SACE Board SACE 2023-style practice: 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 0. Determine the probability that a randomly chosen customer receives their order in less than 5 minutes.

Probability is the area under the density: P(T < 5) = _0^5 0.3e^-0.3t\,dt. The antiderivative of 0.3e^-0.3t is -e^-0.3t. Evaluate: $P(T < 5) = [-e^-0.3t]_0^5 = -e^-1.5 + e^0 = 1 - e^-1.5 = 1 - 0.2231 ≈ 0.777.$ So P(T < 5) ≈ 0.777. Marks: one for the integral set-up, one for evaluating 1 - e^-1.5.

§-Syllabus dot point

SAMath MethodsSyllabus dot point

How do we describe probability for a variable that can take any value in an interval?

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

What a probability density function is, why probability equals area under the curve, the two conditions a valid PDF must satisfy, and how to find probabilities and the mean by integration.

Generated by Claude Opus 4.88 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
Probability is area
Mean and variance by integration
The median and other percentiles
The cumulative distribution function
Common errors
Why it matters

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 is area

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

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 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

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 practice questions

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

SACE 20225 marksCalculator-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 20232 marksCalculator-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 20213 marksCalculator-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 $P(X \le 2)$ .