Continuous random variables and PDFs (SACE Stage 2 Mathematical Methods)

Syllabus dot point

“A continuous random variable is described by a probability density function, where probability is the area under the curve found by integration.”

8 min readUpdated 2 Jun 2026

Quick answer

A continuous random variable can take any value in an interval, so probability is described by a probability density function $f(x)$ . Probabilities are areas under the curve: $P(a\le X\le b)=\int_a^b f(x)\,dx$ . A valid PDF satisfies $f(x)\ge 0$ everywhere and $\int_{-\infty}^{\infty} f(x)\,dx=1$ .

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What this dot point is asking
Probability is area
Mean and variance by integration
The median and other percentiles
The cumulative distribution function
Common errors
Why it matters
Exam-style questions

What this dot point is asking

A continuous random variable can take any value in a range - a height, a time, a mass - not just a list of separate values. Because there are infinitely many possible values, the probability of any single exact value is zero, and we describe probability by a density function instead. This is the key conceptual shift from Topic 2: where a discrete variable had a probability attached to each value, a continuous variable has probability spread smoothly across an interval, and you recover any probability by measuring area rather than reading off a table entry.

Probability density function (PDF)

A function $f(x)$ is a probability density function for a continuous random variable $X$ if $f(x)\ge 0$ for all $x$ and the total area under it is $1$ :

\int_{-\infty}^{\infty} f(x)\,dx=1.

Probability is area

The defining idea of Topic 5 is that probability equals area under the curve, computed by integration:

Probability from a PDF

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

Because a single point has zero width,

P(X=a)=0

, and so

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

- strict and inclusive inequalities give the same answer.

Checking a PDF and finding a probability

$X$ has density $f(x)=kx$ for $0\le x\le 2$ and $f(x)=0$ elsewhere. This example finds $k$ and then the probability $P(X\le 1)$ .

Step 1: find $k$ using the total-area condition

Every valid PDF must have total area equal to $1$ over its full range. Setting the integral of $f$ equal to $1$ lets us solve for the unknown constant:

\int_0^2 kx\,dx=k\Big[\tfrac{x^2}{2}\Big]_0^2=k\cdot 2=1 \quad\Rightarrow\quad k=\tfrac12.

Step 2: find $P(X\le 1)$ as area under the curve

Probability for a continuous variable is the area under the density between the limits. Now that $k=\tfrac12$ is known, we integrate $f(x)=\tfrac12 x$ from $0$ to $1$ :

P(X\le 1)=\int_0^1 \tfrac12 x\,dx=\tfrac12\Big[\tfrac{x^2}{2}\Big]_0^1=\tfrac12\cdot\tfrac12=\frac14.

Final answer: $k=\tfrac12$ and $P(X\le 1)=\tfrac14$ .

Mean and variance by integration

The discrete sums become integrals for a continuous variable:

Mean and variance of a continuous variable

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

Mean of a density

For the density $f(x)=\tfrac12 x$ on $[0,2]$ , this example finds $E(X)$ using the integral definition of the mean.

Step 1: set up the mean integral

The mean of a continuous random variable is $E(X)=\int x\,f(x)\,dx$ over the variable's range. Substituting $f(x)=\tfrac12 x$ gives the integrand $x\cdot\tfrac12 x=\tfrac12 x^2$ , which we integrate from $0$ to $2$ :

E(X)=\int_0^2 x\cdot\tfrac12 x\,dx=\int_0^2 \tfrac12 x^2\,dx.

Step 2: evaluate the integral

Using the power rule, the antiderivative of $\tfrac12 x^2$ is $\tfrac12\cdot\tfrac{x^3}{3}=\tfrac{x^3}{6}$ . Evaluating between $0$ and $2$ :

\tfrac12\Big[\tfrac{x^3}{3}\Big]_0^2=\tfrac12\cdot\tfrac{8}{3}=\frac{4}{3}\approx 1.33.

Final answer: $E(X)=\tfrac43\approx 1.33$ .

The median and other percentiles

The median $m$ of a continuous variable splits the area in half: it is the value with $\int_{-\infty}^{m} f(x)\,dx=0.5$ . More generally, the $p$ th percentile is the value with cumulative area $\dfrac{p}{100}$ below it. For the density $f(x)=\tfrac12 x$ on $[0,2]$ , the median solves $\int_0^m \tfrac12 x\,dx=0.5$ , that is $\tfrac{m^2}{4}=0.5$ , giving $m=\sqrt{2}\approx 1.41$ . Notice the median is larger than you might first guess: because the density rises toward $x=2$ , more probability sits on the right, pulling the halfway point above the midpoint of the interval. Comparing the mean ( $\tfrac43\approx 1.33$ ) with the median ( $1.41$ ) describes the skew of the distribution.

The cumulative distribution function

Just as discrete variables have a running total, a continuous variable has a cumulative distribution function $F(x)=\int_{-\infty}^{x} f(t)\,dt$ , which gives $P(X\le x)$ directly. By the Fundamental Theorem of Calculus, differentiating the CDF returns the density: $F'(x)=f(x)$ . This is why probability questions can be answered either by integrating the density between limits or by subtracting two values of the CDF, $P(a\le X\le b)=F(b)-F(a)$ - whichever the question makes easier.

Common errors

Treating f(x) as a probability or ignoring the function's range

The value $f(x)$ is a density, not a probability - it can exceed 1, and only the area under it between limits is a probability. Also, the PDF is zero outside its stated interval, so integrate only over the part of $[a,b]$ where $f$ is non-zero; integrating $kx$ beyond $[0,2]$ would wrongly add area that does not exist.

Always verify total area equals 1

Before computing any probability, confirm (or use) $\int f(x)\,dx=1$ over the variable's range. This both finds unknown constants like $k$ and is a quick sanity check that your density is valid.

Why it matters

Probability density functions are the continuous counterpart of the discrete distributions in Topic 2 and the foundation for the normal distribution that follows. The "probability = area = integral" principle ties Topic 5 directly back to the integral calculus of Topic 3.

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 2022-style5 marks

Calculator-assumed. A continuous probability density function is

f(x) = a

(a constant) defined for

0 \le x \le 4

. (a) Find the value of

a

. (b) Write an integral expression for the mean and evaluate it.

Show worked answer →

(a) A valid PDF has total area $1$ , so $\int_0^4 a\,dx = 1$ . The rectangle of width $4$ and height $a$ gives $4a = 1$ , so $a = \tfrac{1}{4}$ . (2 marks)

(b) The mean is $\int_0^4 x f(x)\,dx$ :

\mu = \int_0^4 x \cdot \tfrac{1}{4}\,dx = \tfrac{1}{4}\Big[\tfrac{x^2}{2}\Big]_0^4 = \tfrac{1}{4}\cdot 8 = 2.

So $a = \tfrac{1}{4}$ and the mean is $2$ . This is a uniform distribution, so the mean sits at the midpoint, a useful check. (3 marks: 1 for the expression, 2 for evaluation.)

SACE 2023-style2 marks

Calculator-assumed. The time

T

(in minutes) for a customer to receive an order is modelled by

f(t) = 0.3e^{-0.3t}

for

t \ge 0

. Determine the probability that a randomly chosen customer receives their order in less than 5 minutes.

Show worked answer →

Probability is the area under the density: $P(T < 5) = \int_0^5 0.3e^{-0.3t}\,dt$ .

The antiderivative of $0.3e^{-0.3t}$ is $-e^{-0.3t}$ . Evaluate:

P(T < 5) = \Big[-e^{-0.3t}\Big]_0^5 = -e^{-1.5} + e^0 = 1 - e^{-1.5} = 1 - 0.2231 \approx 0.777.

So $P(T < 5) \approx 0.777$ . Marks: one for the integral set-up, one for evaluating $1 - e^{-1.5}$ .

SACE 2021-style3 marks

Calculator-assumed. The lifetime

X

(years) of a component has density

f(x) = kx(4 - x)

for

0 \le x \le 4

, zero otherwise. Find

k

, then

P(X \le 2)

Show worked answer →

Total area is $1$ : $\int_0^4 kx(4 - x)\,dx = k\int_0^4 (4x - x^2)\,dx = k\Big[2x^2 - \tfrac{x^3}{3}\Big]_0^4 = k\left(32 - \tfrac{64}{3}\right) = k\cdot\tfrac{32}{3} = 1$ , so $k = \tfrac{3}{32}$ .

Then $P(X \le 2) = \tfrac{3}{32}\int_0^2 (4x - x^2)\,dx = \tfrac{3}{32}\Big[2x^2 - \tfrac{x^3}{3}\Big]_0^2 = \tfrac{3}{32}\left(8 - \tfrac{8}{3}\right) = \tfrac{3}{32}\cdot\tfrac{16}{3} = \tfrac{1}{2}$ .

So $k = \tfrac{3}{32}$ and $P(X \le 2) = 0.5$ (which is expected by the symmetry of the density about $x = 2$ ). Marks: one for the area integral, one for $k$ , one for $P(X \le 2)$ .

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

What this dot point is asking

Probability is area

Step 1: find kkk using the total-area condition

Step 2: find P(X≤1)P(X\le 1)P(X≤1) as area under the curve

Mean and variance by integration

Step 1: set up the mean integral

Step 2: evaluate the integral

The median and other percentiles

The cumulative distribution function

Common errors

Why it matters

Exam-style questions

Practise this

Sources & how we know this

Step 1: find $k$ using the total-area condition

Step 2: find $P(X\le 1)$ as area under the curve