Cumulative distribution functions: Worked examples

This page is a practice arena. We take the ideas from Cumulative distribution functions and throw every kind of case at them, one worked example per case-type. Nothing here is new theory — but by the end you will have seen every scenario a CDF can throw at you, so nothing on an exam surprises you.

Before we start, one reminder in plain words. The letters we use:

Everything below is just those three characters in different costumes.

The scenario matrix

Every CDF problem falls into one of these case classes. The whole point of this page is that we hit all of them.

Cell	Case class	What makes it tricky	Covered by
A	Discrete, value between jumps	staircase: only landings below the target count	Ex. 1
B	Discrete, exact point mass	$P (X = x) = F_{X} (x) - F_{X} (x^{-})$ , the jump height	Ex. 2
C	Continuous ramp, interval	subtract two CDF values	Ex. 3
D	Degenerate / boundary inputs	$x <$ support, $x >$ support, $x =$ edge	Ex. 4
E	Continuous, inverting the CDF (quantile)	solve $F_{X} (x) = p$ for $x$	Ex. 5
F	Limiting / tail behaviour	$x \to \infty$ , complement $P (X > x)$	Ex. 6
G	Real-world word problem	translate English → CDF	Ex. 7
H	Exam twist: mixed / sign trap	negative "probability" trap, mixed variable	Ex. 8

Read the Forecast line first and guess before revealing each solution — that is where learning actually happens.

Ex. 1 — Cell A: discrete value between the steps

Worked example Rolling a fair die, target not a whole number

Let $X$ be a fair six-sided die, so $p_{X} (k) = \frac{1}{6}$ for $k \in {1, \dots, 6}$ . Find $F_{X} (4.2)$ .

Forecast: guess a fraction with denominator $6$ . Bigger or smaller than $\frac{1}{2}$ ?

Step 1 — List which outcomes are $\leq 4.2$ . The die can only land on whole numbers. The ones $\leq 4.2$ are $1, 2, 3, 4$ . Why this step? $F_{X} (x) = P (X \leq x)$ sums the mass of every outcome at or below $x$ . A die never shows $4.2$ , so we round down to the last real landing.

Step 2 — Add their probabilities. $F_{X} (4.2) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \approx 0.6667$ Why this step? Disjoint outcomes' probabilities add.

Verify: $\frac{2}{3} > \frac{1}{2}$ ✓ (matches forecast — we've passed the midpoint value $3.5$ ). The staircase in the figure below sits at height $\frac{4}{6}$ across the whole flat run $4 \leq x < 5$ .

Ex. 2 — Cell B: the exact point mass as a jump height

Ex. 3 — Cell C: continuous ramp, interval probability

Ex. 4 — Cell D: degenerate / boundary inputs

Worked example Feeding the CDF values outside and on the edge of the support

With the same Uniform $(0, 1)$ , evaluate $F_{X} (- 2)$ , $F_{X} (0)$ , $F_{X} (1)$ , and $F_{X} (3.5)$ .

Forecast: two of these are $0$ -or-edge, two are saturated. Which?

Step 1 — Recall the piecewise CDF. $F_{X} (x) = ⎩ ⎨ ⎧ 0 x 1 x < 0 0 \leq x \leq 1 x > 1$ Why this step? Outside the support no probability has accumulated (left) or all of it has (right). The formula bakes in these limits.

Step 2 — Read each off the right branch.

$F_{X} (- 2) = 0$ — nothing below the support yet (this is the $lim_{x \to - \infty} = 0$ rule in action).
$F_{X} (0) = 0$ — the very left edge; still nothing accumulated.
$F_{X} (1) = 1$ — the right edge; all probability collected.
$F_{X} (3.5) = 1$ — past the support; the running total can't exceed $1$ (the $lim_{x \to \infty} = 1$ rule). Why this step? These four values check the monotone, bounded, limit properties all at once.

Verify: all outputs lie in $[0, 1]$ ✓, and they never decrease as $x$ grows: $0 \leq 0 \leq 1 \leq 1$ ✓.

Ex. 5 — Cell E: inverting the CDF (quantile)

Worked example 90th percentile of an exponential

$X \sim Exp (λ)$ with $F_{X} (x) = 1 - e^{- λ x}$ for $x \geq 0$ , and $λ = 2$ . Find the value below which $90%$ of outcomes fall — i.e. $F_{X}^{- 1} (0.9)$ .

Forecast: should be a small positive number (rate $2$ means events happen fast). Guess whether it's above or below $1$ .

Step 1 — Set the CDF equal to the target fraction. $F_{X} (x) = 0.9 \Rightarrow 1 - e^{- 2 x} = 0.9$ Why this step? The quantile function answers "which $x$ has fraction $p$ below it?" — that is literally solving $F_{X} (x) = p$ . This is the engine behind Inverse Transform Sampling.

Step 2 — Isolate the exponential. $e^{- 2 x} = 0.1$ Why this step? We want the unknown out of the exponent; first get $e^{- 2 x}$ alone.

Step 3 — Take the natural log (the tool that undoes $e^{(\cdot)}$ ). $- 2 x = ln (0.1) \Rightarrow x = \frac{- l n ( 0.1 )}{2} = \frac{l n ( 10 )}{2} \approx \frac{2.3026}{2} \approx 1.1513$ Why $ln$ and not another tool? $ln$ is the exact inverse of $e^{x}$ , so it peels the variable out of the exponent in one clean move.

Verify: plug back: $1 - e^{- 2 (1.1513)} = 1 - e^{- 2.3026} = 1 - 0.1 = 0.9$ ✓. It's just above $1$ , matching a "fast" distribution's tail.

Ex. 6 — Cell F: limiting / tail behaviour (complement)

Worked example Survival past a threshold

Same $X \sim Exp (0.5)$ . Find $P (X > 4)$ , and confirm the tail $\to 0$ as the threshold grows.

Forecast: the complement — should be small. Above or below $0.2$ ?

Step 1 — Use the complement rule. $P (X > x) = 1 - F_{X} (x)$ Why this step? "Greater than $x$ " is everything not "at or below $x$ ". Total probability is $1$ , so subtract. This tail is exactly the survival function of Survival Analysis.

Step 2 — Substitute $x = 4, λ = 0.5$ . $P (X > 4) = 1 - (1 - e^{- 0.5 \cdot 4}) = e^{- 2} \approx 0.1353$ Why this step? The $1$ 's cancel, leaving the clean tail $e^{- λ x}$ .

Step 3 — Check the limit. $lim_{x \to \infty} P (X > x) = lim_{x \to \infty} e^{- 0.5 x} = 0$ Why this step? As the threshold marches right, eventually no probability remains beyond it — the CDF property $lim_{x \to \infty} F_{X} (x) = 1$ restated for the tail.

Verify: $0.1353 < 0.2$ ✓, and $e^{- 0.5 x}$ shrinks monotonically to $0$ ✓.

Ex. 7 — Cell G: real-world word problem

Worked example Bus waiting time

Buses arrive so that your waiting time $X$ (minutes) is $Exp (λ)$ with average wait $10$ minutes. What fraction of the time do you wait between $5$ and $15$ minutes?

Forecast: a chunk around the average — guess roughly a third?

Step 1 — Turn "average $= 10$ " into $λ$ . For an exponential, the mean is $\frac{1}{λ}$ (see Expected Value and Variance). So $\frac{1}{λ} = 10 \Rightarrow λ = 0.1$ . Why this step? The word "average" pins down the one parameter the exponential has.

Step 2 — Translate the English into a CDF difference. "Between $5$ and $15$ minutes" $= P (5 < X \leq 15) = F_{X} (15) - F_{X} (5)$ . Why this step? An interval of a continuous variable is always a difference of running totals.

Step 3 — Compute. $F_{X} (15) - F_{X} (5) = (1 - e^{- 0.1 \cdot 15}) - (1 - e^{- 0.1 \cdot 5}) = e^{- 0.5} - e^{- 1.5}$ $\approx 0.6065 - 0.2231 = 0.3834$ Why this step? Same subtraction machinery, just with real numbers plugged in.

Verify: $0.3834$ is about a third ✓ (matches forecast). Units: it's a fraction of time, dimensionless in $[0, 1]$ ✓. (Same value the parent note's exponential example got — good consistency check.)

Ex. 8 — Cell H: exam twist (the sign trap + a mixed variable)

Worked example The negative-"probability" trap, then a mixed piecewise CDF

Part (a): A student computes $F_{X} (5) - F_{X} (8) = 0.4 - 0.7 = - 0.3$ and reports $P = - 0.3$ . What went wrong, and what's the right answer for the probability of $X$ between $5$ and $8$ ?

Part (b): A variable has the mixed CDF $F_{X} (x) = ⎩ ⎨ ⎧ 0 0.5 x 0.8 1 x < 0 0 \leq x < 1 1 \leq x < 2 x \geq 2$ Find $P (X = 1)$ and $P (0.5 < X \leq 2)$ .

Forecast: part (a) — spot the flipped subtraction. Part (b) — one flat stretch hides a point mass; where?

Step 1 (a) — Diagnose the sign. The CDF is non-decreasing, so $F_{X} (8) \geq F_{X} (5)$ . A probability can never be negative. The student subtracted in the wrong order. Why this step? Monotonicity is a hard rule; a negative result is an instant red flag.

Step 2 (a) — Correct it. $P (5 < X \leq 8) = F_{X} (8) - F_{X} (5) = 0.7 - 0.4 = 0.3$ Why this step? Larger bound minus smaller bound isolates the slice $(5, 8]$ .

Step 3 (b) — Point mass at $x = 1$ = the jump there. Just below $1$ : $F_{X} (1^{-}) = 0.5 (1) = 0.5$ . At $1$ : $F_{X} (1) = 0.8$ . $P (X = 1) = F_{X} (1) - F_{X} (1^{-}) = 0.8 - 0.5 = 0.3$ Why this step? A vertical jump in an otherwise-continuous CDF means a lump of probability sits exactly at that point — this is why the variable is "mixed".

Step 4 (b) — Interval across the jump. $P (0.5 < X \leq 2) = F_{X} (2) - F_{X} (0.5) = 1 - 0.5 (0.5) = 1 - 0.25 = 0.75$ Why this step? The interval rule works for any valid CDF, ramp or staircase or mixed — it automatically includes the $0.3$ jump at $x = 1$ .

Verify: (a) $0.3 \geq 0$ ✓ and equals $- (- 0.3)$ ✓. (b) $P (X = 1) = 0.3$ lies in $[0, 1]$ ✓; $0.75 \geq P (X = 1)$ ✓ since the wider interval contains the point mass.

Recap of the matrix

Recall Which example hit which cell?

Discrete between steps ::: Ex. 1 (Cell A) Discrete exact point / jump height ::: Ex. 2 (Cell B) Continuous interval on a ramp ::: Ex. 3 (Cell C) Boundary / outside-support inputs ::: Ex. 4 (Cell D) Inverting the CDF for a percentile ::: Ex. 5 (Cell E) Tail / complement / limiting ::: Ex. 6 (Cell F) Real-world word problem ::: Ex. 7 (Cell G) Sign trap + mixed variable ::: Ex. 8 (Cell H)

See also: Empirical Distribution (the CDF built from data), Kolmogorov-Smirnov Test (comparing two CDFs), Copulas (stitching CDFs together), Probability Distributions (where these CDFs come from).