What is wrong variance formula?

Var = E(X^2) - [E(X)]^2, not E(X^2) - E(X).

What is wrong support?

If f = 0 outside [a, b], the integrals must be on [a, b], not (-∞, ∞).

QCAA QCAA 2024-style practice: Paper 2 (complex familiar). A continuous random variable X has pdf f(x) = k(4 - x^2) on [-2, 2] and 0 elsewhere. (a) Determine k. (b) Determine E(X). (c) Determine Var(X).

**(a) Find k.** Normalisation: int f = 1. _-2^2 k(4 - x^2) \, dx = 2k _0^2 (4 - x^2) \, dx (by symmetry). = 2k [ 4x - (x^3)/(3) ]_0^2 = 2k ( 8 - (8)/(3) ) = 2k · (16)/(3) = (32 k)/(3). Set (32k)/(3) = 1: k = (3)/(32). **(b) Expected value.** E(X) = _-2^2 x · (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) = _-2^2 x^2 · (3)/(32)(4 - x^2) \, dx = (3)/(32) _-2^2 (4 x^2 - x^4) \, dx. By symmetry: = (3)/(16) _0^2 (4 x^2 - x^4) \, dx = (3)/(16) [ (4 x^3)/(3) - (x^5)/(5) ]_0^2. Evaluate: = (3)/(16) ( (32)/(3) - (32)/(5) ) = (3)/(16) · (160 - 96)/(15) = (3)/(16) · (64)/(15) = (192)/(240) = (4)/(5). Var(X) = E(X^2) - [E(X)]^2 = (4)/(5) - 0 = (4)/(5). Markers reward the normalisation calculation with symmetry shortcut, the odd-integrand observation for E(X), and the variance formula with correct arithmetic.

QCAA QCAA 2023-style practice: Paper 2 (complex familiar). A waiting time X (minutes) has pdf f(x) = x8 on [0, 4] and 0 elsewhere. (a) Verify that f is a valid pdf. (b) Determine P(X ≤ 2). (c) Determine the median waiting time.

(a) _0^4 x8\,dx = [x^216]_0^4 = 1616 = 1, and f(x) ≥ 0 on [0, 4], so f is a valid pdf. (b) P(X ≤ 2) = _0^2 x8\,dx = [x^216]_0^2 = 416 = 14. (c) The median m solves _0^m x8\,dx = 12, so m^216 = 12, m^2 = 8, m = 2 2 ≈ 2.83 minutes. Markers reward the normalisation check, the probability integral, and solving for the median.

§-Syllabus dot point

QLDMath MethodsSyllabus dot point

Topic 3: Continuous random variables, the normal distribution, and statistical inference

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

A focused answer to the QCE Maths Methods Unit 4 dot point on continuous random variables. Defines the pdf, cdf, mean, variance and standard deviation as integrals, including the normalisation condition and a worked PSMT-style example.

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
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

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.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

QCAA 20245 marksPaper 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.

QCAA 20234 marksPaper 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.