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.

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 4.79 min answerUpdated 2026-05-29

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What this dot point is asking
Probability density functions
The normal distribution
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

Find $k$ and a probability

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

The total area must equal 1.

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

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

Now find the required probability.

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}

So $P(X \le 1) = 0.25$ . Markers reward setting total area equal to 1 to solve for $k$ , then integrating over the correct interval.

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.

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.