Continuous random variables and pdf: QCE Maths Methods Unit 4

Syllabus dot point

“Define a continuous random variable, its probability density function (pdf), cumulative distribution function (cdf), and compute probabilities, expected value (mean), variance and standard deviation as definite integrals”

9 min readUpdated 7 Oct 2026

Quick answer

A continuous random variable $X$ is described by a probability density function $f(x)$ that integrates to 1 over its support and is non-negative, with probabilities computed as definite integrals $P(a \leq X \leq b) = \int f$ , expected value as $E(X) = \int x f \, dx$ , and variance as $E(X^2) - [E(X)]^2$ where $E(X^2) = \int x^2 f \, dx$ .

Jump to a section

What this dot point is asking
Continuous random variable
Probability density function (pdf)
Cumulative distribution function (cdf)
Expected value (mean)
Variance and standard deviation
Median and quartiles
The uniform distribution
PSMT and EA contexts
Exam-style questions

What this dot point is asking

QCAA wants you to define a continuous random variable through its probability density function, compute probabilities as definite integrals, and compute expected value (mean), variance and standard deviation as integrals. The dot point bridges Unit 4 calculus to Unit 4 statistics.

Continuous random variable

A continuous random variable $X$ takes values in a continuum (an interval of real numbers). Examples: a waiting time, a length, a temperature. Because $X$ has uncountably many possible values, $P(X = x) = 0$ for any single value; probabilities are computed for intervals.

Probability density function (pdf)

A continuous random variable $X$ is described by its probability density function $f(x)$ , with:

P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx

Two conditions for a valid pdf:

Non-negative. $f(x) \geq 0$ everywhere.
Integrates to 1. $\int_{-\infty}^{\infty} f(x) \, dx = 1$ .

For $X$ supported on $[a, b]$ (and $f = 0$ outside), condition 2 becomes $\int_{a}^{b} f(x) \, dx = 1$ .

Finding a normalising constant

A common PSMT and EA question gives $f(x) = k g(x)$ on $[a, b]$ and asks for $k$ . Set $\int k g = 1$ ; solve for $k$ .

Cumulative distribution function (cdf)

The cdf is $F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) \, dt$ .

Properties:

$F$ is non-decreasing.
$F(-\infty) = 0$ , $F(\infty) = 1$ .
$P(a \leq X \leq b) = F(b) - F(a)$ .
$F'(x) = f(x)$ where $f$ is continuous (fundamental theorem of calculus).

Expected value (mean)

The expected value of $X$ is:

E(X) = \mu = \int_{-\infty}^{\infty} x f(x) \, dx

(For $X$ on $[a, b]$ , the limits reduce to $a$ and $b$ .)

Interpretation: the centre of mass of the pdf; the long-run average of independent observations.

Linearity: $E(aX + b) = a E(X) + b$ .

Variance and standard deviation

\text{Var}(X) = E[(X - \mu)^2] = E(X^2) - [E(X)]^2

where $E(X^2) = \int x^2 f(x) \, dx$ .

The right-hand identity is the working formula.

Standard deviation $\sigma = \sqrt{\text{Var}(X)}$ .

Property: $\text{Var}(aX + b) = a^2 \text{Var}(X)$ .

Median and quartiles

Median: $m$ such that $P(X \leq m) = 1/2$ . Solve $\int_{a}^{m} f \, dx = 1/2$ .

Quartiles: $\int = 1/4$ and $\int = 3/4$ .

For symmetric pdfs the median equals the mean; for skewed pdfs they differ.

The uniform distribution

The simplest continuous random variable. $X \sim U(a, b)$ has $f(x) = \frac{1}{b-a}$ on $[a, b]$ , $0$ elsewhere.

$E(X) = (a+b)/2$ .
$\text{Var}(X) = (b-a)^2 / 12$ .
$P(c \leq X \leq d) = (d-c)/(b-a)$ for $a \leq c \leq d \leq b$ .

Triangular pdf

The pdf $f(x) = kx$ on $[0, 2]$ and $0$ elsewhere forms a triangular shape. We find $k$ , then compute the mean, variance and standard deviation.

Step 1: Find the normalising constant $k$

Every valid pdf must integrate to $1$ over its support. Setting up the normalisation condition gives an equation in $k$ .

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

Solving gives $k = \dfrac{1}{2}$ , so the pdf is $f(x) = \dfrac{x}{2}$ on $[0, 2]$ .

Step 2: Compute the mean $E(X)$

The expected value is $E(X) = \int x f(x) \, dx$ . We substitute $f(x) = \tfrac{x}{2}$ and integrate $\tfrac{x^2}{2}$ .

E(X) = \int_0^2 x \cdot \frac{x}{2} \, dx = \int_0^2 \frac{x^2}{2} \, dx = \frac{1}{2} \cdot \left[\frac{x^3}{3}\right]_0^2 = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}

Step 3: Compute the variance and standard deviation

To use the shortcut formula $\mathrm{Var}(X) = E(X^2) - [E(X)]^2$ , we first compute $E(X^2) = \int x^2 f(x) \, dx$ .

E(X^2) = \int_0^2 x^2 \cdot \frac{x}{2} \, dx = \int_0^2 \frac{x^3}{2} \, dx = \frac{1}{2} \cdot \left[\frac{x^4}{4}\right]_0^2 = \frac{1}{2} \cdot \frac{16}{4} = 2

Now apply the variance shortcut.

\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = 2 - \left(\frac{4}{3}\right)^2 = 2 - \frac{16}{9} = \frac{18 - 16}{9} = \frac{2}{9}

Final answer: $k = \dfrac{1}{2}$ , $E(X) = \dfrac{4}{3}$ , $\mathrm{Var}(X) = \dfrac{2}{9}$ , and $\sigma = \sqrt{\dfrac{2}{9}} = \dfrac{\sqrt{2}}{3}$ .

PSMT and EA contexts

The IA3 PSMT often models a continuous distribution arising from a real-world variable (waiting time, lifetime, error magnitude). Typical questions:

Find the normalising constant for a given pdf shape.
Compute the probability of a specific event.
Compute and interpret the mean and standard deviation.
Compute the median or a percentile.

The EA Paper 2 short response examines the formulas explicitly.

Common errors

Forgetting normalisation: A pdf must integrate to 1. Forgetting this in finding $k$ is the most common slip.
Using $f(x)$ as probability: $f(2) = 0.3$ does not mean $P(X = 2) = 0.3$ . The pdf is a density, not a probability.
Wrong variance formula: $\text{Var} = E(X^2) - [E(X)]^2$ , not $E(X^2) - E(X)$ .
Forgetting the pdf inside $E(X)$: $E(X) = \int x f(x) \, dx$ , not $\int x \, dx$ .
Wrong support: If $f = 0$ outside $[a, b]$ , the integrals must be on $[a, b]$ , not $(-\infty, \infty)$ .

Exam-style questions

Questions in the style of QCAA 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.

Practice5 marks

Paper 2 (complex familiar). A continuous random variable

X

has pdf

f(x) = k(4 - x^2)

[-2, 2]

and

0

elsewhere. (a) Determine

k

. (b) Determine

E(X)

. (c) Determine

\text{Var}(X)

Show worked answer →

(a) Find $k$ . Normalisation: $\int f = 1$ .

$\int_{-2}^{2} k(4 - x^2) \, dx = 2k \int_{0}^{2} (4 - x^2) \, dx$ (by symmetry).

$= 2k \left[ 4x - \frac{x^3}{3} \right]_{0}^{2} = 2k \left( 8 - \frac{8}{3} \right) = 2k \cdot \frac{16}{3} = \frac{32 k}{3}$ .

Set $\frac{32k}{3} = 1$ : $k = \frac{3}{32}$ .

(b) Expected value. $E(X) = \int_{-2}^{2} x \cdot \frac{3}{32}(4 - x^2) \, dx$ .

The integrand $x(4 - x^2)$ is an odd function on a symmetric interval $[-2, 2]$ , so the integral is 0. $E(X) = 0$ .

(c) Variance. Need $E(X^2)$ .

$E(X^2) = \int_{-2}^{2} x^2 \cdot \frac{3}{32}(4 - x^2) \, dx = \frac{3}{32} \int_{-2}^{2} (4 x^2 - x^4) \, dx$ .

By symmetry: $= \frac{3}{16} \int_{0}^{2} (4 x^2 - x^4) \, dx = \frac{3}{16} \left[ \frac{4 x^3}{3} - \frac{x^5}{5} \right]_{0}^{2}$ .

Evaluate: $= \frac{3}{16} \left( \frac{32}{3} - \frac{32}{5} \right) = \frac{3}{16} \cdot \frac{160 - 96}{15} = \frac{3}{16} \cdot \frac{64}{15} = \frac{192}{240} = \frac{4}{5}$ .

$\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{4}{5} - 0 = \frac{4}{5}$ .

Markers reward the normalisation calculation with symmetry shortcut, the odd-integrand observation for $E(X)$ , and the variance formula with correct arithmetic.

Practice4 marks

Paper 2 (complex familiar). A waiting time

X

(minutes) has pdf

f(x) = \dfrac{x}{8}

[0, 4]

and

0

elsewhere. (a) Verify that

f

is a valid pdf. (b) Determine

P(X \leq 2)

. (c) Determine the median waiting time.

Show worked answer →

(a) $\displaystyle\int_0^4 \dfrac{x}{8}\,dx = \left[\dfrac{x^2}{16}\right]_0^4 = \dfrac{16}{16} = 1$ , and $f(x) \geq 0$ on $[0, 4]$ , so $f$ is a valid pdf.

(b) $P(X \leq 2) = \displaystyle\int_0^2 \dfrac{x}{8}\,dx = \left[\dfrac{x^2}{16}\right]_0^2 = \dfrac{4}{16} = \dfrac{1}{4}.$

(c) The median $m$ solves $\displaystyle\int_0^m \dfrac{x}{8}\,dx = \dfrac{1}{2}$ , so $\dfrac{m^2}{16} = \dfrac{1}{2}$ , $m^2 = 8$ , $m = 2\sqrt 2 \approx 2.83$ minutes.

Markers reward the normalisation check, the probability integral, and solving for the median.

Practise this

Sources & how we know this

Mathematical Methods General senior syllabus 2025 — QCAA