Continuous random variables — PDF, CDF, percentiles: Worked examples

This page is the workout for Continuous random variables — PDF, CDF, percentiles. The parent note built the ideas; here we drill every kind of question the topic can hand you, one worked example per case. Read the parent's [!definition] boxes first if any symbol below feels new.

Recall the vocabulary from the parent, in one breath:

== $f (x)$ == = the density — how thickly probability is spread at the point $x$ . Height, not probability.
== $F (x) = P (X \leq x)$ == = the cumulative — the total area swept from the far left up to $x$ .
The link: $F (x) = \int_{- \infty}^{x} f (t) d t$ and (backwards) $f (x) = F^{'} (x)$ by the Fundamental Theorem of Calculus.
The == $p$ -th percentile== $x_{p}$ solves $F (x_{p}) = p$ .

The scenario matrix

Each row is a case class — a kind of situation that behaves differently and can trip you if you never saw it. The final column names the example that covers it.

#	Case class	What's tricky about it	Covered by
C1	Given $f$ , finite support, find $F$ + interval prob	must state $F = 0$ below, $F = 1$ above	Ex 1
C2	Given $f$ , half-line support (waiting time)	integral runs to $\infty$ ; percentile via logs	Ex 2
C3	Given $F$ , recover $f$ by differentiating	reverse direction; check total area	Ex 3
C4	Piecewise $f$ (triangle) — probability crosses a corner	must split the integral at the corner	Ex 4
C5	Normalising constant unknown ( $f = c g$ )	solve for $c$ first from total-area $= 1$	Ex 5
C6	Degenerate / zero inputs: $P (X = c)$ , $P (X \leq$ below support $)$ , empty interval	answers are exactly $0$ or $1$	Ex 6
C7	Uniform — constant density, every percentile is linear	density $> 1$ is legal; check the " $f \leq 1$ " myth	Ex 7
C8	Word problem with units + a real decision	translate "at least", carry units	Ex 8
C9	Exam twist: mixed conditional probability $P (X > a ∣ X > b)$	use $F$ ratios, not raw $f$	Ex 9
C10	Limiting behaviour: what $f, F$ do as $x \to$ edges	sanity checks, not a full solve	Ex 10

C1 — Given $f$ , finite support

Worked example Example 1 · linear density on

$[0, 2]$ Let $f (x) = \frac{1}{2} x$ for $0 \leq x \leq 2$ , and $f (x) = 0$ elsewhere. Find $F (x)$ (all $x$ ), then $P (0.5 \leq X \leq 1.5)$ .

Forecast: guess whether $F (2) = 1$ , and whether the answer to the interval is bigger or smaller than $\frac{1}{2}$ .

Step 1 — Confirm it's a valid PDF. $\int_{0}^{2} \frac{1}{2} x d x = [\frac{x ^{2}}{4}]_{0}^{2} = 1$ , and $\frac{1}{2} x \geq 0$ on $[0, 2]$ . ✓ Why this step? Nothing downstream is meaningful unless total area is $1$ and density is non-negative.

Step 2 — Accumulate for $0 \leq x \leq 2$ . $F (x) = \int_{0}^{x} \frac{1}{2} t d t = [\frac{t ^{2}}{4}]_{0}^{x} = \frac{x ^{2}}{4} .$ Why this step? $F$ is the running area from the left edge of the support (here $0$ ).

Step 3 — State $F$ on the WHOLE line. $F (x) = 0$ for $x < 0$ ; $F (x) = \frac{x ^{2}}{4}$ for $0 \leq x \leq 2$ ; $F (x) = 1$ for $x > 2$ . Why this step? The #4 parent mistake: forgetting $F$ must reach $0$ below and $1$ above the support.

Step 4 — Interval by subtraction. $P (0.5 \leq X \leq 1.5) = F (1.5) - F (0.5) = \frac{1. 5 ^{2}}{4} - \frac{0. 5 ^{2}}{4} = \frac{2.25 - 0.25}{4} = 0.5.$ Why this step? $\int_{a}^{b} f = F (b) - F (a)$ — no need to re-integrate.

Verify: $F (2) = \frac{4}{4} = 1$ ✓, $F (0) = 0$ ✓. The answer $0.5$ is a probability in $[0, 1]$ ✓, and since more density sits on the right, the middle strip grabbing half seems reasonable.

The figure below draws this density and shades the interval — look at how the shaded red strip is exactly the "area" our subtraction computed. The dashed mint lines mark where the support begins and ends (outside them $f = 0$ ), a visual reminder of Step 3.

Notice the density rises from left to right: that slope is why a middle strip still captures half the probability — the extra height on the right compensates for its distance from the peak.

C2 — Half-line support (exponential)

Worked example Example 2 · exponential waiting time,

$λ = \frac{1}{3}$ Bus gaps follow $f (x) = \frac{1}{3} e^{- x /3}$ for $x \geq 0$ (minutes). Find $F$ , the median, and the 90th percentile.

Forecast: the median of a waiting time is usually less than the mean ( $= 3$ min here). Guess above or below $3$ .

Step 1 — CDF by integrating to $x$ . $F (x) = \int_{0}^{x} \frac{1}{3} e^{- t /3} d t = [- e^{- t /3}]_{0}^{x} = 1 - e^{- x /3}$ for $x \geq 0$ . Why this step? $\int λ e^{- λ t} d t = - e^{- λ t}$ ; the support starts at $0$ so we integrate from $0$ .

Step 2 — Median: solve $F (m) = 0.5$ . $1 - e^{- m /3} = 0.5 \Rightarrow e^{- m /3} = 0.5 \Rightarrow m = 3 ln 2 \approx 2.079$ min. Why this step? Median is the 50th percentile — invert $F$ , don't touch $f$ .

Step 3 — 90th percentile: solve $F (x) = 0.9$ . $1 - e^{- x /3} = 0.9 \Rightarrow e^{- x /3} = 0.1 \Rightarrow x = 3 ln 10 \approx 6.908$ min. Why this step? Same inversion with $p = 0.9$ ; the log un-does the exponential.

Verify: median $2.079 < 3$ ✓ (matches the forecast — waiting times are right-skewed). Plug $x = 6.908$ : $e^{- 6.908/3} = e^{- 2.303} \approx 0.100$ , so $F \approx 0.900$ ✓. See Exponential distribution for the memoryless twist used in Ex 9.

C3 — Given $F$ , recover $f$

Worked example Example 3 · differentiate the CDF

$F (x) = 1 - e^{- x^{2}}$ for $x \geq 0$ and $F (x) = 0$ for $x < 0$ . Find $f$ and verify total area.

Forecast: since $F$ climbs from $0$ toward $1$ , its slope $f$ should be positive and eventually shrink back to $0$ .

Step 1 — Differentiate (FTC). $f (x) = F^{'} (x) = \frac{d}{d x} (1 - e^{- x^{2}}) = 2 x e^{- x^{2}}$ for $x \geq 0$ ; $f = 0$ for $x < 0$ . Why this step? PDF is the rate of accumulation — the derivative of $F$ .

Step 2 — Non-negativity check. For $x \geq 0$ , $2 x \geq 0$ and $e^{- x^{2}} > 0$ , so $f \geq 0$ . ✓ Why this step? A recovered $f$ is only a valid density if it never goes negative.

Step 3 — Total area check. $\int_{0}^{\infty} 2 x e^{- x^{2}} d x = [- e^{- x^{2}}]_{0}^{\infty} = 0 - (- 1) = 1.$ ✓ Why this step? Confirms $F (\infty) = 1$ , i.e. all probability is accounted for.

Verify: $f (0) = 0$ , and as $x \to \infty$ the term $e^{- x^{2}}$ kills the growth so $f \to 0$ — exactly the "starts at 0, peaks, decays" shape the forecast predicted.

C4 — Piecewise density (probability crosses a corner)

Worked example Example 4 · triangle density

$f (x) = x$ for $0 \leq x < 1$ , $f (x) = 2 - x$ for $1 \leq x \leq 2$ , $0$ elsewhere. Find $F (1.5)$ and $P (0.5 \leq X \leq 1.5)$ .

Forecast: by symmetry the peak is at $x = 1$ ; guess whether $F (1) = 0.5$ .

Step 1 — First piece, $0 \leq x \leq 1$ . $F (x) = \int_{0}^{x} t d t = \frac{x ^{2}}{2}$ . So $F (1) = \frac{1}{2}$ . Why this step? Accumulate area over the rising slope.

Step 2 — Second piece, $1 \leq x \leq 2$ . Start from $F (1) = \frac{1}{2}$ and add: $F (x) = \frac{1}{2} + \int_{1}^{x} (2 - t) d t = \frac{1}{2} + [2 t - \frac{t ^{2}}{2}]_{1}^{x}$ . Evaluate: $F (x) = \frac{1}{2} + (2 x - \frac{x ^{2}}{2}) - (2 - \frac{1}{2}) = 2 x - \frac{x ^{2}}{2} - 1.$ Why this step? The corner at $x = 1$ forces a split; we carry the area already collected.

Step 3 — Plug in. $F (1.5) = 2 (1.5) - \frac{1. 5 ^{2}}{2} - 1 = 3 - 1.125 - 1 = 0.875.$ $P (0.5 \leq X \leq 1.5) = F (1.5) - F (0.5) = 0.875 - \frac{0. 5 ^{2}}{2} = 0.875 - 0.125 = 0.75.$ Why this step? Interval prob straddles the corner but the CDF has already "healed" the split — one subtraction finishes it.

Verify: $F (1) = 0.5$ ✓ (matches symmetry forecast), $F (2) = 4 - 2 - 1 = 1$ ✓. Answer $0.75$ lies in $[0, 1]$ ✓.

The figure below shows why this problem needs two integrals, not one: follow the dashed mint line at the corner $x = 1$ — to its left the density climbs as $f = x$ , to its right it falls as $f = 2 - x$ , so the shaded red region (our $P = 0.75$ ) is really two differently-shaped pieces glued at the corner. Trying to integrate one formula across the corner would miss the switch.

C5 — Unknown normalising constant

C6 — Degenerate & zero inputs

Worked example Example 6 · the "trick" answers

Using $f (x) = \frac{1}{2} x$ on $[0, 2]$ from Ex 1, evaluate: (a) $P (X = 1)$ , (b) $P (X \leq - 4)$ , (c) $P (X \leq 100)$ , (d) $P (1.3 \leq X \leq 1.3)$ .

Forecast: three of these are exactly $0$ or $1$ with no integration needed. Spot which.

Step 1 — (a) single point. $P (X = 1) = \int_{1}^{1} f = 0.$ Why this step? A point has zero width, so zero area — the parent's core rule.

Step 2 — (b) below the support. $- 4 < 0$ , and $F (x) = 0$ for $x < 0$ , so $P (X \leq - 4) = 0.$ Why this step? No probability lives before the support begins.

Step 3 — (c) above the support. $100 > 2$ , and $F (x) = 1$ for $x > 2$ , so $P (X \leq 100) = 1.$ Why this step? All the probability was collected by $x = 2$ ; nothing new can be added.

Step 4 — (d) empty-width interval. $P (1.3 \leq X \leq 1.3) = F (1.3) - F (1.3) = 0.$ Why this step? Same width-zero logic — the $<$ vs $\leq$ distinction never matters here.

Verify: all four answers are exactly $0, 0, 1, 0$ — pure boundary behaviour, no integral evaluated ✓.

C7 — Uniform (constant density, the " $f > 1$ " myth)

Worked example Example 7 · density above 1 is fine

$X$ is uniform on $[0.2, 0.6]$ : $f (x) = \frac{1}{0.6 - 0.2} = 2.5$ on that interval, $0$ elsewhere. Find $F$ , $P (X \leq 0.5)$ , and the 25th percentile.

Forecast: the density is $2.5 > 1$ . Does that break anything? (No — watch.)

Step 1 — Height check via area. Area $= 2.5 \times (0.6 - 0.2) = 2.5 \times 0.4 = 1$ . ✓ Why this step? Only area must equal $1$ ; a tall thin box is perfectly legal. This is the parent's #1 mistake, live.

Step 2 — CDF (a straight ramp). For $0.2 \leq x \leq 0.6$ : $F (x) = 2.5 (x - 0.2)$ . Outside: $0$ below $0.2$ , $1$ above $0.6$ . Why this step? Constant density $\Rightarrow$ area grows linearly $\Rightarrow$ $F$ is a straight line (see Uniform distribution).

Step 3 — $P (X \leq 0.5)$ . $F (0.5) = 2.5 (0.5 - 0.2) = 2.5 \times 0.3 = 0.75.$ 25th percentile: solve $2.5 (x - 0.2) = 0.25 \Rightarrow x - 0.2 = 0.1 \Rightarrow x = 0.3.$ Why this step? Read the CDF for the probability; invert it for the percentile.

Verify: $F (0.6) = 2.5 \times 0.4 = 1$ ✓, $F (0.2) = 0$ ✓. Percentile $0.3 \in [0.2, 0.6]$ ✓, and $f = 2.5 > 1$ caused no trouble ✓.

C8 — Word problem with units and a decision

Worked example Example 8 · battery lifetime

A battery's life (hours) has $f (t) = \frac{1}{4} e^{- t /4}$ for $t \geq 0$ . What fraction last at least 6 hours? What lifetime do only the top $5%$ exceed?

Forecast: mean life is $4$ h, so "at least 6 h" should be a minority — guess under $25%$ .

Step 1 — CDF. $F (t) = 1 - e^{- t /4}$ , $t \geq 0$ (same form as Ex 2 with $λ = \frac{1}{4}$ ). Why this step? We need accumulated probability to answer any "how much lasts up to / beyond" question.

Step 2 — "At least 6 h" = upper tail. $P (T \geq 6) = 1 - F (6) = e^{- 6/4} = e^{- 1.5} \approx 0.223.$ Why this step? "At least" means the complement of " $\leq$ ": $P (T \geq t) = 1 - F (t)$ . Units: hours in, pure probability out.

Step 3 — "Top $5%$ exceed" = 95th percentile. Solve $F (t) = 0.95$ : $1 - e^{- t /4} = 0.95 \Rightarrow e^{- t /4} = 0.05 \Rightarrow t = 4 ln 20 \approx 11.98$ h. Why this step? The value only $5%$ beat is the $95$ th percentile — invert $F$ at $p = 0.95$ .

Verify: $0.223 < 0.25$ ✓ (matches forecast). Check $t = 11.98$ : $e^{- 11.98/4} = e^{- 2.996} \approx 0.050$ so $F \approx 0.95$ ✓. Units are hours throughout ✓.

C9 — Exam twist: conditional probability (memorylessness)

Worked example Example 9 ·

$P (X > a ∣ X > b)$ With the battery of Ex 8 ( $f (t) = \frac{1}{4} e^{- t /4}$ ), given a battery already survived $2$ h, what is $P (T > 5 ∣ T > 2)$ ?

Forecast: exponential is memoryless. Guess $P (T > 5 ∣ T > 2) = P (T > 3)$ .

Step 1 — Conditional-probability definition. $P (T > 5 ∣ T > 2) = \frac{P ( T > 5 and T > 2 )}{P ( T > 2 )} = \frac{P ( T > 5 )}{P ( T > 2 )}$ . Why this step? " $T > 5$ " already implies " $T > 2$ ", so the joint event is just $T > 5$ .

Step 2 — Use tail probabilities. $P (T > t) = e^{- t /4}$ , so $\frac{e ^{- 5/4}}{e ^{- 2/4}} = e^{- (5 - 2) /4} = e^{- 3/4} \approx 0.472.$ Why this step? Ratios of the CDF/tail, never raw $f$ — the #3 parent mistake dodged.

Step 3 — Recognise the pattern. $e^{- 3/4} = P (T > 3)$ : the extra time needed forgot the $2$ h already served. Why this step? Confirms the memoryless property numerically.

Verify: $e^{- 0.75} \approx 0.4724$ , and $P (T > 3) = e^{- 3/4} \approx 0.4724$ ✓ — identical, matching the forecast.

C10 — Limiting behaviour (sanity, not a solve)

Worked example Example 10 · read the edges

For any continuous RV, describe what $F$ and $f$ do as $x \to - \infty$ and $x \to + \infty$ , using $F (x) = 1 - e^{- x^{2}}$ (Ex 3) as the concrete case.

Forecast: $F$ must sandwich between $0$ and $1$ ; $f$ must fade to $0$ at both far ends (else area would be infinite).

Step 1 — Far left. $F (x) = 0$ for $x < 0$ , so $lim_{x \to - \infty} F (x) = 0$ and $f = 0$ there. Why this step? No probability precedes the support; $F$ starts flat at $0$ .

Step 2 — Far right. $lim_{x \to \infty} (1 - e^{- x^{2}}) = 1 - 0 = 1$ , so $F \to 1$ ; and $f (x) = 2 x e^{- x^{2}} \to 0$ because $e^{- x^{2}}$ crushes the linear $2 x$ . Why this step? $F$ must saturate at $1$ (total probability), and $f$ must vanish at the far edge for the area to stay finite.

Step 3 — General rule. Every CDF is a non-decreasing curve rising from $0$ to $1$ ; every PDF must integrate to a finite total of $1$ , which forces $f \to 0$ in both tails — if $f$ stayed above some fixed positive height forever, the tail area would keep growing without bound and the total could never settle at $1$ . Why this step? This turns the two concrete limits of Steps 1–2 into a universal law you can use as an instant sanity check on any density: a PDF that does not decay to $0$ in its tails cannot have total area $1$ , so it is not a valid density.

Verify: $F (- \infty) = 0$ , $F (+ \infty) = 1$ , $f (\pm \infty) = 0$ ✓ — the universal boundary conditions, matching the forecast.

Recall Which tool for which question?

Given $f$ , asked for a probability ::: integrate / use $F (b) - F (a)$ . Given $f$ , asked for a percentile ::: build $F$ , then solve $F (x_{p}) = p$ . Given $F$ , asked for $f$ ::: differentiate: $f = F^{'}$ . Asked $P (X = c)$ or a zero-width interval ::: it is exactly $0$ . Asked $P (X \leq$ below support $)$ / $P (X \leq$ above support $)$ ::: $0$ / $1$ . Asked "at least $a$ " ::: $P (X \geq a) = 1 - F (a)$ . Asked $P (X > a ∣ X > b)$ ::: ratio of tails $\frac{1 - F ( a )}{1 - F ( b )}$ .

Connections

Continuous random variables — PDF, CDF, percentiles (parent — theory this page drills)
Exponential distribution (Ex 2, 8, 9 — half-line support & memorylessness)
Uniform distribution (Ex 7 — constant density, linear CDF)
Fundamental Theorem of Calculus (Ex 3, 10 — $f \leftrightarrow F$ )
Quantile function and inverse-transform sampling (every percentile solve $= F^{- 1}$ )
Normal distribution (percentiles via $Φ$ , same recipe as Ex 8)
Expectation and Variance of continuous RVs (next step once $f$ is known)

Worked examples — Continuous random variables — PDF, CDF, percentiles