What is a continuous random variable in the context of Edexcel International A Levels Statistics 2?

In Edexcel International A Levels Statistics 2, a continuous random variable is a variable that can take any value within a given range. Unlike discrete random variables, which have distinct and separate values, continuous random variables are associated with measurements and can assume an infinite number of possible values within an interval. Examples include height, weight, and temperature.

How is the probability density function (PDF) used with continuous random variables?

The probability density function (PDF) is used to describe the likelihood of a continuous random variable taking on a particular value. For continuous random variables, the probability of the variable taking an exact value is zero. Instead, the PDF helps determine the probability that the variable falls within a certain range by calculating the area under the curve of the PDF over that range.

What is the difference between a probability density function (PDF) and a cumulative distribution function (CDF) for continuous random variables?

For continuous random variables, the probability density function (PDF) represents the relative likelihood of the variable taking on a specific value, while the cumulative distribution function (CDF) represents the probability that the variable will take a value less than or equal to a specific value. The CDF is obtained by integrating the PDF from the lower bound of the variable's range up to the value of interest.

How do you calculate the expected value of a continuous random variable?

The expected value of a continuous random variable is calculated by integrating the product of the variable and its probability density function (PDF) over the entire range of the variable. Mathematically, it is expressed as E(X) = ∫ x * f(x) dx, where f(x) is the PDF of the random variable X. This integral provides a measure of the central tendency of the distribution.

Why is understanding continuous random variables important in Statistics 2 for Edexcel International A Levels?

Understanding continuous random variables is crucial in Statistics 2 for Edexcel International A Levels because they form the basis for modeling and analyzing real-world phenomena that involve measurements. Mastery of this concept allows students to apply statistical methods to a wide range of fields, including science, engineering, and economics, where data is often continuous in nature.

Why is "Forgetting to find the normalisation constant $k$" a common mistake in Continuous random variables?

Why it happens: Students treat the PDF as already valid. How to avoid it: Always check: if f involves an unknown constant, the FIRST step is f\,dx = 1 to find it. Source: WST02/01 examiner reports

Why is "CDF written only on the support, not as a piecewise function" a common mistake in Continuous random variables?

Why it happens: Students stop after the 'interesting' interval. How to avoid it: Write F(x) = 0 for x b. ALL three pieces. Source: WST02/01 examiner reports

Why is "Using $\text{Var}(X) = \int (x - \mu)^{2} f(x)\,dx$ when it's awkward" a common mistake in Continuous random variables?

Why it happens: Definition is correct but harder to compute. How to avoid it: Always use the computational version: Var(X) = E(X^2) - ^2. Easier integration.

Why is "Missing endpoint modes" a common mistake in Continuous random variables?

Why it happens: Students automatically differentiate. How to avoid it: If f is monotonic on its support, the mode is at the endpoint where f is largest — NOT from f' = 0. Always check the behaviour on the support before differentiating.

Why is "Computing $E(X)$ by integration when symmetry gives it directly" a common mistake in Continuous random variables?

Why it happens: Habit. How to avoid it: If f is symmetric about c on its support, E(X) = c. Use the shortcut and save time; the variance still needs integration.

Why is "Forgetting to state '$f(x) = 0$ otherwise'" a common mistake in Continuous random variables?

Why it happens: Students focus on the formula. How to avoid it: PDFs MUST be defined for all real x. Always close with 'and zero otherwise'.

Statistics 2Edexcel International A Levels Mathematics (XMA01-YMA01)

Continuous random variables

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Continuous Random Variables Study Notes — Edexcel IAL Mathematics S2 (WST02/01, 2018 spec — 2026 onwards)

Probability is AREA under a density. PDF $f(x) \ge 0$ with $\int f = 1$ . CDF $F(x) = P(X \le x)$ links to PDF via $f = F'$ . Mean, variance, median, quartiles, and mode all flow from integration. The framework underpins the continuous uniform distribution, the normal distribution, and sampling theory.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

S2 4.1 — Understand the concept of a continuous random variable and its PDF.
S2 4.2 — Verify a PDF; find unknown constants by normalisation.
S2 4.3 — Compute probabilities by integration.
S2 4.4 — Find and use the CDF including piecewise representation.
S2 4.5 — Compute mean, variance, median, quartiles, and mode.

PDF — the density function

$f(x) \ge 0$ with $\int f = 1$ . Probability is area.

Definition. A continuous random variable $X$ has a probability density function $f(x)$ if:

$f(x) \ge 0$ for all $x$ .
$\int_{-\infty}^{\infty} f(x)\,dx = 1$ .
$P(a \le X \le b) = \int_a^b f(x)\,dx$ for any $a \le b$ .

Critical fact. $P(X = c) = \int_c^c f(x)\,dx = 0$ for any single $c$ . So:

$P(a < X < b) = P(a \le X \le b) = P(a < X \le b) = P(a \le X < b).$

Strict and non-strict inequalities give the SAME probability for continuous variables.

Piecewise PDFs. Typically $f(x)$ is defined by a formula on an interval $[a, b]$ and equals zero elsewhere. The 'and zero otherwise' clause is non-negotiable.

Worked example. $f(x) = kx(4 - x)$ for $0 \le x \le 4$ , zero otherwise.

Find $k$ : $\int_0^4 kx(4-x)\,dx = 1 \Rightarrow k \cdot 32/3 = 1 \Rightarrow k = 3/32$ .
$P(1 \le X \le 3) = (3/32)\int_1^3 x(4-x)\,dx = 11/16$ .

$f \ge 0$ everywhere.
$\int f = 1$ over all reals.
$P(X = c) = 0$ for any single $c$ .
Strict / non-strict inequalities equivalent.
Always state 'zero otherwise'.

CDF — cumulative distribution

$F(x) = P(X \le x)$ . Piecewise: $0$ below support, integrand on support, $1$ above.

Definition.

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

Properties.

$F$ is non-decreasing.
$F(-\infty) = 0$ , $F(\infty) = 1$ .
$F$ is continuous (no jumps — the variable is continuous).
Relationship: $f(x) = F'(x)$ .

Piecewise structure. If $f$ is non-zero only on $[a, b]$ , then:

$F(x) = 0$ for $x < a$ .
$F(x) = \int_a^x f(t)\,dt$ for $a \le x \le b$ .
$F(x) = 1$ for $x > b$ .

ALL three pieces must be written out — examiners deduct for partial answers.

Worked example. $f(x) = x/8$ on $[0, 4]$ .

$x < 0$ : $F(x) = 0$ .
$0 \le x \le 4$ : $F(x) = \int_0^x (t/8)\,dt = x^{2}/16$ .
$x > 4$ : $F(x) = 1$ .
Check continuity: $F(4) = 16/16 = 1$ . Good.

Using the CDF.

$P(X \le k) = F(k)$ .
$P(X > k) = 1 - F(k)$ .
$P(a \le X \le b) = F(b) - F(a)$ .
Inverse for median / quartiles: solve $F(x) = p$ .

$F(x) = P(X \le x)$ .
$F$ piecewise: $0$ , integrand, $1$ .
$f = F'$ .
Check $F(\text{upper bound}) = 1$ .

Mean and variance

$E(X) = \int x f\,dx$ , $\text{Var}(X) = E(X^{2}) - \mu^{2}$ .

Mean.

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

In practice, integrate only over the support of $X$ .

Second moment.

$E(X^{2}) = \int_{-\infty}^{\infty} x^{2} f(x)\,dx.$

Variance (computational form — preferred).

$\text{Var}(X) = E(X^{2}) - [E(X)]^{2}.$

Definitional form (rarely used in S2 working).

$\text{Var}(X) = \int (x - \mu)^{2} f(x)\,dx.$

Both give the same answer, but the computational form involves easier integration.

Standard deviation. $\sigma = \sqrt{\text{Var}(X)}$ .

Worked example. $f(x) = (3/32) x(4 - x)$ , $0 \le x \le 4$ .

$E(X) = (3/32)\int_0^4 x \cdot x(4 - x)\,dx = (3/32) \cdot 64/3 = 2$ .
$E(X^{2}) = (3/32)\int_0^4 x^{2} \cdot x(4 - x)\,dx = (3/32) \cdot 256/5 = 24/5 = 4.8$ .
$\text{Var}(X) = 4.8 - 4 = 0.8$ .

Symmetry shortcut. If $f$ is symmetric about $c$ on its support, $E(X) = c$ directly. The variance still needs integration.

Expectation of a function. $E(g(X)) = \int g(x) f(x)\,dx$ . Used for $E(X^{2})$ , but generalises.

$E(X) = \int x f\,dx$ .
$E(X^{2}) = \int x^{2} f\,dx$ .
$\text{Var}(X) = E(X^{2}) - \mu^{2}$ .
Symmetric $f$ $\Rightarrow E(X)$ = centre.

Median, quartiles, percentiles

Solve $F(x) = p$ for the $p$ -quantile. Median is $p = 0.5$ .

Median. Solve $F(m) = 1/2$ .

Quartiles.

Lower quartile $Q_1$ : $F(Q_1) = 1/4$ .
Upper quartile $Q_3$ : $F(Q_3) = 3/4$ .
Interquartile range IQR $= Q_3 - Q_1$ .

General percentile. $p$ th percentile $x_p$ : $F(x_p) = p/100$ .

Worked example. $F(x) = x^{2}/16$ on $[0, 4]$ .

Median: $m^{2}/16 = 0.5 \Rightarrow m = \sqrt{8} = 2\sqrt 2 \approx 2.83$ .
$Q_1$ : $Q_1^{2}/16 = 0.25 \Rightarrow Q_1 = 2$ .
$Q_3$ : $Q_3^{2}/16 = 0.75 \Rightarrow Q_3 = \sqrt{12} = 2\sqrt 3 \approx 3.46$ .
IQR $= 2\sqrt 3 - 2 \approx 1.46$ .

Sanity check. Median, $Q_1$ , $Q_3$ must all be in the support — here $[0, 4]$ . Confirmed.

Mean vs median. For a symmetric distribution, mean = median. For a skewed distribution, they differ. A PDF skewed right has mean $>$ median; skewed left has mean $<$ median.

Median: $F(m) = 1/2$ .
$Q_1$ : $F(Q_1) = 1/4$ . $Q_3$ : $F(Q_3) = 3/4$ .
Always check the solution lies in the support.
Skewness sign: mean vs median.

Mode — maximum of $f$

Differentiate $f$ , set to zero, confirm max. Or take endpoint if monotonic.

Definition. Mode = value(s) of $x$ that maximise $f(x)$ on its support.

Method (smooth PDF).

Differentiate $f$ : find $f'(x)$ .
Set $f'(x) = 0$ and solve.
Confirm maximum: $f''(x) < 0$ at the critical point, or check behaviour at boundaries.

Method (monotonic on support). If $f$ is strictly increasing on $[a, b]$ , mode is $b$ . If strictly decreasing, mode is $a$ . NO differentiation needed.

Worked example 1 (smooth). $f(x) = (3/32) x(4 - x)$ , $0 \le x \le 4$ .

$f'(x) = (3/32)(4 - 2x)$ .
$f'(x) = 0 \Rightarrow x = 2$ .
$f''(x) = -3/16 < 0$ , so max.
Mode $= 2$ .

Worked example 2 (monotonic). $f(x) = x/8$ , $0 \le x \le 4$ .

$f$ increasing. Mode is at $x = 4$ (the right endpoint).

Bimodal / multimodal. If $f$ has multiple maxima of equal height, all are modes. Rare in S2.

No mode. If $f$ is uniform (constant), every point in the support is a 'mode' — usually reported as 'no unique mode' or 'all values equally likely'.

Mode = $\arg\max_x f(x)$ .
Smooth: $f'(x) = 0$ , confirm max.
Monotonic: take the endpoint where $f$ is larger.
Uniform: no unique mode.

How it’s examined

Continuous random variables appear on every WST02/01 paper — typically a $12$ - $15$ mark multi-part question covering: find $k$ ( $2$ - $3$ marks), find a probability ( $2$ - $3$ marks), derive the CDF ( $3$ - $4$ marks), find mean / variance ( $4$ - $5$ marks), find median / quartiles or mode ( $2$ - $4$ marks). The CDF question is where the most marks are dropped — examiners insist on the full piecewise statement. Symmetry of the PDF often gives $E(X)$ without integration — exploit it. Variance always via $E(X^{2}) - \mu^{2}$ (NEVER the direct definition).

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WST02/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

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Step-by-step worked examples — Continuous random variables

Step-by-step solutions to past-paper-style questions on continuous random variables, written exactly the way a tutor would explain them at the board.

1Verify a PDF and compute probability (5 marks, S2)
Core• Adapted from WST02/01 January 2024 Q3• pdf, verification, probability
Question
The continuous random variable $X$ has PDF $f(x) = kx(4 - x)$ for $0 \le x \le 4$ , and zero otherwise. (a) Find $k$ . (b) Find $P(1 \le X \le 3)$ .
Step-by-step solution
1. Step 1
  (a) Total probability is $1$ . Integrate $f$ over its support:
  $\int_0^4 k x(4 - x)\,dx = 1$
2. Step 2
  Expand and integrate:
  $k \int_0^4 (4x - x^{2})\,dx = k\left[2x^{2} - \dfrac{x^{3}}{3}\right]_0^4 = k\left(32 - \dfrac{64}{3}\right) = \dfrac{32k}{3}$
3. Step 3
  Set equal to $1$ :
  $\dfrac{32k}{3} = 1 \Rightarrow k = \dfrac{3}{32}$
4. Step 4
  (b) Probability is the area under $f$ between $1$ and $3$ :
  $P(1 \le X \le 3) = \dfrac{3}{32}\int_1^3 (4x - x^{2})\,dx$
5. Step 5
  Compute the integral:
  $\int_1^3 (4x - x^{2})\,dx = \left[2x^{2} - \dfrac{x^{3}}{3}\right]_1^3 = (18 - 9) - (2 - \tfrac{1}{3}) = 9 - \tfrac{5}{3} = \tfrac{22}{3}$
6. Step 6
  Multiply:
  $P(1 \le X \le 3) = \dfrac{3}{32} \times \dfrac{22}{3} = \dfrac{22}{32} = \dfrac{11}{16}$
Answer
(a) $k = 3/32$ . (b) $P(1 \le X \le 3) = 11/16 = 0.6875$ .
Examiner tip
(a) M1 for $\int_0^4 f\,dx = 1$ . M1 for the antiderivative. A1 for $k = 3/32$ . (b) M1 for setting up $\int_1^3 f\,dx$ . A1 for $11/16$ . Decimal answer also acceptable.
2Mean and variance of a continuous RV (6 marks, S2)
Core• Adapted from WST02/01 June 2024 Q4• expectation, variance
Question
$X$ has PDF $f(x) = \dfrac{3}{32}x(4 - x)$ for $0 \le x \le 4$ , zero otherwise. Find (a) $E(X)$ , (b) $\text{Var}(X)$ .
Step-by-step solution
1. Step 1
  (a) Mean:
  $E(X) = \int_0^4 x \cdot \dfrac{3}{32}x(4 - x)\,dx = \dfrac{3}{32}\int_0^4 (4x^{2} - x^{3})\,dx$
2. Step 2
  Integrate:
  $\int_0^4 (4x^{2} - x^{3})\,dx = \left[\dfrac{4x^{3}}{3} - \dfrac{x^{4}}{4}\right]_0^4 = \dfrac{256}{3} - 64 = \dfrac{64}{3}$
3. Step 3
  So
  $E(X) = \dfrac{3}{32} \times \dfrac{64}{3} = 2$
4. Step 4
  By symmetry of the PDF about $x = 2$ , $E(X) = 2$ is expected — confirmed.
5. Step 5
  (b) Variance via $E(X^{2}) - \mu^{2}$ :
  $E(X^{2}) = \dfrac{3}{32}\int_0^4 x^{2} \cdot x(4 - x)\,dx = \dfrac{3}{32}\int_0^4 (4x^{3} - x^{4})\,dx$
6. Step 6
  Integrate:
  $\int_0^4 (4x^{3} - x^{4})\,dx = \left[x^{4} - \dfrac{x^{5}}{5}\right]_0^4 = 256 - \dfrac{1024}{5} = \dfrac{256}{5}$
7. Step 7
  So $E(X^{2}) = \dfrac{3}{32} \times \dfrac{256}{5} = \dfrac{24}{5} = 4.8$ . Then
  $\text{Var}(X) = 4.8 - 2^{2} = 0.8$
Answer
(a) $E(X) = 2$ . (b) $\text{Var}(X) = 0.8$ .
Examiner tip
(a) M1 for the integral set-up. M1 for antiderivative. A1 for $2$ . (b) M1 for $E(X^{2})$ set-up. A1 for $4.8$ . M1 for $E(X^{2}) - \mu^{2}$ . A1 for $0.8$ . Common error: integrating $x \cdot f$ but forgetting the $k$ factor.
3Find and use the CDF (6 marks, S2)
Extended• Adapted from WST02/01 January 2024 Q5• cdf, piecewise
Question
$X$ has PDF $f(x) = \dfrac{1}{8}x$ for $0 \le x \le 4$ , and zero otherwise. (a) Find the CDF $F(x)$ for all $x$ . (b) Hence find $P(X \le 2)$ and $P(X > 3)$ .
Step-by-step solution
1. Step 1
  (a) $F(x) = \int_{-\infty}^x f(t)\,dt$ . Three pieces:
2. Step 2
  For $x < 0$ : $F(x) = 0$ .
3. Step 3
  For $0 \le x \le 4$ :
  $F(x) = \int_0^x \dfrac{1}{8}t\,dt = \dfrac{1}{16}t^{2}\Big|_0^x = \dfrac{x^{2}}{16}$
4. Step 4
  For $x > 4$ : $F(x) = 1$ (cumulative probability saturates).
5. Step 5
  Check: $F(4) = 16/16 = 1$ — continuity at the upper bound. Good.
6. Step 6
  (b) From CDF: $P(X \le 2) = F(2) = 4/16 = 1/4$ .
7. Step 7
  And $P(X > 3) = 1 - F(3) = 1 - 9/16 = 7/16$ .
Answer
(a) $F(x) = 0$ for $x < 0$ ; $F(x) = x^{2}/16$ for $0 \le x \le 4$ ; $F(x) = 1$ for $x > 4$ . (b) $P(X \le 2) = 1/4$ , $P(X > 3) = 7/16$ .
Examiner tip
(a) M1 for $\int_0^x f\,dt$ . M1 for antiderivative. A1 for $F(x) = x^{2}/16$ . B1 for piecewise statement covering all three regions. (b) B1 each. Failing to state $F(x) = 0$ and $F(x) = 1$ for the outer regions loses a mark — the CDF must be defined everywhere.
4Find the median (4 marks, S2)
Extended• Adapted from WST02/01 June 2024 Q5• median, cdf inverse
Question
The PDF of $X$ is $f(x) = \dfrac{1}{8}x$ for $0 \le x \le 4$ . Find the median $m$ .
Step-by-step solution
1. Step 1
  Median $m$ satisfies $F(m) = 0.5$ .
2. Step 2
  From previous: $F(x) = x^{2}/16$ for $0 \le x \le 4$ . Set:
  $\dfrac{m^{2}}{16} = \dfrac{1}{2}$
3. Step 3
  Solve:
  $m^{2} = 8 \Rightarrow m = \sqrt{8} = 2\sqrt{2} \approx 2.83$
4. Step 4
  Check $m$ is in the support: $0 \le 2\sqrt 2 \le 4$ . Yes.
Answer
$m = 2\sqrt{2} \approx 2.83$ .
Examiner tip
M1 for $F(m) = 0.5$ . M1 for the quadratic. A1 for $m^{2} = 8$ . A1 for $m = 2\sqrt 2$ or $2.83$ . The PDF is increasing on $[0, 4]$ , so the median is above the midpoint $2$ — quick sanity check.
5Find the mode (3 marks, S2)
Extended• Adapted from WST02/01 January 2024 Q6• mode, differentiation
Question
$f(x) = \dfrac{3}{32}x(4 - x)$ for $0 \le x \le 4$ . Find the mode of $X$ .
Step-by-step solution
1. Step 1
  Mode = location of maximum of $f$ . Differentiate:
  $f'(x) = \dfrac{3}{32}(4 - 2x)$
2. Step 2
  Set $f'(x) = 0$ :
  $4 - 2x = 0 \Rightarrow x = 2$
3. Step 3
  Confirm maximum: $f''(x) = -\dfrac{3}{16} < 0$ , so $x = 2$ is a max. (Or note $f$ is a downward-opening parabola.)
Answer
Mode $= 2$ .
Examiner tip
M1 for differentiating $f$ . M1 for $f' = 0$ . A1 for $x = 2$ . B1 (in some mark schemes) for justifying max — usually implicit if $f$ is clearly a downward parabola. For PDFs that are monotonic on their support, the mode is at the endpoint where $f$ is largest.

Key Formulae — Continuous random variables

The formulae you need to memorise for continuous random variables on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

PDF properties
$f(x) \ge 0 \text{ everywhere}, \quad \int_{-\infty}^{\infty} f(x)\,dx = 1$
$f(x)$
probability density function
When to use
Whenever you need to find an unknown constant in a PDF, or to verify a candidate PDF. Properties NOT in the booklet — fundamental definitions.
Example
$f(x) = kx(4 - x)$ on $[0, 4]$ : $\int_0^4 kx(4-x)\,dx = 1 \Rightarrow k = 3/32$ .
Probability as area
$P(a < X < b) = \int_a^b f(x)\,dx$
$a, b$
bounds (real numbers)
When to use
All interval probabilities. NOT in booklet — definition.
Example
$P(1 \le X \le 3) = \int_1^3 \dfrac{3}{32}x(4-x)\,dx = 11/16$ .
CDF definition
$F(x) = P(X \le x) = \int_{-\infty}^{x} f(t)\,dt$
$F(x)$
cumulative distribution function
When to use
To find probabilities of half-infinite intervals; to find median / quartiles by solving $F(x) = p$ .
Example
$f(x) = x/8$ on $[0, 4]$ : $F(x) = x^{2}/16$ for $0 \le x \le 4$ .
Link between PDF and CDF
$f(x) = F'(x)$
$F'(x)$
derivative of CDF
When to use
When the question gives the CDF and asks for the PDF, or to verify consistency.
Example
$F(x) = x^{2}/16$ $\Rightarrow$ $f(x) = x/8$ .
Expectation
$E(X) = \int_{-\infty}^{\infty} x f(x)\,dx$
$E(X)$
mean of $X$
When to use
Whenever a question asks for $E(X)$ , mean, or expected value. Integral is over the support of $X$ .
Example
$f(x) = (3/32)x(4-x)$ : $E(X) = 2$ .
Variance
$\text{Var}(X) = E(X^{2}) - [E(X)]^{2}, \quad E(X^{2}) = \int x^{2} f(x)\,dx$
$\text{Var}(X)$
variance of $X$
When to use
Variance and SD. Compute $E(X^{2})$ first, then subtract $\mu^{2}$ .
Example
$f(x) = (3/32)x(4-x)$ : $E(X^{2}) = 24/5$ , $\text{Var}(X) = 24/5 - 4 = 4/5 = 0.8$ .
Median
$F(m) = \dfrac{1}{2}$
$m$
median
When to use
Median, quartiles ( $Q_1 = F^{-1}(0.25)$ , $Q_3 = F^{-1}(0.75)$ ), percentiles. Solve the equation for $m$ .
Example
$F(x) = x^{2}/16$ : $m^{2}/16 = 0.5 \Rightarrow m = 2\sqrt 2$ .
Mode
$\text{Mode} = \arg\max_x f(x)$
$f$
PDF
When to use
Mode of a continuous RV is the maximum point of the PDF. Differentiate $f$ and set to zero.
Example
$f(x) = (3/32)x(4-x)$ : $f'(x) = (3/32)(4 - 2x) = 0 \Rightarrow x = 2$ .

Key Definitions and Keywords — Continuous random variables

Definitions to memorise and the exact keywords mark schemes credit for continuous random variables answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Continuous random variable
Examiner keyword
A random variable $X$ whose distribution is described by a PDF $f(x)$ (rather than a PMF). $X$ takes uncountably many values, and $P(X = x) = 0$ for any single $x$ .
Probability density function (PDF)
Examiner keyword
Function $f(x) \ge 0$ with $\int_{-\infty}^{\infty} f(x)\,dx = 1$ , such that $P(a < X < b) = \int_a^b f(x)\,dx$ .
Cumulative distribution function (CDF)
Examiner keyword
$F(x) = P(X \le x) = \int_{-\infty}^{x} f(t)\,dt$ . Non-decreasing; $F(-\infty) = 0$ , $F(\infty) = 1$ .
Mean (expectation) $E(X)$
Examiner keyword
$E(X) = \int x f(x)\,dx$ . Long-run average value of $X$ .
Variance $\text{Var}(X)$
Examiner keyword
$\text{Var}(X) = E(X^{2}) - [E(X)]^{2}$ . Measures spread about the mean. SD is $\sqrt{\text{Var}(X)}$ .
Median
Examiner keyword
Value $m$ with $F(m) = 0.5$ . Splits the distribution into two halves of equal probability.
Mode
Examiner keyword
Value(s) where $f(x)$ is maximised. For differentiable PDFs, found by $f'(x) = 0$ .

Common Mistakes and Misconceptions — Continuous random variables

The traps other students keep falling into on continuous random variables questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to find the normalisation constant $k$
WST02/01 examiner reports
Why it happens
Students treat the PDF as already valid.
How to avoid it
Always check: if $f$ involves an unknown constant, the FIRST step is $\int f\,dx = 1$ to find it.
✕CDF written only on the support, not as a piecewise function
WST02/01 examiner reports
Why it happens
Students stop after the 'interesting' interval.
How to avoid it
Write $F(x) = 0$ for $x < a$ , $F(x) = \ldots$ for $a \le x \le b$ , $F(x) = 1$ for $x > b$ . ALL three pieces.
✕Using $\text{Var}(X) = \int (x - \mu)^{2} f(x)\,dx$ when it's awkward
Why it happens
Definition is correct but harder to compute.
How to avoid it
Always use the computational version: $\text{Var}(X) = E(X^{2}) - \mu^{2}$ . Easier integration.
✕Missing endpoint modes
Why it happens
Students automatically differentiate.
How to avoid it
If $f$ is monotonic on its support, the mode is at the endpoint where $f$ is largest — NOT from $f' = 0$ . Always check the behaviour on the support before differentiating.
✕Computing $E(X)$ by integration when symmetry gives it directly
Why it happens
Habit.
How to avoid it
If $f$ is symmetric about $c$ on its support, $E(X) = c$ . Use the shortcut and save time; the variance still needs integration.
✕Forgetting to state ' $f(x) = 0$ otherwise'
Why it happens
Students focus on the formula.
How to avoid it
PDFs MUST be defined for all real $x$ . Always close with 'and zero otherwise'.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Continuous random variables — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Continuous random variables

Short Study Notes in the form of Slides

Detailed Notes

What you’ll learn

How it’s examined

1Verify a PDF and compute probability (5 marks, S2)

2Mean and variance of a continuous RV (6 marks, S2)

3Find and use the CDF (6 marks, S2)

4Find the median (4 marks, S2)

5Find the mode (3 marks, S2)

PDF properties

Probability as area

CDF definition

Link between PDF and CDF

Expectation

Variance

Median

Mode

Continuous random variable

Probability density function (PDF)

Cumulative distribution function (CDF)

Mean (expectation) E(X)E(X)E(X)

Variance Var(X)\text{Var}(X)Var(X)

Median

Mode

✕Forgetting to find the normalisation constant kkk

✕CDF written only on the support, not as a piecewise function

✕Using Var(X)=∫(x−μ)2f(x) dx\text{Var}(X) = \int (x - \mu)^{2} f(x)\,dxVar(X)=∫(x−μ)2f(x)dx when it's awkward

✕Missing endpoint modes

✕Computing E(X)E(X)E(X) by integration when symmetry gives it directly

✕Forgetting to state 'f(x)=0f(x) = 0f(x)=0 otherwise'

Practice questions

Past paper style quiz

Continuous random variables — frequently asked questions

Mean (expectation) $E(X)$

Variance $\text{Var}(X)$

✕Forgetting to find the normalisation constant $k$

✕Using $\text{Var}(X) = \int (x - \mu)^{2} f(x)\,dx$ when it's awkward

✕Computing $E(X)$ by integration when symmetry gives it directly

✕Forgetting to state ' $f(x) = 0$ otherwise'