WAMath MethodsSyllabus dot point

How does a probability density function describe a continuous random variable, and how do we find probabilities from it?

Use probability density functions to find probabilities as areas, determine an unknown constant, and compute the mean and variance by integration

WACE Year 12 Mathematics Methods Unit 4 probability density functions: the validity conditions, probability as area, finding an unknown constant, the median, and mean and variance by integration, with worked SCSA-style examples.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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What this dot point is asking
Validity conditions
Finding an unknown constant
Mean and variance by integration
The median

What this dot point is asking

SCSA Unit 4 introduces continuous random variables, where outcomes fill an interval rather than a list. This dot point asks you to verify a probability density function (pdf), find probabilities and an unknown constant, and compute the mean and variance by integration. It is examined in both sections.

Validity conditions

Because area over a single point is zero, $P(X=a)=0$ for any value $a$ , so $P(a\le X\le b)=P(a<X<b)$ . The inequality endpoints do not change a continuous probability.

Finding an unknown constant

If a pdf contains an unknown coefficient, the total-area condition determines it.

Finding the constant and a probability

A pdf on $[0,3]$

$X$ has $f(x)=k x^{2}$ for $0\le x\le 3$ and $0$ otherwise. Find $k$ , then $P(X\le 2)$ .

Set the total area to $1$ .

\int_0^3 k x^{2}\,dx = k\left[\frac{x^{3}}{3}\right]_0^3 = k\cdot 9 = 1,

so $k=\dfrac{1}{9}$ .

Now the probability.

P(X\le 2) = \int_0^2 \frac{1}{9}x^{2}\,dx = \frac{1}{9}\left[\frac{x^{3}}{3}\right]_0^2 = \frac{1}{9}\cdot\frac{8}{3} = \frac{8}{27}.

So $P(X\le 2)=\dfrac{8}{27}\approx 0.296$ . Markers reward the area-equals-one equation and integrating over the correct interval.

Mean and variance by integration

The discrete sums become integrals for a continuous variable.

The median

The median $m$ splits the area in half: $\int_{-\infty}^{m} f(x)\,dx = 0.5$ . For a symmetric pdf the median equals the mean, but for a skewed density such as the one above they differ, and each must be found from its own integral equation.