ .

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

WACE 20236 marksCalculator-assumed. A continuous random variable has pdf

f(x)=\dfrac{3}{4}(1-x^{2})

for

-1\le x\le 1

. (a) Verify the total area is

1

. (b) Find

P(0\le X\le 0.5)

. (c) State the mean, justifying by symmetry.

Show worked answer →

A pdf question combining integration and symmetry.

(a) $\displaystyle\int_{-1}^{1}\dfrac{3}{4}(1-x^{2})\,dx=\dfrac{3}{4}\left[x-\dfrac{x^{3}}{3}\right]_{-1}^{1}=\dfrac{3}{4}\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{3}{4}\cdot\dfrac{4}{3}=1$ .

(b) $\displaystyle\int_{0}^{0.5}\dfrac{3}{4}(1-x^{2})\,dx=\dfrac{3}{4}\left[x-\dfrac{x^{3}}{3}\right]_{0}^{0.5}=\dfrac{3}{4}\left(0.5-\dfrac{0.125}{3}\right)\approx 0.344$ .

Markers reward the area check, the definite integral, and a symmetry justification for the mean.