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

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:

Two conditions for a valid pdf:

1. Non-negative. $f(x) \geq 0$ everywhere.
2. 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:

• IMATH_20 is non-decreasing.
• IMATH_21 , $F(\infty) = 1$.
• IMATH_23 .
• IMATH_24 where $f$ is continuous (fundamental theorem of calculus).

Expected value (mean)

The expected value of $X$ is:

(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

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.

• IMATH_44 .
• IMATH_45 .
• IMATH_46 for $a \leq c \leq d \leq b$.

Worked example. Triangular pdf

$f(x) = k x$ on $[0, 2]$, 0 elsewhere.

Find $k$. $\int_{0}^{2} k x \, dx = [\frac{k x^2}{2}]_{0}^{2} = 2k = 1$, so $k = 1/2$.

Find $E(X)$. $E(X) = \int_{0}^{2} x \cdot \frac{x}{2} \, dx = \int_{0}^{2} \frac{x^2}{2} \, dx = \frac{8}{6} = \frac{4}{3}$.

Find $E(X^2)$. $E(X^2) = \int_{0}^{2} \frac{x^3}{2} \, dx = \frac{16}{8} = 2$.

Find $\text{Var}(X)$. $\text{Var}(X) = 2 - (4/3)^2 = 2 - 16/9 = (18-16)/9 = 2/9$.

Find SD. $\sigma = \sqrt{2/9} = \sqrt{2}/3$.

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 $(-\infty, \infty)$.

In one sentence

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