Continuous probability distributions

CIE A-Level Further Mathematics· Unit 4: Further Probability & Statistics· 15 min read

1. Valid Probability Density Functions★★☆☆☆⏱ 5 min

A continuous random variable can take any value within an interval of real numbers, unlike discrete variables which only take distinct, separate values. The probability distribution of a continuous variable is described by a probability density function (pdf).

📘 Definition

Probability Density Function (pdf)

$f_{X} (x)$

A non-negative function $f (x)$ that describes the relative likelihood for the continuous random variable $X$ to take a given value. The total area under the curve of $f (x)$ over the full range of $X$ must equal 1.

Example:

$f (x) = 2 x$ for $0 \leq x \leq 1$ is a valid pdf.

📐 Worked Example

Show that $f (x) = \frac{1}{8} (x + 1)$ for $2 \leq x \leq 4$ is a valid pdf.

1
Check the two required conditions for validity: non-negativity and total area = 1.
2
For $2 \leq x \leq 4$ , $x + 1$ is always positive, so $f (x) > 0$ , satisfying the non-negativity condition.
3
Integrate $f (x)$ over the full range of $x$ :
4
$\int_{2}^{4} \frac{1}{8}(x+1) dx = \frac{1}{8} \left[ \frac{x^2}{2} + x \right]_{2}^{4}$
5
Evaluate the definite integral:
6
$\frac{1}{8} \left( (8 + 4) - (2 + 2) \right) = \frac{1}{8} (12 - 4) = 1$
7
Both conditions are satisfied, so $f (x)$ is a valid pdf.

Exam tip:

Always check both conditions for validity: non-negativity AND total area = 1. Examiners regularly penalise candidates who only check one condition.

2. Calculating Probabilities from pdfs★★☆☆☆⏱ 5 min

For continuous distributions, probability equals the area under the pdf curve between the bounds of the interval. A key property of continuous random variables is that the probability of $X$ taking any single specific value is always zero.

P(a \leq X \leq b) = \int_{a}^{b} f(x) dx

📐 Worked Example

For the valid pdf $f (x) = \frac{1}{8} (x + 1)$ for $2 \leq x \leq 4$ , calculate $P (2.5 < X < 3.5)$ .

1
Set up the integral of $f (x)$ between the given bounds:
2
$P(2.5 < X < 3.5) = \int_{2.5}^{3.5} \frac{1}{8}(x+1) dx$
3
Compute the antiderivative and evaluate:
4
$= \frac{1}{8} \left[ \frac{x^2}{2} + x \right]_{2.5}^{3.5}$
5
$= \frac{1}{8} \left( (\frac{12.25}{2} + 3.5) - (\frac{6.25}{2} + 2.5) \right) = 0.5$

3. Cumulative Distribution Functions (cdfs)★★★☆☆⏱ 6 min

The cumulative distribution function $F (x)$ gives the probability that $X$ is less than or equal to a given value $x$ , and is derived by integrating the pdf from the lower bound of $X$ up to $x$ .

📘 Definition

Cumulative Distribution Function (cdf)

$F_{X} (x)$

$F (x) = P (X \leq x) = \int_{- \infty}^{x} f (t) d t$ . For a pdf defined over $a \leq x \leq b$ , $F (x) = 0$ for $x < a$ and $F (x) = 1$ for $x > b$ .

📐 Worked Example

Find the cumulative distribution function for the pdf $f (x) = \frac{1}{8} (x + 1)$ , $2 \leq x \leq 4$ .

1
For $x < 2$ , there is no probability, so $F (x) = 0$ .
2
For $2 \leq x \leq 4$ , integrate the pdf from 2 to $x$ :
3
$F(x) = \int_{2}^{x} \frac{1}{8}(t+1) dt = \frac{1}{8} \left[ \frac{t^2}{2} + t \right]_{2}^{x}$
4
Simplify the expression:
5
$= \frac{1}{8} \left( \frac{x^2}{2} + x - 4 \right) = \frac{x^2 + 2x - 8}{16}$
6
For $x > 4$ , $F (x) = 1$ , which checks out at $x = 4$ : $\frac{16 + 8 - 8}{16} = 1$ . The full cdf is:
7
$F(x) = \begin{cases} 0 & x < 2 \\ \frac{x^2 + 2x - 8}{16} & 2 \leq x \leq 4 \\ 1 & x > 4 \end{cases}$

4. Descriptive Statistics for Continuous Distributions★★★☆☆⏱ 7 min

We calculate common descriptive statistics for continuous distributions using integration, following these standard rules:

Expectation (mean): $E (X) = \int_{- \infty}^{\infty} x f (x) d x$
Variance: $V a r (X) = E (X^{2}) - [E (X)]^{2}$ , where $E (X^{2}) = \int_{- \infty}^{\infty} x^{2} f (x) d x$
Median: The value $m$ that satisfies $F (m) = 0.5$
Mode: The value of $x$ that maximises $f (x)$ , found via differentiation

📐 Worked Example

Calculate the median of the distribution with pdf $f (x) = \frac{1}{8} (x + 1)$ , $2 \leq x \leq 4$ .

1
Use the cdf derived earlier, and set $F (m) = 0.5$ :
2
$\frac{m^2 + 2m - 8}{16} = 0.5$
3
Rearrange to form a quadratic equation:
4
$m^2 + 2m - 16 = 0$
5
Solve the quadratic, take the root in the interval $[2, 4]$ :
6
$m = -1 + \sqrt{17} \approx 3.12$
7
The median is approximately 3.12 (3 significant figures).

5. Common Pitfalls

Wrong move:

Forgetting to check the non-negativity condition when testing for a valid pdf

Why:

Many students only check that the total integral equals 1, but a function can integrate to 1 and be negative over part of its range

Correct move:

Always confirm that $f (x) \geq 0$ for all $x$ in the range of $X$ before checking the integral condition

Wrong move:

Calculating $P (X = a)$ as $f (a)$ for continuous distributions

Why:

Confusion between discrete probability mass functions and continuous probability density functions

Correct move:

Remember that $P (X = a) = 0$ for any continuous random variable, probability is only non-zero over intervals

Wrong move:

Only writing the expression for cdf over the interval where $f (x)$ is non-zero, omitting the other cases

Why:

Skipping the cases for $x$ below the lower bound or above the upper bound of $X$

Correct move:

Always specify $F (x) = 0$ for $x < min (X)$ and $F (x) = 1$ for $x > max (X)$

Wrong move:

Calculating variance as $[E (X)]^{2} - E (X^{2})$

Why:

Mixing up the order of terms in the variance formula, leading to negative variance

Correct move:

Memorise that $V a r (X) = E (X^{2}) - [E (X)]^{2}$ , which always gives a non-negative result

Wrong move:

Stopping at the first stationary point when finding the mode, not checking endpoints

Why:

Assuming the mode must be an interior critical point, when the maximum could be at an endpoint

Correct move:

Check all critical points and endpoints of the range to find the maximum value of $f (x)$

6. Quick Reference Cheatsheet

Quantity	Formula
Valid pdf conditions	$f (x) \geq 0 \forall x, \int_{- \infty}^{\infty} f (x) d x = 1$
$P (a \leq X \leq b)$	$\int_{a}^{b} f (x) d x$
cdf from pdf	$F (x) = \int_{- \infty}^{x} f (t) d t$
Expectation $E (X)$	$\int_{- \infty}^{\infty} x f (x) d x$
Variance $V a r (X)$	$E (X^{2}) - [E (X)]^{2}$
Median $m$	$F (m) = 0.5$
Mode	Value of $x$ maximising $f (x)$

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.

2022 · 1
Valid pdf + probability calculation
2023 · 2
Find cdf + median
2021 · 1
Calculate expectation and variance

What's Next

Continuous probability distributions form the foundation of all further work in statistics for CIE A-Level Further Maths. You will apply these core integration and manipulation skills to specific named continuous distributions, including uniform, exponential, normal and chi-squared distributions. These skills are also critical for more advanced topics like probability generating functions, continuous hypothesis testing and linear combinations of random variables later in the unit.