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.
Reviewed by: AI editorial process; not yet individually human-reviewed
Have a quick question? Jump to the Q&A page
Jump to a section
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.
Probability is area
The defining idea of Topic 5 is that probability equals area under the curve, computed by integration:
Mean and variance by integration
The discrete sums become integrals for a continuous variable:
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.
2017 SACE Stage 25 marksA continuous probability density function is f(x) = a (a constant) defined for 0 <= x <= 4. (i) Find the value of a. (ii) Write an integral expression for the mean and evaluate it to find the mean of the distribution.Show worked answer →
(i) A valid pdf must have total area 1, so the integral of f(x) over its domain equals 1. Here the region is a rectangle of width 4 and height a:
integral from 0 to 4 of a dx = 4a = 1, so a = 1/4. (2 marks)
(ii) The mean is the integral of x times f(x) over the domain:
mean = integral from 0 to 4 of x times (1/4) dx. (1 mark for the expression)
Evaluate: (1/4) integral from 0 to 4 of x dx = (1/4) [x^2 / 2] from 0 to 4 = (1/4)(16/2 - 0) = (1/4)(8) = 2. (2 marks)
So a = 1/4 and the mean is 2. This is a uniform distribution, so the mean sits at the midpoint of the interval, which is a useful check. Markers reward using "area = 1" for part (i) and the mean = integral of x f(x) formula for part (ii).
2023 SACE Stage 21 marksThe time T (in minutes) for a customer to receive an order is modelled by the probability density function f(t) = 0.3 e^(-0.3t) for t >= 0. Determine the probability that a randomly chosen customer receives their order in less than 5 minutes, according to the model.Show worked answer →
Probability is the area under the density function, found by integration. P(T < 5) is the integral of f(t) from 0 to 5:
P(T < 5) = integral from 0 to 5 of 0.3 e^(-0.3t) dt.
The antiderivative of 0.3 e^(-0.3t) is -e^(-0.3t) (since d/dt[-e^(-0.3t)] = 0.3 e^(-0.3t)). Evaluate between 0 and 5:
= [-e^(-0.3t)] from 0 to 5 = -e^(-1.5) - (-e^0) = 1 - e^(-1.5) = 1 - 0.2231 = 0.777 (to three significant figures).
So P(T < 5) = 0.777. The single mark is for the correct probability, obtained by integrating the density from 0 to 5 (or equivalently using the calculator's cumulative function).