Cumulative Distribution Functions

The cumulative distribution function (CDF), written $F$F, is the single most versatile object attached to a random variable. Where the density $f$f describes how probability is spread, the CDF describes how it accumulates: $F(x)$F(x) is the total probability sitting at or below $x$x. From it you read off any probability without further integration, locate the median and quartiles by solving a single equation, recover the density by differentiation, and — crucially for 7367/3S — generate the distribution of a transformed variable $Y = g(X)$Y=g(X). This lesson builds the CDF from the ground up and turns it into a complete exam toolkit.

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1. Where this sits in AQA 7367

This is Paper 3 Statistics option (7367/3S) content, sitting directly on the continuous-distribution thread (Lessons 4–5). The per-paper AO weighting for the option papers is AO1 40% / AO2 25% / AO3 35%, so the problem-solving (AO3) demand is higher here than in the compulsory pure papers. Routine work — integrating a density to a CDF, or differentiating a CDF back to a density — is AO1. Setting up and solving $F(m) = 0.5$F(m)=0.5 for a median, or building the CDF of $Y = X^2$Y=X2 from first principles, is AO3. Justifying why $F$F must be non-decreasing and continuous, or interpreting a percentile in context, is AO2. The prerequisites are A-Level Mathematics integration/differentiation and the PDF definition and validity condition from Lesson 4.

The CDF sits at the hub of the continuous-distribution lessons: Lesson 5 found summary measures by integrating the density, but many of those same quantities — medians, quartiles, percentiles, "more than"/"less than" probabilities — are found more directly through the CDF. For that reason this lesson is among the most heavily examined in 3S: a single question can ask you to derive $F$F, read several probabilities from it, locate the median and quartiles, and then use the CDF to build the distribution of a transformed variable. Mastering the CDF therefore pays back across the whole option, and the methods here recur in Lesson 7 (the exponential CDF) and underpin the simulation idea touched on in the stretch section.

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2. Core theory

Definition

The cumulative distribution function of a random variable $X$X is

$F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t)\,dt.$F(x)=P(X≤x)=∫−∞x​f(t)dt.

The dummy variable $t$t inside the integral is deliberate: $x$x is the upper limit, so it must not also be the variable of integration. Geometrically, $F(x)$F(x) is the area under the density to the left of $x$x.

The four defining properties

Any function that is to serve as a CDF of a continuous variable must satisfy:

Property	Statement	Why
Limits	$F(-\infty) = 0,\ F(+\infty) = 1$F(−∞)=0, F(+∞)=1	no probability below the support; all probability accounted for
Monotonic	$a < b \implies F(a) \leq F(b)$a<b⟹F(a)≤F(b)	$F'(x) = f(x) \geq 0$F′(x)=f(x)≥0, so $F$F never decreases
Continuity	$F$F is continuous everywhere	a continuous variable has no point masses, so no jumps
Range	$0 \leq F(x) \leq 1$0≤F(x)≤1	it is a probability

The monotonicity follows directly from the Fundamental Theorem of Calculus: since $F'(x) = f(x)$F′(x)=f(x) and a valid density is non-negative, $F$F can only stay level or rise. A CDF that decreased anywhere would correspond to a negative density — impossible. These four properties are not arbitrary book-keeping; together they characterise exactly the functions that can serve as a CDF. If you are handed a candidate function and asked "could this be a cumulative distribution function?", you check the four: does it start at 0, end at 1, never decrease, and stay continuous? A function failing any one of them cannot be a CDF, and spotting the failure is a genuine exam skill — for instance, a function that reached $1.2$1.2 somewhere, or that dipped then rose, is immediately disqualified.

It is also worth contrasting the shape of the CDF with that of the density. The density $f$f can do almost anything non-negative — rise, fall, have several humps, even exceed 1 (it is a density, not a probability). The CDF $F$F, by contrast, is always a smooth, weakly-increasing S-shaped (or ramp-shaped) curve climbing from the floor $0$0 to the ceiling $1$1. Where the density is large, the CDF is steep; where the density is zero, the CDF is flat. This visual dictionary — "tall density ⇒ steep CDF" — is a fast sanity check on any answer.

The PDF–CDF link (the central identity)

The Fundamental Theorem of Calculus ties the two functions together:

$F(x) = \int_{-\infty}^{x} f(t)\,dt \qquad \Longleftrightarrow \qquad f(x) = \frac{dF}{dx} = F'(x).$F(x)=∫−∞x​f(t)dt⟺f(x)=dxdF​=F′(x).

In words: integrate the density to get the CDF; differentiate the CDF to get the density. This two-way street is the engine of almost every CDF question.

Reading probabilities straight off $F$F

$P(X \leq a) = F(a), \qquad P(X > a) = 1 - F(a), \qquad P(a \leq X \leq b) = F(b) - F(a).$P(X≤a)=F(a),P(X>a)=1−F(a),P(a≤X≤b)=F(b)−F(a).

Because $X$X is continuous, $P(X = a) = 0$P(X=a)=0, so the inequalities may be strict or weak interchangeably: $P(X < a) = P(X \leq a)$P(X<a)=P(X≤a). This is the single biggest convenience of the continuous CDF over the discrete case.

Exam Tip: If a question gives you the CDF and asks for a probability, do not re-integrate the density — just subtract two values of $F$F. Re-integrating wastes time and invites slips.

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3. Worked examples with M1/A1 mark scheme

Example 1 — from PDF to CDF (and back)

Let $f(x) = \tfrac{3}{8}x^2$f(x)=83​x2 for $0 \leq x \leq 2$0≤x≤2, and zero otherwise. Find $F(x)$F(x) fully, then recover $f$f by differentiating.

For $x < 0$x<0, nothing has accumulated, so $F(x) = 0$F(x)=0. For $0 \leq x \leq 2$0≤x≤2,

$F(x) = \int_{0}^{x} \tfrac{3}{8}t^2\,dt = \tfrac{3}{8}\!\left[\tfrac{t^3}{3}\right]_0^x = \frac{x^3}{8}. \quad (\textbf{M1}\ \text{integrate};\ \textbf{A1}\ x^3/8)$F(x)=∫0x​83​t2dt=83​[3t3​]0x​=8x3​.(M1 integrate; A1 x3/8)

For $x > 2$x>2 all the probability lies below, so $F(x) = 1$F(x)=1. Hence

$F(x) = \begin{cases} 0 & x < 0 \\[2pt] \dfrac{x^3}{8} & 0 \leq x \leq 2 \\[4pt] 1 & x > 2 \end{cases} \quad (\textbf{A1}\ \text{fully-defined piecewise CDF, all three branches})$F(x)=⎩⎨⎧​08x3​1​x<00≤x≤2x>2​(A1 fully-defined piecewise CDF, all three branches)

Verification (always do this): $F(0) = 0$F(0)=0 and $F(2) = \tfrac{8}{8} = 1$F(2)=88​=1; the join values match the limits, so $F$F is continuous and rises from 0 to 1. Differentiating the middle branch recovers $F'(x) = \tfrac{3x^2}{8} = f(x)$F′(x)=83x2​=f(x) (B1), confirming consistency.

Example 2 — from CDF to PDF with M/A annotation

Given

$F(x) = \begin{cases} 0 & x < 1 \\[2pt] \dfrac{(x-1)^2}{4} & 1 \leq x \leq 3 \\[4pt] 1 & x > 3, \end{cases}$F(x)=⎩⎨⎧​04(x−1)2​1​x<11≤x≤3x>3,​

find the density.

$f(x) = F'(x) = \frac{2(x-1)}{4} = \frac{x-1}{2}, \quad 1 \leq x \leq 3, \quad (\textbf{M1}\ \text{differentiate};\ \textbf{A1})$f(x)=F′(x)=42(x−1)​=2x−1​,1≤x≤3,(M1 differentiate; A1)

and $f(x) = 0$f(x)=0 elsewhere. Check it is a valid density:

$\int_{1}^{3} \frac{x-1}{2}\,dx = \frac{1}{2}\!\left[\frac{(x-1)^2}{2}\right]_1^3 = \frac{1}{2}\cdot\frac{4}{2} = 1. \quad (\textbf{B1})$∫13​2x−1​dx=21​[2(x−1)2​]13​=21​⋅24​=1.(B1)

Also $f \geq 0$f≥0 on $[1,3]$[1,3], so this is a legitimate PDF.

Example 3 — using $F$F for probabilities, median and IQR

With $F(x) = \tfrac{x^3}{8}$F(x)=8x3​ on $[0,2]$[0,2] from Example 1:

$P(1 \leq X \leq 1.5) = F(1.5) - F(1) = \frac{1.5^3}{8} - \frac{1^3}{8} = \frac{3.375 - 1}{8} = \frac{2.375}{8} = 0.296875. \quad (\textbf{M1 A1})$P(1≤X≤1.5)=F(1.5)−F(1)=81.53​−813​=83.375−1​=82.375​=0.296875.(M1 A1)

Median $m$m: solve $F(m) = 0.5$F(m)=0.5:

$\frac{m^3}{8} = 0.5 \implies m^3 = 4 \implies m = \sqrt[3]{4} \approx 1.587. \quad (\textbf{M1}\ \text{set }F=0.5;\ \textbf{A1})$8m3​=0.5⟹m3=4⟹m=34​≈1.587.(M1 set F=0.5; A1)

Quartiles: $\tfrac{Q_1^3}{8} = 0.25 \Rightarrow Q_1 = \sqrt[3]{2} \approx 1.260$8Q13​​=0.25⇒Q1​=32​≈1.260; $\tfrac{Q_3^3}{8} = 0.75 \Rightarrow Q_3 = \sqrt[3]{6} \approx 1.817$8Q33​​=0.75⇒Q3​=36​≈1.817, so

$\text{IQR} = Q_3 - Q_1 = 1.817 - 1.260 = 0.557. \quad (\textbf{A1})$IQR=Q3​−Q1​=1.817−1.260=0.557.(A1)

Example 4 — an exponential CDF on an unbounded support

Densities supported on $[0,\infty)$[0,∞) integrate to a CDF that approaches — never quite reaches — 1. Let $f(x) = 3e^{-3x}$f(x)=3e−3x for $x \ge 0$x≥0. Then for $x \ge 0$x≥0,

$F(x) = \int_0^x 3e^{-3t}\,dt = \Big[-e^{-3t}\Big]_0^x = 1 - e^{-3x}, \quad (\textbf{M1}\ \text{integrate};\ \textbf{A1})$F(x)=∫0x​3e−3tdt=[−e−3t]0x​=1−e−3x,(M1 integrate; A1)

and $F(x) = 0$F(x)=0 for $x < 0$x<0. As $x \to \infty$x→∞, $e^{-3x} \to 0$e−3x→0 so $F(x) \to 1$F(x)→1 (the ceiling is a limit, reached only at infinity). The median solves $1 - e^{-3m} = 0.5$1−e−3m=0.5, i.e. $e^{-3m} = 0.5$e−3m=0.5, giving $m = \tfrac{\ln 2}{3} \approx 0.231$m=3ln2​≈0.231 (M1 A1). Note the median is less than the mean $E(X) = \tfrac13 \approx 0.333$E(X)=31​≈0.333, the signature of the right-skewed exponential — a point we return to in Lesson 7.

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4. Specimen-style exam question

(Specimen-style — not from any real paper.)

A continuous random variable $X$X has probability density function $f(x) = \tfrac{3}{4}x(2-x)$f(x)=43​x(2−x) for $0 \le x \le 2$0≤x≤2, and $f(x) = 0$f(x)=0 otherwise. (a) Find the cumulative distribution function $F(x)$F(x). (b) Hence find $P(0.5 \le X \le 1.5)$P(0.5≤X≤1.5). (c) Find the median of $X$X.

(a) For $0 \le x \le 2$0≤x≤2,

$F(x) = \int_0^x \tfrac{3}{4}t(2-t)\,dt = \tfrac{3}{4}\!\left[t^2 - \tfrac{t^3}{3}\right]_0^x = \tfrac{3}{4}\!\left(x^2 - \tfrac{x^3}{3}\right) = \frac{3x^2}{4} - \frac{x^3}{4}.$F(x)=∫0x​43​t(2−t)dt=43​[t2−3t3​]0x​=43​(x2−3x3​)=43x2​−4x3​.

Check $F(2) = \tfrac{3\cdot 4}{4} - \tfrac{8}{4} = 3 - 2 = 1$F(2)=43⋅4​−48​=3−2=1 ✓. So

$F(x) = \begin{cases} 0 & x<0 \\[2pt] \dfrac{3x^2 - x^3}{4} & 0 \le x \le 2 \\[4pt] 1 & x>2. \end{cases}$F(x)=⎩⎨⎧​043x2−x3​1​x<00≤x≤2x>2.​

(b) $F(1.5) = \tfrac{3(2.25) - 3.375}{4} = \tfrac{6.75 - 3.375}{4} = \tfrac{3.375}{4} = 0.84375$F(1.5)=43(2.25)−3.375​=46.75−3.375​=43.375​=0.84375; $F(0.5) = \tfrac{3(0.25) - 0.125}{4} = \tfrac{0.75 - 0.125}{4} = \tfrac{0.625}{4} = 0.15625$F(0.5)=43(0.25)−0.125​=40.75−0.125​=40.625​=0.15625. Hence

$P(0.5 \le X \le 1.5) = 0.84375 - 0.15625 = 0.6875.$P(0.5≤X≤1.5)=0.84375−0.15625=0.6875.

(c) By the symmetry of $f$f about $x = 1$x=1 the median is $m = 1$m=1; confirm $F(1) = \tfrac{3 - 1}{4} = \tfrac{2}{4} = 0.5$F(1)=43−1​=42​=0.5 ✓.

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5. Synoptic links

• Discrete CDF (Lessons 1–2): for a discrete variable the CDF is a step function, $F(x) = \sum_{k \le x} P(X=k)$F(x)=∑k≤x​P(X=k), jumping at each value; the continuous CDF is its smooth limit. The defining property $F(m) \geq 0.5$F(m)≥0.5 for the discrete median becomes the equation $F(m) = 0.5$F(m)=0.5 here.
• PDF and $E(X)$E(X) (Lessons 4–5): $f = F'$f=F′ closes the loop with the validity/normalisation and moment integrals.
• A-Level Maths: the Fundamental Theorem of Calculus is the PDF–CDF link; solving $F(m) = 0.5$F(m)=0.5 is just root-finding (sometimes needing numerical methods from Pure).
• Transformations (below): building $F_Y$FY​ for $Y = g(X)$Y=g(X) is the gateway to the exponential and other derived distributions (Lesson 7) and underpins the inverse-CDF method of simulation.

The recurring theme across these links is the duality of $f$f and $F$F: differentiation and integration carry you back and forth, and almost every continuous-distribution result can be phrased in terms of either. When a question feels hard via the density, try the CDF, and vice versa — the two viewpoints are equivalent but one is usually much cleaner for a given task. This flexibility is itself a synoptic skill: the ablest candidates choose the representation that minimises the work, rather than mechanically integrating from scratch every time.

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6. Mark-scheme literacy