What is a Continuous Uniform Distribution in the context of Edexcel International A Levels Statistics 2?

In Edexcel International A Levels Statistics 2, a Continuous Uniform Distribution, also known as a rectangular distribution, is a type of probability distribution where all outcomes are equally likely within a certain range. It is defined by two parameters, a and b, which represent the minimum and maximum values of the distribution. The probability density function is constant between these two values and zero outside this range.

How do you calculate the probability of an event in a Continuous Uniform Distribution?

To calculate the probability of an event in a Continuous Uniform Distribution, you need to determine the range of the event and divide it by the total range of the distribution. If X is uniformly distributed between a and b, the probability that X falls between two values c and d (where a ≤ c < d ≤ b) is given by (d - c) / (b - a).

What are the mean and variance of a Continuous Uniform Distribution?

For a Continuous Uniform Distribution defined on the interval [a, b], the mean (expected value) is calculated as (a + b) / 2. The variance is given by (b - a)^2 / 12. These formulas help in understanding the central tendency and the spread of the distribution, which are key concepts in the Edexcel International A Levels Statistics 2 curriculum.

How is a Continuous Uniform Distribution different from a Discrete Uniform Distribution?

A Continuous Uniform Distribution differs from a Discrete Uniform Distribution in that it deals with continuous data, meaning it can take any value within a specified range. In contrast, a Discrete Uniform Distribution deals with discrete data, where outcomes are distinct and finite. In the context of Edexcel International A Levels Statistics 2, understanding these differences is crucial for correctly applying statistical methods to different types of data.

Can you provide an example of a real-world scenario where a Continuous Uniform Distribution might be used?

A real-world example of a Continuous Uniform Distribution could be the time it takes for a bus to arrive at a stop, assuming it can arrive at any time within a 10-minute window with equal likelihood. If the bus is expected to arrive between 8:00 and 8:10, the arrival time is uniformly distributed over this interval. This concept is often explored in Edexcel International A Levels Statistics 2 to illustrate practical applications of probability distributions.

Why is "Stating $f(x) = 1$ instead of $f(x) = 1/(b - a)$" a common mistake in Continuous Uniform Distribution?

Why it happens: Confusion with uniform on [0, 1] (where f = 1). How to avoid it: Always compute 1/(b - a) explicitly. U(2, 10): f = 1/8, not 1.

Why is "Writing $\text{Var}(X) = (b - a)/12$ instead of $(b - a)^{2}/12$" a common mistake in Continuous Uniform Distribution?

Why it happens: Missing the squared. How to avoid it: Memorise: (b - a)^2/12 — SQUARED. The variance has dimensions of X^2. Source: WST02/01 examiner reports

Why is "Writing only the middle piece of the CDF" a common mistake in Continuous Uniform Distribution?

Why it happens: Same issue as continuous random variables generally. How to avoid it: Always state CDF as piecewise on all three regions: 0 below a, integrand on [a, b], 1 above b.

Why is "Computing a probability where the interval is partly outside the support" a common mistake in Continuous Uniform Distribution?

Why it happens: Students apply the formula without checking. How to avoid it: Clip the interval to the support. U(2, 10), P(X > 12) = 0 — not negative. P(0 < X < 5) = (5 - 2)/(10 - 2) — clip 0 to 2.

Why is "Misidentifying a non-uniform variable as uniform" a common mistake in Continuous Uniform Distribution?

Why it happens: Many real-world variables are bounded but not uniform. How to avoid it: Uniform requires CONSTANT density. Waiting times for an event arriving randomly within a fixed cycle are uniform. Heights / weights / exam marks are NOT uniform.

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

Continuous Uniform Distribution

Q: What are the mean and variance of a Continuous Uniform Distribution?

For a Continuous Uniform Distribution defined on the interval [a, b], the mean (expected value) is calculated as (a + b) / 2. The variance is given by (b - a)^2 / 12. These formulas help in understanding the central tendency and the spread of the distribution, which are key concepts in the Edexcel International A Levels Statistics 2 curriculum.

Q: How is a Continuous Uniform Distribution different from a Discrete Uniform Distribution?

A Continuous Uniform Distribution differs from a Discrete Uniform Distribution in that it deals with continuous data, meaning it can take any value within a specified range. In contrast, a Discrete Uniform Distribution deals with discrete data, where outcomes are distinct and finite. In the context of Edexcel International A Levels Statistics 2, understanding these differences is crucial for correctly applying statistical methods to different types of data.

Q: Can you provide an example of a real-world scenario where a Continuous Uniform Distribution might be used?

A real-world example of a Continuous Uniform Distribution could be the time it takes for a bus to arrive at a stop, assuming it can arrive at any time within a 10-minute window with equal likelihood. If the bus is expected to arrive between 8:00 and 8:10, the arrival time is uniformly distributed over this interval. This concept is often explored in Edexcel International A Levels Statistics 2 to illustrate practical applications of probability distributions.

Q: Why is "Stating $f(x) = 1$ instead of $f(x) = 1/(b - a)$" a common mistake in Continuous Uniform Distribution?

Why it happens: Confusion with uniform on [0, 1] (where f = 1). How to avoid it: Always compute 1/(b - a) explicitly. U(2, 10): f = 1/8, not 1.

Q: Why is "Writing $\text{Var}(X) = (b - a)/12$ instead of $(b - a)^{2}/12$" a common mistake in Continuous Uniform Distribution?

Why it happens: Missing the squared. How to avoid it: Memorise: (b - a)^2/12 — SQUARED. The variance has dimensions of X^2. Source: WST02/01 examiner reports

Q: Why is "Writing only the middle piece of the CDF" a common mistake in Continuous Uniform Distribution?

Why it happens: Same issue as continuous random variables generally. How to avoid it: Always state CDF as piecewise on all three regions: 0 below a, integrand on [a, b], 1 above b.

Q: Why is "Computing a probability where the interval is partly outside the support" a common mistake in Continuous Uniform Distribution?

Why it happens: Students apply the formula without checking. How to avoid it: Clip the interval to the support. U(2, 10), P(X > 12) = 0 — not negative. P(0 < X < 5) = (5 - 2)/(10 - 2) — clip 0 to 2.

Q: Why is "Misidentifying a non-uniform variable as uniform" a common mistake in Continuous Uniform Distribution?

Why it happens: Many real-world variables are bounded but not uniform. How to avoid it: Uniform requires CONSTANT density. Waiting times for an event arriving randomly within a fixed cycle are uniform. Heights / weights / exam marks are NOT uniform.

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next.

Previous Next

Learn this Topic

Start with the slides for the quick version, then go deeper with the full study notes.

Short Study Notes in the form of Slides

Read the notes first. If the method in a worked example clicks, you're ready for the questions.

Page 1 / 0

Detailed Notes

Full prose, callouts and a recap — built for A* mastery, not just a quick scan.

Take these study notes with you

Download a branded PDF — full prose, callouts, recap and memorise list for Continuous Uniform Distribution, ready to print or save offline.

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

$X \sim U(a, b)$ : constant density $1/(b - a)$ on $[a, b]$ . The simplest continuous distribution. Mean is the midpoint $(a+b)/2$ , variance $(b-a)^{2}/12$ . Probabilities reduce to length ratios. The natural model for 'random arrival within a fixed window'.

What you’ll learn

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

S2 5.1 — Identify when the uniform distribution is the appropriate model.
S2 5.2 — Write down PDF and CDF of $U(a, b)$ .
S2 5.3 — Compute probabilities via length ratios or CDF.
S2 5.4 — Use $E(X) = (a + b)/2$ and $\text{Var}(X) = (b - a)^{2}/12$ .
S2 5.5 — Solve inverse problems (find $a$ , $b$ , or specific quantiles).

Setup — when is uniform appropriate?

Every value in $[a, b]$ equally likely. PDF is a flat horizontal line.

Definition. $X \sim U(a, b)$ if $X$ takes values in $[a, b]$ with all sub-intervals of equal length having equal probability.

PDF. Constant on the support:

$f(x) = \dfrac{1}{b - a} \text{ for } a \le x \le b, \quad f(x) = 0 \text{ otherwise.}$

The height $1/(b - a)$ comes from the normalisation $\int_a^b f\,dx = 1$ .

When the model fits.

Random arrival time within a fixed cycle (bus schedules, traffic lights, clock-pulse positions).
Random angle on a clock face.
Pseudo-random number generators (computer-generated 'random numbers' are designed to be $U(0, 1)$ ).
'Maximally uninformative' prior in Bayesian statistics (not S2).

When it does NOT fit.

Heights, weights, exam marks — these cluster around a central value (use normal).
Waiting times for a Poisson process — these are exponential, not uniform.
Any data with a clear central tendency.

Symmetry. The uniform PDF is symmetric about its midpoint $(a + b)/2$ , so mean = median = midpoint.

All values equally likely on $[a, b]$ .
PDF height: $1/(b - a)$ .
Symmetric — mean = median.
Random-arrival models, clock-face angles.

PDF, CDF, and probabilities

Length ratios for probabilities. CDF is a straight ramp.

PDF. $f(x) = 1/(b - a)$ on $[a, b]$ .

CDF. Integrate from $a$ :

$F(x) = \int_a^x \dfrac{1}{b - a}\,dt = \dfrac{x - a}{b - a}.$

Piecewise:

Region	$F(x)$
$x < a$	$0$
$a \le x \le b$	$(x - a)/(b - a)$
$x > b$	$1$

Probabilities (length ratios). For any sub-interval $[c, d] \subseteq [a, b]$ :

$P(c \le X \le d) = \dfrac{d - c}{b - a}.$

Tail probabilities.

$P(X \le k) = (k - a)/(b - a)$ for $a \le k \le b$ .
$P(X > k) = (b - k)/(b - a)$ for $a \le k \le b$ .

Outside the support.

$P(X < a) = 0$ , $P(X > b) = 0$ .
If the question asks $P(0 < X < 5)$ but $X \sim U(2, 10)$ , clip $0$ to $2$ : $P(2 \le X \le 5) = 3/8$ .

Inverse problems. $P(X > c) = 0.2 \Rightarrow F(c) = 0.8 \Rightarrow (c - a)/(b - a) = 0.8 \Rightarrow$ solve.

PDF is a flat rectangle of height $1/(b - a)$ .
CDF is a straight ramp from $0$ to $1$ .
Probability = sub-interval length / total length.
Clip to support before computing.

Mean, variance, and standard deviation

$\mu = (a + b)/2$ , $\sigma^{2} = (b - a)^{2}/12$ . Both in the booklet.

Mean. By symmetry:

$E(X) = \dfrac{a + b}{2}.$

Variance. From integration (or memorise):

$\text{Var}(X) = \dfrac{(b - a)^{2}}{12}.$

Derivation sketch.

$E(X^{2}) = \int_a^b x^{2}/(b - a)\,dx = (b^{3} - a^{3})/[3(b - a)] = (a^{2} + ab + b^{2})/3$ .
$\mu^{2} = (a + b)^{2}/4$ .
$\text{Var}(X) = (a^{2} + ab + b^{2})/3 - (a + b)^{2}/4 = (b - a)^{2}/12$ after algebra.

Standard deviation.

$\sigma = \dfrac{b - a}{\sqrt{12}} = \dfrac{b - a}{2\sqrt 3}.$

Worked example. $X \sim U(2, 10)$ .

$E(X) = 6$ .
$\text{Var}(X) = 64/12 = 16/3 \approx 5.333$ .
$\sigma = 4/\sqrt 3 \approx 2.309$ .

Linear transformations. If $Y = aX + b$ then $E(Y) = aE(X) + b$ and $\text{Var}(Y) = a^{2}\text{Var}(X)$ . Useful when the uniform is rescaled.

Mean: $(a + b)/2$ — booklet.
Variance: $(b - a)^{2}/12$ — booklet.
SD: $(b - a)/\sqrt{12}$ .
Both formulae are quoted, not derived, in standard exam working.

Inverse problems — find $a$ , $b$ or quantiles

Two probability statements give two linear equations in $a, b$ — solve.

Setup. Given partial information about $X \sim U(a, b)$ , find the unknown parameters or quantiles.

Type 1: Given mean and variance, find $a, b$ .

$\mu = \dfrac{a + b}{2}, \quad \sigma^{2} = \dfrac{(b - a)^{2}}{12}.$

Two equations: solve for $a + b$ and $b - a$ , then for $a$ and $b$ .

Example. $\mu = 5$ , $\sigma^{2} = 3$ . Then $a + b = 10$ and $(b - a)^{2} = 36 \Rightarrow b - a = 6$ . Solve: $a = 2$ , $b = 8$ .

Type 2: Given $a$ and a probability, find $b$ .

E.g. $X \sim U(0, b)$ with $P(X < 4) = 0.2$ . Then $4/b = 0.2 \Rightarrow b = 20$ .

Type 3: Find a quantile.

$F(x_p) = p \Rightarrow \dfrac{x_p - a}{b - a} = p \Rightarrow x_p = a + p(b - a).$

E.g. $X \sim U(0, 1)$ , $90$ th percentile: $x_{0.9} = 0.9$ .

Two unknowns ( $a, b$ ) need two equations.
Solve for $a + b$ and $b - a$ , then for each.
Quantile: $x_p = a + p(b - a)$ — linear interpolation.

Modelling judgements

Uniform = 'random arrival in a fixed window'. Check constancy of density.

When examiners ask about modelling fitness:

Is the support FINITE? Uniform requires bounded support. Heights, weights cannot be uniform — they have no hard upper bound (or a very arbitrary one).
Is the density CONSTANT? Equal probability per unit length. Real data with a peak (mode) is not uniform.
Is there a SYMMETRY? Uniform is symmetric. Skewed data is not uniform.

Classic 'YES' examples.

Time until next bus, IF buses run on a strict $T$ -minute cycle AND the passenger arrives at a random time.
Position of a randomly placed point on a finite line segment.
Angle of a needle thrown randomly onto a circle.

Classic 'NO' examples.

Waiting time for a Poisson arrival process — these are EXPONENTIAL, not uniform.
Heights of adults — heights cluster around a mean (normal).
Exam scores — clustering, often skewed.

Caveat for buses. If buses run on a strict schedule (every $15$ minutes), waiting times are $U(0, 15)$ . If buses run irregularly (Poisson arrivals), waiting times are exponential. The question must specify a fixed schedule.

Uniform needs finite support and constant density.
Random arrivals in a fixed cycle: uniform.
Poisson arrivals: exponential, NOT uniform.
Heights / weights: normal, not uniform.

How it’s examined

Uniform appears on roughly half of WST02/01 papers — usually a $5$ - $8$ mark question covering: state PDF / CDF, compute a probability, find mean / variance, and a small modelling comment. Examiners frequently test the SQUARE in the variance formula — writing $(b - a)/12$ instead of $(b - a)^{2}/12$ is a flagged error. Modelling questions ('comment on whether $U$ is appropriate') reward specific scenario references over generic statements.

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.

Master this Topic

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

Take this whole topic with you

Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Continuous Uniform Distribution, ready to print or save as PDF.

Step-by-step worked examples — Continuous Uniform Distribution

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

1Probabilities from a continuous uniform (4 marks, S2)
Core• Adapted from WST02/01 January 2024 Q4• uniform, probability, length ratio
Question
$X \sim U(2, 10)$ . Find (a) $P(X < 5)$ , (b) $P(3 < X < 8)$ , (c) $P(X > 9)$ .
Step-by-step solution
1. Step 1
  PDF $f(x) = 1/(10 - 2) = 1/8$ on $[2, 10]$ . Probabilities are areas of rectangles, equivalently length ratios.
2. Step 2
  (a) Range $[2, 5]$ has length $3$ :
  $P(X < 5) = \dfrac{5 - 2}{10 - 2} = \dfrac{3}{8}$
3. Step 3
  (b) Range $[3, 8]$ has length $5$ :
  $P(3 < X < 8) = \dfrac{8 - 3}{10 - 2} = \dfrac{5}{8}$
4. Step 4
  (c) Range $[9, 10]$ has length $1$ :
  $P(X > 9) = \dfrac{10 - 9}{10 - 2} = \dfrac{1}{8}$
Answer
(a) $3/8$ . (b) $5/8$ . (c) $1/8$ .
Examiner tip
B1 for $f = 1/8$ (or equivalent recognition). B1 each for the three probabilities. No integration needed for a uniform — just length ratios.
2Mean and variance of a uniform (4 marks, S2)
Core• Adapted from WST02/01 June 2024 Q5• uniform, mean, variance
Question
$X \sim U(2, 10)$ . State $E(X)$ and find $\text{Var}(X)$ and the standard deviation.
Step-by-step solution
1. Step 1
  Mean of uniform is the midpoint of the interval:
  $E(X) = \dfrac{a + b}{2} = \dfrac{2 + 10}{2} = 6$
2. Step 2
  Variance:
  $\text{Var}(X) = \dfrac{(b - a)^{2}}{12} = \dfrac{8^{2}}{12} = \dfrac{64}{12} = \dfrac{16}{3} \approx 5.333$
3. Step 3
  Standard deviation:
  $\sigma = \sqrt{16/3} = \dfrac{4}{\sqrt 3} \approx 2.309$
Answer
$E(X) = 6$ , $\text{Var}(X) = 16/3 \approx 5.33$ , $\sigma \approx 2.31$ .
Examiner tip
B1 for $E(X) = 6$ . M1 for variance formula. A1 for $16/3$ or $5.33$ . B1 for SD if asked. Both formulae are in the IAL formula booklet — no derivation required.
3CDF, median, and inverse problem (6 marks, S2)
Extended• Adapted from WST02/01 January 2024 Q5• uniform, cdf, median, inverse
Question
$X \sim U(-3, 5)$ . (a) Write down the PDF and CDF of $X$ . (b) Find the median. (c) Find $c$ such that $P(X > c) = 0.2$ .
Step-by-step solution
1. Step 1
  (a) PDF: $f(x) = 1/(5 - (-3)) = 1/8$ for $-3 \le x \le 5$ , zero otherwise.
2. Step 2
  CDF (piecewise):
  $F(x) = \begin{cases} 0, & x < -3 \\ \dfrac{x + 3}{8}, & -3 \le x \le 5 \\ 1, & x > 5 \end{cases}$
3. Step 3
  (b) Median: $F(m) = 0.5 \Rightarrow (m + 3)/8 = 0.5 \Rightarrow m + 3 = 4 \Rightarrow m = 1$ .
4. Step 4
  Equivalently, median $= (a + b)/2 = (-3 + 5)/2 = 1$ . Confirmed.
5. Step 5
  (c) $P(X > c) = 0.2 \Rightarrow 1 - F(c) = 0.2 \Rightarrow F(c) = 0.8$ .
  $\dfrac{c + 3}{8} = 0.8 \Rightarrow c + 3 = 6.4 \Rightarrow c = 3.4$
Answer
(a) $f = 1/8$ on $[-3, 5]$ ; CDF piecewise. (b) median $= 1$ . (c) $c = 3.4$ .
Examiner tip
(a) B1 for PDF. M1 for CDF piecewise (all three regions). (b) B1 for median. (c) M1 for $1 - F(c) = 0.2$ . A1 for $c = 3.4$ . Examiners deduct for omitting the outer pieces of the CDF.
4Modelling waiting time as uniform (5 marks, S2)
Extended• Adapted from WST02/01 June 2024 Q6• uniform, modelling, context
Question
Buses run every $15$ minutes. A passenger arrives at a random time and waits $W$ minutes for the next bus. (a) State a suitable model for $W$ . (b) Find $E(W)$ and $\sigma_W$ . (c) Find the probability that the passenger waits more than $10$ minutes.
Step-by-step solution
1. Step 1
  (a) Random arrival within a $15$ -minute cycle: $W \sim U(0, 15)$ .
2. Step 2
  (b) Mean: $E(W) = (0 + 15)/2 = 7.5$ minutes.
3. Step 3
  Variance: $\text{Var}(W) = 15^{2}/12 = 18.75$ .
4. Step 4
  SD: $\sigma_W = \sqrt{18.75} \approx 4.330$ .
5. Step 5
  (c) $P(W > 10) = (15 - 10)/15 = 1/3 \approx 0.333$ .
Answer
(a) $W \sim U(0, 15)$ . (b) $E(W) = 7.5$ , $\sigma_W \approx 4.33$ . (c) $P(W > 10) = 1/3 \approx 0.333$ .
Examiner tip
(a) B1. (b) B1 for mean, M1 for variance, A1 for $\sigma$ . (c) M1 for length ratio, A1 for $1/3$ . The uniform model assumes passengers arrive at a TRULY random time — independent of bus schedule. In rush hour this can fail.
5Derive mean and variance from definition (6 marks, S2)
Challenge• Adapted from WST02/01 January 2023 Q7• uniform, derivation, integration
Question
$X \sim U(a, b)$ . By integration, derive (i) $E(X) = (a + b)/2$ , (ii) $\text{Var}(X) = (b - a)^{2}/12$ .
Step-by-step solution
1. Step 1
  PDF $f(x) = 1/(b - a)$ on $[a, b]$ .
2. Step 2
  (i) Mean:
  $E(X) = \int_a^b \dfrac{x}{b - a}\,dx = \dfrac{1}{b - a} \cdot \dfrac{x^{2}}{2}\Big|_a^b$
3. Step 3
  Simplify:
  $E(X) = \dfrac{b^{2} - a^{2}}{2(b - a)} = \dfrac{(b - a)(b + a)}{2(b - a)} = \dfrac{a + b}{2}$
4. Step 4
  (ii) Second moment:
  $E(X^{2}) = \int_a^b \dfrac{x^{2}}{b - a}\,dx = \dfrac{1}{b - a} \cdot \dfrac{x^{3}}{3}\Big|_a^b = \dfrac{b^{3} - a^{3}}{3(b - a)}$
5. Step 5
  Factor $b^{3} - a^{3} = (b - a)(b^{2} + ab + a^{2})$ :
  $E(X^{2}) = \dfrac{b^{2} + ab + a^{2}}{3}$
6. Step 6
  Variance:
  $\text{Var}(X) = E(X^{2}) - \mu^{2} = \dfrac{b^{2} + ab + a^{2}}{3} - \dfrac{(a + b)^{2}}{4}$
7. Step 7
  Common denominator $12$ :
  $\text{Var}(X) = \dfrac{4(b^{2} + ab + a^{2}) - 3(a + b)^{2}}{12}$
8. Step 8
  Expand $3(a + b)^{2} = 3a^{2} + 6ab + 3b^{2}$ :
  $4b^{2} + 4ab + 4a^{2} - 3a^{2} - 6ab - 3b^{2} = a^{2} - 2ab + b^{2} = (b - a)^{2}$
9. Step 9
  Therefore $\text{Var}(X) = (b - a)^{2}/12$ .
Answer
(i) $E(X) = (a + b)/2$ . (ii) $\text{Var}(X) = (b - a)^{2}/12$ .
Examiner tip
(i) M1 for integral set-up. A1 for $(a + b)/2$ . (ii) M1 for $E(X^{2})$ set-up. A1 for $(b^{2} + ab + a^{2})/3$ . M1 for $E(X^{2}) - \mu^{2}$ . M1 for algebraic simplification. A1 for $(b - a)^{2}/12$ . Derivation is occasionally tested — the algebra is the only tricky bit.

Key Formulae — Continuous Uniform Distribution

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

Uniform PDF
$f(x) = \begin{cases} \dfrac{1}{b - a}, & a \le x \le b \\ 0, & \text{otherwise} \end{cases}$
$a, b$
endpoints of the support, with $a < b$
When to use
Whenever the question says 'uniform on $[a, b]$ ' or 'every value equally likely between $a$ and $b$ '. NOT in the booklet — derive from $\int f = 1$ .
Example
$U(2, 10)$ : $f(x) = 1/8$ on $[2, 10]$ .
Uniform CDF
$F(x) = \begin{cases} 0, & x < a \\ \dfrac{x - a}{b - a}, & a \le x \le b \\ 1, & x > b \end{cases}$
$F(x)$
CDF
When to use
For half-infinite probabilities and inverse problems (median, percentiles).
Example
$U(2, 10)$ : $F(5) = 3/8$ .
Mean
$E(X) = \dfrac{a + b}{2}$
$a, b$
endpoints
When to use
By symmetry, mean = midpoint. IS in the IAL formula booklet.
Example
$U(2, 10)$ : $E(X) = 6$ .
Variance
$\text{Var}(X) = \dfrac{(b - a)^{2}}{12}$
$a, b$
endpoints
When to use
Quoted directly — IS in the IAL formula booklet. SD $= (b - a)/\sqrt{12}$ .
Example
$U(2, 10)$ : $\text{Var}(X) = 64/12 = 16/3$ .
Probability as length ratio
$P(c < X < d) = \dfrac{d - c}{b - a}, \quad a \le c \le d \le b$
$c, d$
sub-interval bounds within the support
When to use
Fastest way to compute uniform probabilities. NOT in the booklet but immediate from PDF.
Example
$U(2, 10)$ : $P(3 < X < 8) = 5/8$ .

Key Definitions and Keywords — Continuous Uniform Distribution

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

Continuous uniform distribution
Examiner keyword
Distribution with constant density on a finite interval $[a, b]$ . Notation $X \sim U(a, b)$ .
Support
The set of $x$ -values where $f(x) > 0$ . For $U(a, b)$ , the support is the interval $[a, b]$ .
Rectangular distribution
Alternative name for the continuous uniform — the PDF graph is a rectangle of width $b - a$ and height $1/(b - a)$ .
Mean of $U(a, b)$
Examiner keyword
$E(X) = (a + b)/2$ — the midpoint of the support.
Variance of $U(a, b)$
Examiner keyword
$\text{Var}(X) = (b - a)^{2}/12$ . SD $= (b - a)/\sqrt{12}$ .

Common Mistakes and Misconceptions — Continuous Uniform Distribution

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

✕Stating $f(x) = 1$ instead of $f(x) = 1/(b - a)$
Why it happens
Confusion with uniform on $[0, 1]$ (where $f = 1$ ).
How to avoid it
Always compute $1/(b - a)$ explicitly. $U(2, 10)$ : $f = 1/8$ , not $1$ .
✕Writing $\text{Var}(X) = (b - a)/12$ instead of $(b - a)^{2}/12$
WST02/01 examiner reports
Why it happens
Missing the squared.
How to avoid it
Memorise: $(b - a)^{2}/12$ — SQUARED. The variance has dimensions of $X^{2}$ .
✕Writing only the middle piece of the CDF
Why it happens
Same issue as continuous random variables generally.
How to avoid it
Always state CDF as piecewise on all three regions: $0$ below $a$ , integrand on $[a, b]$ , $1$ above $b$ .
✕Computing a probability where the interval is partly outside the support
Why it happens
Students apply the formula without checking.
How to avoid it
Clip the interval to the support. $U(2, 10)$ , $P(X > 12) = 0$ — not negative. $P(0 < X < 5) = (5 - 2)/(10 - 2)$ — clip $0$ to $2$ .
✕Misidentifying a non-uniform variable as uniform
Why it happens
Many real-world variables are bounded but not uniform.
How to avoid it
Uniform requires CONSTANT density. Waiting times for an event arriving randomly within a fixed cycle are uniform. Heights / weights / exam marks are NOT uniform.

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 Uniform Distribution — frequently asked questions

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

Continuous Uniform Distribution

Short Study Notes in the form of Slides

Detailed Notes

What you’ll learn

How it’s examined

1Probabilities from a continuous uniform (4 marks, S2)

2Mean and variance of a uniform (4 marks, S2)

3CDF, median, and inverse problem (6 marks, S2)

4Modelling waiting time as uniform (5 marks, S2)

5Derive mean and variance from definition (6 marks, S2)

Uniform PDF

Uniform CDF

Mean

Variance

Probability as length ratio

Continuous uniform distribution

Support

Rectangular distribution

Mean of U(a,b)U(a, b)U(a,b)

Variance of U(a,b)U(a, b)U(a,b)

✕Stating f(x)=1f(x) = 1f(x)=1 instead of f(x)=1/(b−a)f(x) = 1/(b - a)f(x)=1/(b−a)

✕Writing Var(X)=(b−a)/12\text{Var}(X) = (b - a)/12Var(X)=(b−a)/12 instead of (b−a)2/12(b - a)^{2}/12(b−a)2/12

✕Writing only the middle piece of the CDF

✕Computing a probability where the interval is partly outside the support

✕Misidentifying a non-uniform variable as uniform

Practice questions

Past paper style quiz

Continuous Uniform Distribution — frequently asked questions

Mean of $U(a, b)$

Variance of $U(a, b)$

✕Stating $f(x) = 1$ instead of $f(x) = 1/(b - a)$

✕Writing $\text{Var}(X) = (b - a)/12$ instead of $(b - a)^{2}/12$