Continuous Distributions in Depth

A continuous random variable is described not by a list of probabilities but by a probability density function (pdf): a curve whose area gives probability. This lesson works two of the most important named continuous models in depth — the continuous uniform $U(a,b)$U(a,b) and the exponential $\text{Exp}(\lambda)$Exp(λ) — deriving their means and variances from first principles, proving the exponential's defining memoryless property, and tying the exponential to the Poisson process that generates it. These two distributions, and the integration techniques used to handle them, underpin almost everything that follows in Further Statistics 2.

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Where this sits in AQA 7367

This is Paper 3 optional content — Statistics (7367/3S), one of the two applied options a Further-Maths student may choose (the others being Mechanics 7367/3M and Discrete 7367/3D). Paper 3 is 2 hours, 100 marks, weighted AO1 40% / AO2 25% / AO3 35% — markedly more problem-solving-heavy than the compulsory pure papers. This lesson leans on AO1 (carry out integration to find moments and probabilities) and AO2 (interpret the memoryless property; reason about which model fits a context), with AO3 appearing whenever a worded scenario must first be translated into the right distribution. It builds directly on A-Level Maths Statistics (the normal and the idea of a continuous distribution) and on A-Level Maths Pure (definite integration, integration by parts, and improper integrals — the $\int_0^\infty$∫0∞​ you meet in Further Calculus).

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Core theory: continuous random variables

A continuous random variable $X$X has a probability density function $f(x)$f(x) satisfying

$f(x) \ge 0 \quad \text{for all } x, \qquad \int_{-\infty}^{\infty} f(x)\,dx = 1.$f(x)≥0for all x,∫−∞∞​f(x)dx=1.

Probability is area under the curve: $P(a \le X \le b) = \int_a^b f(x)\,dx$P(a≤X≤b)=∫ab​f(x)dx. A single point carries zero area, so $P(X = c) = 0$P(X=c)=0 for any constant $c$c; this is why $\le$≤ and $<$< are interchangeable for continuous variables — a fact worth remembering when you read off table values. The cumulative distribution function (cdf) is

$F(x) = P(X \le x) = \int_{-\infty}^{x} f(t)\,dt, \qquad \text{so} \qquad f(x) = F'(x),$F(x)=P(X≤x)=∫−∞x​f(t)dt,sof(x)=F′(x),

and moments are weighted integrals of the density:

$E(X) = \int_{-\infty}^{\infty} x f(x)\,dx, \qquad E(X^2) = \int_{-\infty}^{\infty} x^2 f(x)\,dx, \qquad \operatorname{Var}(X) = E(X^2) - [E(X)]^2.$E(X)=∫−∞∞​xf(x)dx,E(X2)=∫−∞∞​x2f(x)dx,Var(X)=E(X2)−[E(X)]2.

That last identity — $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$Var(X)=E(X2)−[E(X)]2 — is the engine of almost every variance calculation below. Everything in this lesson is an application of these four formulae to two particular choices of $f(x)$f(x).

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The continuous uniform distribution $U(a, b)$U(a,b)

If $X \sim U(a, b)$X∼U(a,b), all sub-intervals of equal length inside $[a,b]$[a,b] are equally likely, so the density is constant. For the total area to be $1$1 over a base of width $b - a$b−a, the height must be $1/(b-a)$1/(b−a):

$f(x) = \frac{1}{b - a}, \quad a \le x \le b, \qquad f(x) = 0 \text{ otherwise}.$f(x)=b−a1​,a≤x≤b,f(x)=0 otherwise.

Integrating gives the cdf, a straight ramp from $0$0 to $1$1:

$F(x) = \begin{cases} 0 & x < a \\[2pt] \dfrac{x - a}{b - a} & a \le x \le b \\[2pt] 1 & x > b. \end{cases}$F(x)=⎩⎨⎧​0b−ax−a​1​x<aa≤x≤bx>b.​

Deriving the mean and variance

By symmetry the mean is the midpoint, but we derive it for practice:

$E(X) = \int_a^b \frac{x}{b-a}\,dx = \frac{1}{b-a}\left[\frac{x^2}{2}\right]_a^b = \frac{b^2 - a^2}{2(b-a)} = \frac{(b-a)(b+a)}{2(b-a)} = \frac{a + b}{2}.$E(X)=∫ab​b−ax​dx=b−a1​[2x2​]ab​=2(b−a)b2−a2​=2(b−a)(b−a)(b+a)​=2a+b​.

For the variance, first compute the second moment:

$E(X^2) = \int_a^b \frac{x^2}{b-a}\,dx = \frac{1}{b-a}\cdot\frac{b^3 - a^3}{3} = \frac{a^2 + ab + b^2}{3},$E(X2)=∫ab​b−ax2​dx=b−a1​⋅3b3−a3​=3a2+ab+b2​,

using $b^3 - a^3 = (b-a)(a^2 + ab + b^2)$b3−a3=(b−a)(a2+ab+b2). Then

$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \frac{a^2 + ab + b^2}{3} - \left(\frac{a+b}{2}\right)^2 = \frac{4(a^2 + ab + b^2) - 3(a+b)^2}{12} = \frac{(b - a)^2}{12}.$Var(X)=E(X2)−[E(X)]2=3a2+ab+b2​−(2a+b​)2=124(a2+ab+b2)−3(a+b)2​=12(b−a)2​.

Property	Formula
$E(X)$E(X)	$\dfrac{a + b}{2}$2a+b​
$E(X^2)$E(X2)	$\dfrac{a^2 + ab + b^2}{3}$3a2+ab+b2​
$\operatorname{Var}(X)$Var(X)	$\dfrac{(b - a)^2}{12}$12(b−a)2​
$F(x)$F(x) on $[a,b]$[a,b]	$\dfrac{x-a}{b-a}$b−ax−a​

Notice that the variance depends only on the width $b - a$b−a, not on the location of the interval — shifting $U(a,b)$U(a,b) along the axis changes the mean but never the spread.

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The exponential distribution $\text{Exp}(\lambda)$Exp(λ)

The exponential models the waiting time until the next event in a process where events happen at a constant average rate $\lambda$λ per unit time. Its density decays from a peak at $0$0:

$f(x) = \lambda e^{-\lambda x}, \quad x \ge 0, \qquad \lambda > 0.$f(x)=λe−λx,x≥0,λ>0.

Integrating from $0$0 to $x$x gives the cdf, and its complement is the survival function:

$F(x) = 1 - e^{-\lambda x}, \qquad P(X > x) = 1 - F(x) = e^{-\lambda x}, \quad x \ge 0.$F(x)=1−e−λx,P(X>x)=1−F(x)=e−λx,x≥0.

The survival form $P(X > x) = e^{-\lambda x}$P(X>x)=e−λx is the one you will reach for most often — "the probability of still waiting after time $x$x."

Deriving the mean by parts

$E(X) = \int_0^{\infty} x\,\lambda e^{-\lambda x}\,dx.$E(X)=∫0∞​xλe−λxdx.

Take $u = x$u=x, $\dfrac{dv}{dx} = \lambda e^{-\lambda x}$dxdv​=λe−λx, so $\dfrac{du}{dx} = 1$dxdu​=1 and $v = -e^{-\lambda x}$v=−e−λx:

$E(X) = \Big[-x e^{-\lambda x}\Big]_0^{\infty} + \int_0^{\infty} e^{-\lambda x}\,dx = 0 + \left[-\frac{1}{\lambda}e^{-\lambda x}\right]_0^{\infty} = \frac{1}{\lambda}.$E(X)=[−xe−λx]0∞​+∫0∞​e−λxdx=0+[−λ1​e−λx]0∞​=λ1​.

The boundary term vanishes because $x e^{-\lambda x} \to 0$xe−λx→0 as $x \to \infty$x→∞ (exponential decay beats linear growth). A second integration by parts gives $E(X^2) = 2/\lambda^2$E(X2)=2/λ2, hence

$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}.$Var(X)=E(X2)−[E(X)]2=λ22​−λ21​=λ21​.

The median $m$m solves $F(m) = \tfrac12$F(m)=21​: $1 - e^{-\lambda m} = \tfrac12 \Rightarrow e^{-\lambda m} = \tfrac12 \Rightarrow m = \dfrac{\ln 2}{\lambda}$1−e−λm=21​⇒e−λm=21​⇒m=λln2​. The density is largest at $x = 0$x=0, so the mode is $0$0. For a right-skewed distribution the order mode $<$< median $<$< mean is exactly what we see: $0 < \dfrac{\ln 2}{\lambda} < \dfrac{1}{\lambda}$0<λln2​<λ1​ (since $\ln 2 \approx 0.693$ln2≈0.693).

Property	Value
$E(X)$E(X)	$1/\lambda$1/λ
$\operatorname{Var}(X)$Var(X)	$1/\lambda^2$1/λ2
Median	$(\ln 2)/\lambda$(ln2)/λ
Mode	$0$0
$P(X > x)$P(X>x)	$e^{-\lambda x}$e−λx

The memoryless property

The exponential is the only continuous distribution that is memoryless: having already waited $s$s, the further wait is distributed exactly as if you had just started.

$P(X > s + t \mid X > s) = P(X > t).$P(X>s+t∣X>s)=P(X>t).

Proof. Using the conditional-probability definition and the survival function,

$P(X > s + t \mid X > s) = \frac{P(X > s + t \cap X > s)}{P(X > s)} = \frac{P(X > s + t)}{P(X > s)} = \frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} = e^{-\lambda t} = P(X > t).$P(X>s+t∣X>s)=P(X>s)P(X>s+t∩X>s)​=P(X>s)P(X>s+t)​=e−λse−λ(s+t)​=e−λt=P(X>t).

The event $\{X > s+t\}${X>s+t} is a subset of $\{X > s\}${X>s}, so their intersection is just $\{X > s+t\}${X>s+t} — that is the only subtle step. Interpretation: a component with an exponential lifetime "does not age." If a bus is equally likely to arrive in any small instant, then having waited ten minutes already tells you nothing about how much longer you must wait.

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Connection: the Poisson process and exponential waiting times

A Poisson process of rate $\lambda$λ is the standard model for events occurring randomly and independently at a constant average rate. It has two complementary descriptions:

• the number of events in a fixed interval of length $t$t follows $\text{Po}(\lambda t)$Po(λt);
• the waiting time between consecutive events follows $\text{Exp}(\lambda)$Exp(λ);
• the waiting time until the $n$nth event follows $\text{Gamma}(n, \lambda)$Gamma(n,λ).

The link between the first two is exact and worth seeing. "No event has occurred by time $x$x" means zero events in $[0,x]$[0,x], and the count there is $\text{Po}(\lambda x)$Po(λx), so

$P(X > x) = P(\text{0 events in }[0,x]) = e^{-\lambda x}\frac{(\lambda x)^0}{0!} = e^{-\lambda x},$P(X>x)=P(0 events in [0,x])=e−λx0!(λx)0​=e−λx,

which is precisely the exponential survival function. So the exponential gap is not an extra assumption — it is forced by the Poisson count. This duality is the single most useful idea in the topic: any question phrased in terms of counts ("how many arrivals in an hour") is Poisson, while the same scenario phrased in terms of time ("how long until the next arrival") is exponential, and you may switch freely between the two descriptions. The Gamma generalises this: the sum of $n$n independent $\text{Exp}(\lambda)$Exp(λ) gaps is the time to the $n$nth event, with density $f(x) = \dfrac{\lambda^n x^{n-1} e^{-\lambda x}}{(n-1)!}$f(x)=(n−1)!λnxn−1e−λx​, mean $n/\lambda$n/λ and variance $n/\lambda^2$n/λ2 (the exponential is the case $n = 1$n=1).

Worked example — switching between Poisson and exponential

Faults occur on a cable in a Poisson process at a rate of $0.5$0.5 per km. (a) Find the probability that a $4$4 km stretch has no faults. (b) Find the probability that the distance to the first fault exceeds $4$4 km. (c) Comment on your two answers.

(a) Count in $4$4 km: $N \sim \text{Po}(0.5 \times 4) = \text{Po}(2)$N∼Po(0.5×4)=Po(2), so

$P(N = 0) = e^{-2}\frac{2^0}{0!} = e^{-2} = 0.1353.$P(N=0)=e−20!20​=e−2=0.1353.

(b) Distance to first fault: $X \sim \text{Exp}(0.5)$X∼Exp(0.5), so

$P(X > 4) = e^{-0.5 \times 4} = e^{-2} = 0.1353.$P(X>4)=e−0.5×4=e−2=0.1353.

(c) The answers are identical, because "no fault in $4$4 km" and "first fault beyond $4$4 km" are the same event described two ways — the exact Poisson–exponential link. Working either route is valid; choose whichever the question makes more natural.

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Worked examples (with mark scheme)

Example 1 — uniform conditional probability

A train is equally likely to arrive at any time between 09:00 and 09:20; let $X \sim U(0, 20)$X∼U(0,20) be the minutes after 09:00. You reach the platform at 09:05. Find the probability you must wait more than a further 10 minutes.

You arrived at $5$5, and "more than 10 further minutes" means the train arrives after $15$15; since you know it has not yet arrived, condition on $X > 5$X>5:

$P(X > 15 \mid X > 5) = \frac{P(X > 15)}{P(X > 5)} \quad (\text{M1 conditional set-up})$P(X>15∣X>5)=P(X>5)P(X>15)​(M1 conditional set-up) $= \frac{(20-15)/20}{(20-5)/20} = \frac{5/20}{15/20} = \frac{1}{3}. \quad (\text{A1})$=(20−5)/20(20−15)/20​=15/205/20​=31​.(A1)

(M1 for the conditional ratio with correct events; A1 for $\tfrac13$31​. Unlike the exponential, the uniform is not memoryless: the answer $\tfrac13$31​ differs from the unconditional $P(X > 10) = \tfrac12$P(X>10)=21​.)

Example 2 — exponential waiting time and the memoryless property

Customers arrive at a help desk in a Poisson process at a rate of $3$3 per hour. Let $T$T (hours) be the time to the next arrival, so $T \sim \text{Exp}(3)$T∼Exp(3).

(a) Find $P(T \le 10\text{ min})$P(T≤10 min).

Convert: $10$10 min $= \tfrac16$=61​ hour.

$P\!\left(T \le \tfrac16\right) = 1 - e^{-3 \times 1/6} = 1 - e^{-0.5}. \quad (\text{M1 use of } F(x)=1-e^{-\lambda x})$P(T≤61​)=1−e−3×1/6=1−e−0.5.(M1 use of F(x)=1−e−λx) $= 1 - 0.6065 = 0.3935. \quad (\text{A1, 4 s.f.})$=1−0.6065=0.3935.(A1, 4 s.f.)

(b) Given that no customer has arrived in the first $20$20 minutes, find the probability of waiting a further $10$10 minutes.

By the memoryless property the extra wait restarts the clock:

$P(T > 30\text{ min} \mid T > 20\text{ min}) = P(T > 10\text{ min}) = e^{-3 \times 1/6} = e^{-0.5} = 0.6065. \quad (\text{M1 memoryless; A1})$P(T>30 min∣T>20 min)=P(T>10 min)=e−3×1/6=e−0.5=0.6065.(M1 memoryless; A1)

(M1 for converting units and applying the correct exponential formula; M1 for invoking memorylessness rather than recomputing a conditional integral; A1 values to 4 s.f.)

Example 3 — finding the rate from a median

The time $X$X (minutes) between successive calls to a switchboard is exponential with median $4$4 minutes. Find $\lambda$λ, the mean time between calls, and $P(X > 6)$P(X>6).

$\text{Median: } \frac{\ln 2}{\lambda} = 4 \;\Rightarrow\; \lambda = \frac{\ln 2}{4} = 0.1733. \quad (\text{M1 median equation; A1 } \lambda)$Median: λln2​=4⇒λ=4ln2​=0.1733.(M1 median equation; A1 λ) $E(X) = \frac{1}{\lambda} = \frac{4}{\ln 2} = 5.771 \text{ min}. \quad (\text{B1})$E(X)=λ1​=ln24​=5.771 min.(B1) $P(X > 6) = e^{-\lambda \times 6} = e^{-6\ln 2 / 4} = e^{-1.5\ln 2} = 2^{-1.5} = 0.3536. \quad (\text{M1 survival fn; A1})$P(X>6)=e−λ×6=e−6ln2/4=e−1.5ln2=2−1.5=0.3536.(M1 survival fn; A1)

(M1/A1 for solving the median equation; B1 for the mean; M1/A1 for the survival probability. Neat check: $e^{-1.5\ln 2} = (e^{\ln 2})^{-1.5} = 2^{-1.5}$e−1.5ln2=(eln2)−1.5=2−1.5.)

Example 4 — a uniform "show that" and a probability

A continuous random variable $X$X has $X \sim U(0, 6)$X∼U(0,6). (a) Show that $\operatorname{Var}(X) = 3$Var(X)=3. (b) Find $P(|X - 3| < 1)$P(∣X−3∣<1).

(a) Here $a = 0$a=0, $b = 6$b=6:

$\operatorname{Var}(X) = \frac{(b-a)^2}{12} = \frac{6^2}{12} = \frac{36}{12} = 3. \quad (\text{M1 formula; A1})$Var(X)=12(b−a)2​=1262​=1236​=3.(M1 formula; A1)

For a "show that," quoting the formula and substituting is enough; you need not re-derive it from $E(X^2) - [E(X)]^2$E(X2)−[E(X)]2 unless asked.

(b) The event $|X - 3| < 1$∣X−3∣<1 means $2 < X < 4$2<X<4, a sub-interval of length $2$2:

$P(2 < X < 4) = \frac{4 - 2}{6 - 0} = \frac{2}{6} = \frac{1}{3}. \quad (\text{M1 interpret modulus; A1})$P(2<X<4)=6−04−2​=62​=31​.(M1 interpret modulus; A1)

(M1/A1 for the variance; M1 turning the modulus inequality into an interval, A1 for $\tfrac13$31​. The mean is $3$3, so the event is "within $1$1 of the mean.")

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Specimen-style exam question

(specimen-style — not from any past paper)

The lifetime $X$X, in thousands of hours, of a certain electronic component is modelled by $X \sim \text{Exp}(0.4)$X∼Exp(0.4). (a) State the mean lifetime. (b) Find the probability a component lasts more than 3000 hours. (c) Given that a component has already lasted 2000 hours, find the probability it lasts at least 3000 hours more. (d) Comment on what your answers to (b) and (c) reveal about this model.

(a) $E(X) = 1/0.4 = 2.5$E(X)=1/0.4=2.5 thousand hours, i.e. $2500$2500 hours.

(b) $P(X > 3) = e^{-0.4 \times 3} = e^{-1.2} = 0.3012$P(X>3)=e−0.4×3=e−1.2=0.3012.

(c) By the memoryless property, $P(X > 5 \mid X > 2) = P(X > 3) = e^{-1.2} = 0.3012$P(X>5∣X>2)=P(X>3)=e−1.2=0.3012.

(d) The two probabilities are identical: under the exponential model the component "does not age," so prior use gives no information about remaining life. This is a known limitation — real components usually wear out, so an exponential lifetime is realistic only for failures caused by random external shocks rather than gradual wear.

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Synoptic links

• A-Level Maths Statistics — the normal distribution and continuous-variable basics. The pdf/cdf machinery here is the same; only the choice of $f(x)$f(x) changes.
• A-Level Maths Pure — integration by parts and improper integrals. $E(X)$E(X) for the exponential is a textbook $\int_0^\infty$∫0∞​ by parts; the boundary-term vanishing argument is pure Further Calculus.
• The Poisson distribution (Statistics 1). Counts in a fixed interval are Poisson; the gaps between those counts are exponential — two views of one process, linked exactly by $P(X>x) = P(\text{0 events})$P(X>x)=P(0 events).
• The geometric distribution. The exponential is the continuous analogue of the geometric: both are the unique memoryless distributions on their domains (continuous and discrete respectively).
• Hypothesis testing (later in this course). Survival probabilities $e^{-\lambda x}$e−λx become the building blocks of tests and confidence statements about a rate $\lambda$λ.

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Mark-scheme literacy

• "Show that the variance is $(b-a)^2/12$(b−a)2/12" requires every line: state $E(X^2)$E(X2), state $E(X)$E(X), then $E(X^2)-[E(X)]^2$E(X2)−[E(X)]2. A bare final answer scores the A but loses the method marks.
• "Write down the mean of $\text{Exp}(\lambda)$Exp(λ)" is one mark for $1/\lambda$1/λ — quote it, don't re-derive.
• "Hence" links parts: in the specimen question, part (c) expects you to reuse (b) via memorylessness, not to recompute from scratch — recomputing can forfeit the "hence" mark.
• Probabilities are conventionally given to 3 significant figures unless told otherwise; carry full accuracy through and round only at the end.
• Always convert time units before substituting into $e^{-\lambda x}$e−λx; a units mismatch is the single most common way to lose all the marks on an otherwise correct method.

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Grade-band model answers

Question: "The waiting time $T$T (minutes) for a lift follows $\text{Exp}(0.2)$Exp(0.2). Find $P(T > 10)$P(T>10) and interpret it."

Mid-band response. $P(T > 10) = e^{-0.2 \times 10} = e^{-2} = 0.135.$P(T>10)=e−0.2×10=e−2=0.135. Examiner-style commentary: Correct method and value to 3 s.f. The numerical answer earns the marks, but there is no interpretation, so any "comment/interpret" mark is lost.

Stronger response. As above, plus: "So about 13.5% of the time the wait exceeds 10 minutes; since the mean wait is $1/0.2 = 5$1/0.2=5 minutes, a wait beyond 10 minutes (twice the mean) is fairly unusual." Examiner-style commentary: The candidate links the probability to the mean and gives a contextual reading — the AO2/AO3 communication examiners reward.

Top-band response. As Stronger, plus: "Because the exponential is memoryless, this $0.135$0.135 is also the probability of waiting another 10 minutes given any wait so far; the model therefore assumes the lift's arrival is equally likely in every instant — reasonable for random demand but not if lifts run to a timetable." Examiner-style commentary: Connecting the numerical answer to the defining property and critiquing the modelling assumption is exactly the structured reasoning that separates the top band.

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Common misconceptions

1. "The height of a pdf is a probability." It is a density; only the area $\int f$∫f is a probability. A density can exceed $1$1 (e.g. $U(0,0.5)$U(0,0.5) has height $2$2).
2. "$P(X = c)$P(X=c) is small but non-zero." For a continuous variable it is exactly $0$0; hence $P(X \le c) = P(X < c)$P(X≤c)=P(X<c).
3. "The mean of $\text{Exp}(\lambda)$Exp(λ) is $\lambda$λ." It is $1/\lambda$1/λ. A large rate $\lambda$λ means short waits, so the mean is small.
4. "The uniform distribution is memoryless." It is not — Example 1 gives $\tfrac13 \ne P(X>10) = \tfrac12$31​=P(X>10)=21​. Only the exponential has this property among continuous distributions.
5. "$P(X > x) = 1 - e^{-\lambda x}$P(X>x)=1−e−λx." That is the cdf $F(x)$F(x). The survival function is its complement, $P(X > x) = e^{-\lambda x}$P(X>x)=e−λx.
6. "Variance of $U(a,b)$U(a,b) depends on where the interval sits." It depends only on the width: $(b-a)^2/12$(b−a)2/12. Shifting the interval leaves the spread unchanged.
7. "The exponential mode is at the mean." The density is largest at $x = 0$x=0, so the mode is $0$0; the distribution is right-skewed with mode $<$< median $<$< mean.

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Common errors

• Forgetting to convert time units before substituting into $e^{-\lambda x}$e−λx (mixing minutes and hours).
• Using $\lambda$λ where $1/\lambda$1/λ is meant (or vice versa) for the mean — always sanity-check that a higher rate gives a shorter mean wait.
• Dropping the boundary term in the by-parts derivation, or asserting it vanishes without the "exponential beats linear" justification.
• Writing the cdf where the survival function is wanted, giving $1 - e^{-\lambda x}$1−e−λx for $P(X > x)$P(X>x).
• Mis-stating the uniform height as $1$1 instead of $1/(b-a)$1/(b−a), so the total area is not $1$1.
• Computing a conditional exponential probability by integration when the memoryless shortcut is intended (wastes time and invites slips).

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Going further (stretch)

The memoryless property does not just hold for the exponential — it characterises it. Suppose a continuous lifetime $X$X on $[0,\infty)$[0,∞) satisfies $P(X > s+t) = P(X > s)\,P(X > t)$P(X>s+t)=P(X>s)P(X>t) for all $s, t \ge 0$s,t≥0. Write $g(x) = P(X > x)$g(x)=P(X>x); then $g(s+t) = g(s)g(t)$g(s+t)=g(s)g(t). This is Cauchy's multiplicative functional equation, and its only monotone solutions are $g(x) = e^{-\lambda x}$g(x)=e−λx for some $\lambda \ge 0$λ≥0. Hence any memoryless continuous lifetime must be exponential — there is no other choice. (A STEP-style exercise: set $h(x) = \ln g(x)$h(x)=lng(x), deduce $h(s+t) = h(s) + h(t)$h(s+t)=h(s)+h(t), and argue that a monotone additive function is linear, so $h(x) = -\lambda x$h(x)=−λx.)

A second thread: the sum of $n$n independent $\text{Exp}(\lambda)$Exp(λ) variables is Gamma$(n, \lambda)$(n,λ), the waiting time to the $n$nth event of a Poisson process — the continuous cousin of "sum of geometrics is negative binomial." At university you would prove this with moment generating functions: the MGF of $\text{Exp}(\lambda)$Exp(λ) is $M(t) = \lambda/(\lambda - t)$M(t)=λ/(λ−t) for $t < \lambda$t<λ, and the MGF of a sum of independents is the product, giving $[\lambda/(\lambda-t)]^n$[λ/(λ−t)]n — the Gamma MGF. The same tool instantly recovers $E(X) = M'(0) = 1/\lambda$E(X)=M′(0)=1/λ and $E(X^2) = M''(0) = 2/\lambda^2$E(X2)=M′′(0)=2/λ2, confirming our by-parts results without any integration.

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Additional A* practice (with worked answers)

P1. $X \sim U(2, 8)$X∼U(2,8). Find $P(3 \le X \le 5)$P(3≤X≤5) and $\operatorname{Var}(X)$Var(X). Answer: $P = (5-3)/(8-2) = 2/6 = 1/3$P=(5−3)/(8−2)=2/6=1/3; $\operatorname{Var}(X) = (8-2)^2/12 = 36/12 = 3$Var(X)=(8−2)2/12=36/12=3.

P2. $X \sim \text{Exp}(0.5)$X∼Exp(0.5). Find the median and $P(X < 2)$P(X<2). Answer: median $= \ln 2 / 0.5 = 2\ln 2 = 1.386$=ln2/0.5=2ln2=1.386; $P(X < 2) = 1 - e^{-0.5\times 2} = 1 - e^{-1} = 0.632$P(X<2)=1−e−0.5×2=1−e−1=0.632.

P3. Buses arrive in a Poisson process, one every $15$15 minutes on average. Find the probability the next bus takes more than $20$20 minutes. Answer: rate $\lambda = 1/15$λ=1/15 per min, so $P(X > 20) = e^{-20/15} = e^{-4/3} = 0.264$P(X>20)=e−20/15=e−4/3=0.264.

P4. For $X \sim \text{Exp}(\lambda)$X∼Exp(λ), show that $P(X > 2/\lambda) = e^{-2}$P(X>2/λ)=e−2 and comment on why this is independent of $\lambda$λ. Answer: $P(X > 2/\lambda) = e^{-\lambda \cdot 2/\lambda} = e^{-2} = 0.135$P(X>2/λ)=e−λ⋅2/λ=e−2=0.135. The threshold "two means" scales with $1/\lambda$1/λ, so the exceedance probability is the same for every exponential — the distribution has a fixed shape.

P5. $X \sim U(0, a)$X∼U(0,a) has variance $12$12. Find $a$a and $E(X)$E(X). Answer: $(a-0)^2/12 = 12 \Rightarrow a^2 = 144 \Rightarrow a = 12$(a−0)2/12=12⇒a2=144⇒a=12; $E(X) = a/2 = 6$E(X)=a/2=6.

P6. A machine's time-to-failure is $\text{Exp}(\lambda)$Exp(λ) with $P(X > 100) = 0.8$P(X>100)=0.8 (hours). Find $\lambda$λ and the mean lifetime. Answer: $e^{-100\lambda} = 0.8 \Rightarrow \lambda = -\tfrac{1}{100}\ln 0.8 = 0.00223$e−100λ=0.8⇒λ=−1001​ln0.8=0.00223; mean $= 1/\lambda = 448$=1/λ=448 hours (3 s.f.).

P7. Calls arrive in a Poisson process at $4$4 per hour. Using the count description, find the probability of exactly $2$2 calls in the next $30$30 minutes; using the time description, find the probability the first call comes within $15$15 minutes. Answer: count in $30$30 min: $\text{Po}(4 \times 0.5) = \text{Po}(2)$Po(4×0.5)=Po(2), $P(N = 2) = e^{-2}2^2/2! = 2e^{-2} = 0.271$P(N=2)=e−222/2!=2e−2=0.271. Time to first: $\text{Exp}(4)$Exp(4), $P(X < 0.25) = 1 - e^{-4 \times 0.25} = 1 - e^{-1} = 0.632$P(X<0.25)=1−e−4×0.25=1−e−1=0.632.

P8. The time to the $3$3rd arrival in a Poisson process of rate $2$2 per hour follows $\text{Gamma}(3, 2)$Gamma(3,2). State its mean and variance. Answer: mean $= n/\lambda = 3/2 = 1.5$=n/λ=3/2=1.5 hours; variance $= n/\lambda^2 = 3/4 = 0.75$=n/λ2=3/4=0.75.

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Board-alignment footer

Aligned to AQA A-Level Further Mathematics 7367, Paper 3 Statistics option (7367/3S): continuous random variables, the continuous uniform $U(a,b)$U(a,b), the exponential $\text{Exp}(\lambda)$Exp(λ) with its memoryless property, and the Poisson-process link. The content is equivalent to the corresponding sections of Edexcel Further Statistics 1 (9FM0/3B) and OCR (H245) Statistics — the distributions and moment formulae are identical across boards; only the option codes differ.

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Visual summary

$\boxed{\; U(a,b):\ E = \tfrac{a+b}{2},\ \operatorname{Var} = \tfrac{(b-a)^2}{12} \qquad \text{Exp}(\lambda):\ E = \tfrac1\lambda,\ \operatorname{Var} = \tfrac1{\lambda^2},\ P(X>x) = e^{-\lambda x} \;}$U(a,b): E=2a+b​, Var=12(b−a)2​Exp(λ): E=λ1​, Var=λ21​, P(X>x)=e−λx​

Feature	Uniform $U(a,b)$U(a,b)	Exponential $\text{Exp}(\lambda)$Exp(λ)
Support	$[a,b]$[a,b] (bounded)	$[0,\infty)$[0,∞) (unbounded)
Shape	flat (rectangular)	decreasing, right-skewed
Mean	$(a+b)/2$(a+b)/2	$1/\lambda$1/λ
Variance	$(b-a)^2/12$(b−a)2/12	$1/\lambda^2$1/λ2
Memoryless?	No	Yes (the defining property)
Typical use	random point in an interval	waiting time between events

Recap. Probability for a continuous variable is area under its pdf; $\operatorname{Var}(X) = E(X^2)-[E(X)]^2$Var(X)=E(X2)−[E(X)]2 drives every variance derivation. The uniform spreads probability evenly with variance set purely by interval width; the exponential models constant-rate waiting times, has mean $1/\lambda$1/λ, survival $e^{-\lambda x}$e−λx, and is the unique memoryless continuous distribution — the bridge from the Poisson process to all the inference that follows.