Continuous probability distributions and PDFs

IB Mathematics AA HL· 4.8· 15 min read

1. Discrete vs Continuous Random Variables★★☆☆☆⏱ 5 min

Discrete random variables take countable distinct values (e.g. number of heads in 10 coin flips), while continuous random variables can take any value within one or more intervals (e.g. height of a randomly selected adult, wait time for a bus). Probabilities for discrete distributions are assigned to individual outcomes, but this is not true for continuous distributions.

📘 Definition

Continuous random variable

A random variable $X$ is continuous if its set of possible values is an uncountable interval, and the probability that $X$ equals any single value is 0.

Example:

The mass of an apple, measured to any precision, is a continuous random variable.

📐 Worked Example

Explain why $P (X = 2.5) = 0$ if $X$ is the height of a person in meters.

1
The probability of $X$ falling into an interval decreases as the interval width shrinks. For a single point, the interval width approaches 0.
2
$\lim_{\delta \to 0} P(x - \delta/2 < X < x + \delta/2) = 0$
3
Therefore, the probability that $X$ equals exactly 2.5 is 0.

2. Properties of a Valid Probability Density Function★★☆☆☆⏱ 5 min

Instead of assigning probability to single points, we use a function called the probability density function (PDF) to find the probability that a continuous random variable falls in an interval. The probability equals the area under the PDF curve over the interval.

📘 Definition

Probability Density Function (PDF)

f(x)

For a continuous random variable $X$ defined over $(- \infty, \infty)$ , $f (x)$ is the PDF of $X$ if it satisfies two core conditions: (1) $f (x) \geq 0$ for all $x$ , and (2) $\int_{- \infty}^{\infty} f (x) d x = 1$ . Any probability $P (a < X < b)$ is given by $\int_{a}^{b} f (x) d x$ .

📐 Worked Example

Verify if $f (x) = {2 x 0 0 \leq x \leq 1 otherwise$ is a valid PDF.

1
First check the non-negativity condition. For $0 \leq x \leq 1$ , $2 x \geq 0$ , and $f (x) = 0$ elsewhere. So $f (x) \geq 0$ for all $x$ , first condition satisfied.
2
Next, integrate $f (x)$ over the entire domain to check total area is 1:
3
$\int_{-\infty}^{\infty} f(x) dx = \int_0^1 2x dx$
4
$= \left[ x^2 \right]_0^1 = 1^2 - 0^2 = 1$
5
Both conditions are satisfied, so this is a valid PDF.

3. Calculating Probabilities from a PDF★★★☆☆⏱ 6 min

Once you confirm a PDF is valid, you calculate the probability that $X$ lies between two values by integrating the PDF between those bounds. This corresponds directly to finding the area under the PDF curve over the interval of interest.

📐 Worked Example

For the valid PDF $f (x) = {2 x 0 0 \leq x \leq 1 otherwise$ , find $P (0.2 < X < 0.6)$ .

1
We integrate the PDF from 0.2 to 0.6, since $f (x) = 0$ outside this interval:
2
$P(0.2 < X < 0.6) = \int_{0.2}^{0.6} 2x dx$
3
$= \left[ x^2 \right]_{0.2}^{0.6} = (0.6)^2 - (0.2)^2$
4
$= 0.36 - 0.04 = 0.32$
5
So the required probability is 0.32.

✓ Quick check

Check your understanding of core PDF properties

What is $P (X = 0.5)$ for the PDF in the worked example above?
- 0
- 0.5
- 1
- Cannot be determined
Reveal answer
0 —
For any continuous distribution, the probability of $X$ taking a single exact value is always 0, regardless of the PDF value at that point.

4. Mode of a Continuous Distribution★★★☆☆⏱ 4 min

For discrete distributions, the mode is the most likely individual value. For continuous distributions, the mode is the value of $x$ where the PDF $f (x)$ reaches its maximum value. We use calculus to find this maximum.

📘 Definition

Mode of a continuous distribution

The mode is the value of $x$ at which the PDF $f (x)$ attains its global maximum value. A distribution can be unimodal (one mode), bimodal (two modes), or have no unique mode.

📐 Worked Example

Find the mode of $f (x) = {12 x^{2} (1 - x) 0 0 \leq x \leq 1 otherwise$ .

1
Expand and differentiate $f (x)$ to find critical points:
2
$f(x) = 12x^2 - 12x^3$
3
$f'(x) = 24x - 36x^2 = 12x(2 - 3x)$
4
Set derivative equal to 0 to find critical points:
5
$12x(2 - 3x) = 0 \implies x = 0 \text{ or } x = \frac{2}{3}$
6
Use the second derivative test to confirm a maximum:
7
$f''(x) = 24 - 72x. \quad At \ x = \frac{2}{3}, f''(\frac{2}{3}) = 24 - 48 = -24 < 0$
8
Endpoints $x = 0$ and $x = 1$ give $f (x) = 0$ , which is the minimum. So the global maximum is at $x = \frac{2}{3}$ , so the mode is $\frac{2}{3}$ .

5. Common Pitfalls

Wrong move:

Claiming $P (X = a) = f (a)$

Why:

For continuous distributions, $f (a)$ is the density at $a$ , not a probability. Probability is always area over an interval.

Correct move:

Always state $P (X = a) = 0$ for any continuous random variable, regardless of $f (a)$ .

Wrong move:

Using $f (b) - f (a)$ instead of integrating to find $P (a < X < b)$

Why:

Many students mix up PDF properties with the fundamental theorem of calculus for CDFs, and incorrectly use the difference of PDF values.

Correct move:

Always calculate $P (a < X < b)$ as the definite integral of $f (x)$ from $a$ to $b$ .

Wrong move:

Only checking that the total integral equals 1 when verifying a PDF, ignoring non-negativity

Why:

A function with a negative region cannot be a valid PDF, even if its total integral equals 1. Examiners frequently test this.

Correct move:

Always explicitly check both conditions: $f (x) \geq 0$ for all $x$ , and $\int_{- \infty}^{\infty} f (x) d x = 1$ .

Wrong move:

Stopping at the first internal critical point when finding the mode, ignoring endpoints

Why:

The maximum of $f (x)$ can occur at the endpoint of the domain, not just at a turning point inside the interval.

Correct move:

Always evaluate $f (x)$ at all critical points and endpoints to confirm the global maximum for the mode.

6. Quick Reference Cheatsheet

PDF Property	Rule
Non-negativity	$f (x) \geq 0$ for all $x$
Total area under curve	$\int_{- \infty}^{\infty} f (x) d x = 1$
Interval probability	$P (a < X < b) = \int_{a}^{b} f (x) d x$
Single point probability	$P (X = a) = 0$
Mode	Value of $x$ where $f (x)$ is maximized

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 2
Verify PDF validity, calculate probability
2022 · 1
Find mode of a triangular PDF

What's Next

PDFs are the foundation of all continuous probability topics you will encounter in IB AA HL, and mastery of these core rules is essential for exam success. Next, you will extend this knowledge to cumulative distribution functions (CDFs), which let you calculate cumulative probabilities, percentiles, and quantiles. From there, you will study common named continuous distributions including the uniform, exponential, and normal distribution, all of which are heavily tested in IB exams. You will also learn to calculate expectation and variance for continuous distributions, building on the PDF integration skills you practiced here.