Why is P(X=a) = 0 for a continuous random variable? It seems like the variable must take *some* value.

While the variable will indeed take a specific value, the probability of it being *exactly* that one value out of an infinite number of possibilities is infinitesimally small, and thus is defined as zero. Probability for a CRV is the area under the PDF curve over an interval. The area corresponding to a single point (an interval of width zero) is zero. We can only talk about the probability of X being in a range, e.g., P(a ≤ X ≤ b).

Can the probability density function f(x) have a value greater than 1?

Yes, it can. Unlike a probability, the value of f(x) is a measure of 'probability density', not probability itself. For example, a uniform distribution on the interval [0, 0.5] has a PDF of f(x) = 2 for 0 ≤ x ≤ 0.5. The key constraint is that the *area* under the f(x) curve, which represents the total probability, must integrate to 1.

When finding the CDF, F(x), why do I need to define it for values outside the main range of the PDF?

The cumulative distribution function, F(x) = P(X ≤ x), must be defined for all real numbers x from -∞ to +∞. For a PDF defined on an interval [a, b], the probability of X being less than 'a' is zero, so F(x) = 0 for x < a. The probability of X being less than or equal to any value greater than 'b' is one, as all possible values are within that range. Therefore, F(x) = 1 for x ≥ b. A complete piecewise definition is required for a full description of the distribution.

Where can I learn Continuous random variables for Cambridge 9231?

This free lesson on MarkScheme covers syllabus point 4.1 (Continuous random variables) for Cambridge A-Level Further Mathematics. It includes visual revision, key points, and links to real past-paper questions with mark-scheme marking at markscheme.app/courses/9231/4-1-continuous-random-variables.

Is Continuous random variables (4.1) on 9231 free to study?

Yes. The Further Mathematics 9231 course on MarkScheme is 100% free — including premium visual lessons, exam tips, and past-paper marking. No paywall for Continuous random variables.

How do I revise Continuous random variables for 9231 exams?

Start with the visual steps on this page, note the key exam points, then attempt a 9231 past-paper question on Continuous random variables using MarkScheme's /mark tool. Mark strictly against the official scheme and re-read any marks you lost.

Continuous random variables 9231 — Free…

Continuous random variables

Welcome to the study of continuous random variables. Unlike discrete variables which you've met before (like the score on a die), continuous variables can take any value in a range. Think of the exact time it takes for a bus to arrive or the precise height of a student. In this topic, we move from summing probabilities to integrating functions, opening up a powerful way to model real-world phenomena.

BY THE END, YOU CAN…

1Understand the properties of a probability density function (PDF) and a cumulative distribution function (CDF).
2Calculate probabilities by integrating a PDF over a specified interval.
3Find the expectation E(X) and variance Var(X) of a continuous random variable.
4Use the CDF to find probabilities and understand its relationship with the PDF.

In simple terms

A friendly intro before the formal notes — no formulas yet.

From Histograms to Curves

A continuous random variable can take any value in a given range, like height or time. We use a function called a Probability Density Function (PDF) to describe the likelihood of the variable falling within a particular interval, where probability is found by calculating the area under the curve.

THINK OF IT LIKE…

Imagine you have a very fine, smooth pile of sand representing all possible outcomes. The height of the sand pile at any point is the PDF - it tells you where values are more likely to occur. To find the probability of a value falling in a certain range, you don't pick a single grain of sand (which has zero width and zero probability); instead, you scoop up a section of sand and measure its volume (the area under the curve). The total volume of sand is always 1.

1
Define the Probability Density Function (PDF), $f(x)$ , and verify its two key properties: $f(x) \ge 0$ and the total area $\int_{-\infty}^{\infty} f(x) \, dx = 1$ .
2
To find the probability $P(a < X < b)$ , calculate the definite integral of the PDF between the limits $a$ and $b$ : $\int_{a}^{b} f(x) \, dx$ .
3
Calculate the expectation (mean) using $E(X) = \int_{-\infty}^{\infty} x f(x) \, dx$ .
4
Calculate the variance using $Var(X) = E(X^2) - [E(X)]^2$ , where $E(X^2) = \int_{-\infty}^{\infty} x^2 f(x) \, dx$ .

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

f(x)

Tap a symbol — great for exam definitions

$F(x) = P(X \le x) = \int_{-\infty}^{x} f(t) \, dt$

Tap a symbol — great for exam definitions

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

Tap a symbol — great for exam definitions

At a glance — side by side

Compare key properties side by side — ideal for exam contrasts.

Discrete vs. Continuous Random Variables

Feature	Discrete Random Variable (DRV)	Continuous Random Variable (CRV)
Possible Values	Countable set of distinct values (e.g., 0, 1, 2, ...)	Any value within a given continuous range (e.g., 0 ≤ x ≤ 1)
Probability Function	Probability Mass Function (PMF), P(X=x)	Probability Density Function (PDF), f(x)
Probability of a single point	P(X=x) can be > 0	P(X=x) = 0
Total Probability	Sum of all P(X=x) equals 1 (Σ P(X=x) = 1)	Total area under f(x) equals 1 (∫ f(x) dx = 1)
Calculating E(X)	E(X) = Σ x * P(X=x)	E(X) = ∫ x * f(x) dx
Interval Probability	P(a ≤ X ≤ b) ≠ P(a < X < b) in general	P(a ≤ X ≤ b) = P(a < X < b)

Possible Values

Discrete Random Variable (DRV)

Countable set of distinct values (e.g., 0, 1, 2, ...)

Continuous Random Variable (CRV)

Any value within a given continuous range (e.g., 0 ≤ x ≤ 1)

Probability Function

Discrete Random Variable (DRV)

Probability Mass Function (PMF), P(X=x)

Continuous Random Variable (CRV)

Probability Density Function (PDF), f(x)

Probability of a single point

Discrete Random Variable (DRV)

P(X=x) can be > 0

Continuous Random Variable (CRV)

P(X=x) = 0

Total Probability

Discrete Random Variable (DRV)

Sum of all P(X=x) equals 1 (Σ P(X=x) = 1)

Continuous Random Variable (CRV)

Total area under f(x) equals 1 (∫ f(x) dx = 1)

Calculating E(X)

Discrete Random Variable (DRV)

E(X) = Σ x * P(X=x)

Continuous Random Variable (CRV)

E(X) = ∫ x * f(x) dx

Interval Probability

Discrete Random Variable (DRV)

P(a ≤ X ≤ b) ≠ P(a < X < b) in general

Continuous Random Variable (CRV)

P(a ≤ X ≤ b) = P(a < X < b)

Full topic notes

Formal explanation with the rigour you need for the exam.

Understanding Continuous Random Variables and the PDF

A continuous random variable (CRV) is a variable that can take any value within a given range or interval. Unlike discrete random variables which have countable outcomes (like the score on a die), a CRV's possible values are infinite (like the exact height of a person). The behaviour of a CRV is described by its probability density function (PDF), denoted f(x). It is crucial to understand that for a CRV, the value of f(x) itself is not a probability. Instead, the area under the graph of f(x) between two points represents the probability that the variable falls within that interval. A direct consequence of this is that the probability of a CRV taking any single exact value is zero, i.e., P(X = c) = 0.

A CRV can take any value in a continuous interval.
The probability density function, f(x), describes the distribution.
Probability is the area under the PDF curve: P(a ≤ X ≤ b) = ∫[a,b] f(x) dx.
The probability of a CRV equalling a specific value is always zero: P(X = c) = 0.

EXAM TIP

Since P(X=a) = 0 and P(X=b) = 0, remember that P(a ≤ X ≤ b), P(a < X ≤ b), P(a ≤ X < b), and P(a < X < b) are all identical for a continuous random variable. Do not waste time considering the endpoints.

Verifying a Probability Density Function

For a function f(x) to be a valid PDF, it must satisfy two fundamental conditions. Firstly, the function must be non-negative for all possible values of x, meaning f(x) ≥ 0. This makes intuitive sense, as probability density cannot be negative. Secondly, the total area under the curve over its entire domain must be equal to 1. This represents the certainty that the variable will take some value within its range. This condition is expressed mathematically as ∫[-∞,∞] f(x) dx = 1, though in practice the integral is taken over the interval where f(x) is non-zero. Exam questions frequently require you to use this second property to find the value of an unknown constant within a given PDF.

Condition 1: f(x) ≥ 0 for all x in the domain.
Condition 2: The total integral of f(x) over its entire domain is 1.
These two properties are used to validate a function as a PDF.
Finding an unknown constant 'k' in a PDF usually involves setting the total integral equal to 1.

EXAM TIP

When showing that f(x) ≥ 0, consider the nature of the function. For example, if f(x) = kx² for x > 0, you only need to show that the constant k is positive. If the function is more complex, a sketch or consideration of its minimum value may be required.

The Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF), denoted F(x), gives the probability that the random variable X takes a value less than or equal to x. That is, F(x) = P(X ≤ x). It is obtained by integrating the PDF, f(t), from the lower limit of the domain up to x: F(x) = ∫[-∞,x] f(t) dt. The CDF is a non-decreasing function that starts at 0 and ends at 1. It provides a convenient way to calculate probabilities for intervals, using the property P(a < X ≤ b) = F(b) - F(a). Conversely, the PDF can be found by differentiating the CDF: f(x) = F'(x). A complete definition of F(x) must cover all real numbers, including being 0 below the domain and 1 above it.

F(x) = P(X ≤ x).
F(x) is found by integrating the PDF: F(x) = ∫[lower bound, x] f(t) dt.
The PDF is the derivative of the CDF: f(x) = F'(x).
Probabilities of intervals are found easily: P(a < X ≤ b) = F(b) - F(a).
F(x) must be defined for all real x, typically as a piecewise function.

EXAM TIP

When constructing a piecewise CDF, be meticulous. Define F(x) = 0 for x less than the lower bound, F(x) = 1 for x greater than the upper bound, and the integral expression for x within the bounds. Check that your function is continuous at the boundaries.

Expectation and Variance of a CRV

The expectation or mean of a CRV, E(X), represents its long-term average value. It is calculated by integrating the product of x and the PDF f(x) over the variable's domain: E(X) = ∫ x * f(x) dx. The variance, Var(X), measures the spread of the distribution around the mean. It is calculated using the formula Var(X) = E(X²) - [E(X)]². To use this, you must first calculate E(X) and E(X²). The term E(X²) is found using the general formula for the expectation of a function of X, E(g(X)) = ∫ g(x) * f(x) dx, where here g(x) = x². Thus, E(X²) = ∫ x² * f(x) dx. The standard deviation is the square root of the variance.

Expectation (Mean): E(X) = ∫ x * f(x) dx.
Expectation of a function g(X): E(g(X)) = ∫ g(x) * f(x) dx.
Specifically, E(X²) = ∫ x² * f(x) dx.
Variance: Var(X) = E(X²) - [E(X)]².
Standard Deviation: SD(X) = √Var(X).

EXAM TIP

A common mistake is confusing E(X²) with [E(X)]². Calculate E(X) and E(X²) as two separate, distinct steps before substituting them into the variance formula. This systematic approach reduces the chance of calculation errors.

The Probability Density Function (PDF)

For a continuous random variable $X$ , we can't assign a probability to a specific outcome. Instead, we use a Probability Density Function, $f(x)$ , to describe the distribution of probabilities. The function $f(x)$ itself is not a probability, but the area under its curve between two points gives the probability that $X$ lies in that interval. For $f(x)$ to be a valid PDF, it must satisfy two crucial conditions.

A function $f(x)$ is a valid PDF if and only if:

$f(x) \ge 0$ for all $x$ in the domain.
$\int_{-\infty}^{\infty} f(x) \, dx = 1$ (The total area under the curve is 1).

The probability of $X$ lying in the interval $[a, b]$ is given by $P(a \le X \le b) = \int_{a}^{b} f(x) \, dx$ .
For any single point $c$ , $P(X=c) = \int_{c}^{c} f(x) \, dx = 0$ .
This means $P(a \le X \le b) = P(a < X < b) = P(a \le X < b) = P(a < X \le b)$ .

Cumulative Distribution Function (CDF)

The Cumulative Distribution Function, $F(x)$ , gives the probability that the random variable $X$ is less than or equal to a particular value $x$ . It is the 'running total' of the probability. The CDF is found by integrating the PDF from the lower bound of its domain up to $x$ . This is particularly useful for finding probabilities, as $P(a < X < b) = F(b) - F(a)$ .

CDF: $F(x) = P(X \le x) = \int_{-\infty}^{x} f(t) \, dt$ Relationship to PDF: $f(x) = F'(x)$

EXAM TIP

When asked to find the CDF, remember it's a piecewise function. For the example $f(x) = \frac{3}{32}x(4-x)$ on $[0, 4]$ , the CDF is $F(x)=0$ for $x<0$ , $F(x)=1$ for $x>4$ , and $F(x) = \int_0^x \frac{3}{32}t(4-t) \, dt$ for $0 \le x \le 4$ . Don't forget to define the function for all real numbers.

Expectation and Variance

Just as with discrete variables, we can calculate the mean (or expectation) and variance of a continuous random variable. The expectation, $E(X)$ , is the long-run average value of $X$ , and the variance, $Var(X)$ , measures the spread or dispersion of the distribution. The calculations involve integrating $x f(x)$ and $x^2 f(x)$ over the domain of the variable.

Expectation: $E(X) = \int_{-\infty}^{\infty} x f(x) \, dx$ Variance: $Var(X) = E(X^2) - [E(X)]^2$ , where $E(X^2) = \int_{-\infty}^{\infty} x^2 f(x) \, dx$

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

A continuous random variable $X$ has the probability density function given by $f(x) = \begin{cases} kx(4-x) & 0 \le x \le 4 \\ 0 & \text{otherwise} \end{cases}$ (a) Show that $k = \frac{3}{32}$ . (b) Find $P(1 < X < 3)$ .

1
(a) For $f(x)$ to be a valid PDF, the total area under the curve must be 1. $\int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{4} kx(4-x) \, dx = 1$ $k \int_{0}^{4} (4x - x^2) \, dx = 1$ $k \left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4} = 1$ $k \left( (2(4^2) - \frac{4^3}{3}) - (0) \right) = 1$ $k \left( 32 - \frac{64}{3} \right) = 1$ $k \left( \frac{96-64}{3} \right) = 1$ $k \left( \frac{32}{3} \right) = 1$ $k = \frac{3}{32}$ [M1 for integration, A1 for correct value]

EXAMPLE 2

The lifetime, $T$ years, of a particular electronic component has PDF given by $f(t) = \begin{cases} \frac{1}{4}e^{-\frac{t}{4}} & t \ge 0 \\ 0 & \text{otherwise} \end{cases}$ Find the mean lifetime, $E(T)$ , and the variance of the lifetime, $Var(T)$ .

1
First, we find the expectation, $E(T)$ . $E(T) = \int_{0}^{\infty} t f(t) \, dt = \int_{0}^{\infty} t \cdot \frac{1}{4}e^{-\frac{t}{4}} \, dt$ This requires integration by parts: $\int u \frac{dv}{dt} dt = uv - \int v \frac{du}{dt} dt$ . Let $u=t \implies \frac{du}{dt}=1$ . Let $\frac{dv}{dt}=\frac{1}{4}e^{-\frac{t}{4}} \implies v = -e^{-\frac{t}{4}}$ . [M1 for setting up correct integral and identifying integration by parts] $E(T) = \left[ -t e^{-\frac{t}{4}} \right]_{0}^{\infty} - \int_{0}^{\infty} (-e^{-\frac{t}{4}}) \, dt$ $= \left[ -t e^{-\frac{t}{4}} \right]_{0}^{\infty} + \int_{0}^{\infty} e^{-\frac{t}{4}} \, dt$ As $t \to \infty$ , $t e^{-\frac{t}{4}} \to 0$ . At $t=0$ , the term is 0. So the first part is 0. $E(T) = \left[ -4e^{-\frac{t}{4}} \right]_{0}^{\infty} = (0) - (-4e^0) = 4$ . The mean lifetime is 4 years. [A1 for E(T)]

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAContinuous random variables

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 17

What is a continuous random variable (CRV)?

TAP OR SPACE TO FLIP

ANSWER

A random variable that can take any value within a given range or interval. Examples include height, weight, and time.

TAP OR SPACE TO FLIP BACK

1 / 17 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
A CRV can take any value in a continuous interval.
✓
The probability density function, f(x), describes the distribution.
✓
Probability is the area under the PDF curve: P(a ≤ X ≤ b) = ∫[a,b] f(x) dx.
✓
The probability of a CRV equalling a specific value is always zero: P(X = c) = 0.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Practice Questions on Continuous Random Variables

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Practice Questions on Continuous Random Variables on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS3.6 Momentum NEXT →4.2 Inference using normal and t-distributions

Continuous random variables

BY THE END, YOU CAN…

1Understand the properties of a probability density function (PDF) and a cumulative distribution function (CDF).
2Calculate probabilities by integrating a PDF over a specified interval.
3Find the expectation E(X) and variance Var(X) of a continuous random variable.
4Use the CDF to find probabilities and understand its relationship with the PDF.

Feature

Discrete Random Variable (DRV)

Continuous Random Variable (CRV)

Possible Values

Countable set of distinct values (e.g., 0, 1, 2, ...)

Any value within a given continuous range (e.g., 0 ≤ x ≤ 1)

Probability Function

Probability Mass Function (PMF), P(X=x)

Probability Density Function (PDF), f(x)

Probability of a single point

P(X=x) can be > 0

P(X=x) = 0

Total Probability

Sum of all P(X=x) equals 1 (Σ P(X=x) = 1)

Total area under f(x) equals 1 (∫ f(x) dx = 1)

Calculating E(X)

E(X) = Σ x * P(X=x)

E(X) = ∫ x * f(x) dx

Interval Probability

P(a ≤ X ≤ b) ≠ P(a < X < b) in general

P(a ≤ X ≤ b) = P(a < X < b)

Revise Continuous random variables for Cambridge 9231 (4.1)

In simple terms

Explore the concept

Key formulas

At a glance — side by side

Possible Values

Probability Function

Probability of a single point

Total Probability

Calculating E(X)

Interval Probability

Full topic notes

Understanding Continuous Random Variables and the PDF

Verifying a Probability Density Function

The Cumulative Distribution Function (CDF)

Expectation and Variance of a CRV

The Probability Density Function (PDF)

Cumulative Distribution Function (CDF)

Expectation and Variance

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Practice Questions on Continuous Random Variables

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

At a glance — side by side

Possible Values

Probability Function

Probability of a single point

Total Probability

Calculating E(X)

Interval Probability

Full topic notes

Understanding Continuous Random Variables and the PDF

Verifying a Probability Density Function

The Cumulative Distribution Function (CDF)

Expectation and Variance of a CRV

The Probability Density Function (PDF)

Cumulative Distribution Function (CDF)

Expectation and Variance

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Practice Questions on Continuous Random Variables

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Continuous random variables