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  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Learn7367Paper 3 · Statistics optionExponential distributionExponential distribution
Exponential distribution · Subtopic 1 of 1

The Exponential Distribution

If events occur randomly at rate λ\lambda (a Poisson process), the waiting time XX to the next one has pdf λe−λx\lambda e^{-\lambda x} for x≥0x\ge0, with P(X>x)=e−λxP(X>x)=e^{-\lambda x}, mean 1λ\frac1\lambda and variance 1λ2\frac1{\lambda^2}. It is memoryless: having waited doesn't change how much longer you'll wait.

AQA 7367 Paper 3≈ 35 minRevision 1 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use the pdf and cdf of Exp(λ)\text{Exp}(\lambda).
  • Find the mean, variance and median.
  • Link with Poisson and use the memoryless property.
Pause and think

Buses arrive at random, on average every 10 minutes. You've waited 10 minutes already. How long do you expect still to wait?

Section 2 of 14

Core idea

Start hereWaiting for the next call

A help desk gets 3 calls an hour, at random. The number of calls in an hour is Poisson. But what about the time until the next call? That's a continuous quantity, and it has its own distribution.

The big idea

If events occur at random at rate λ\lambda, the waiting time XX satisfies P(X>x)=e−λxP(X>x)=e^{-\lambda x}. So X∼Exp(λ)X\sim\text{Exp}(\lambda) with pdf λe−λx\lambda e^{-\lambda x}, mean 1λ\frac1\lambda, variance 1λ2\frac1{\lambda^2}.

1. From Poisson to exponential

The survival function
  1. X>x  ⟺  no events in [0,x]X>x\iff\text{no events in }[0,x]
    Waiting longer than xx means nothing happened yet.
  2. N∼Po(λx)N\sim\text{Po}(\lambda x)
    Number of events in time xx.
  3. P(X>x)=P(N=0)=e−λxP(X>x)=P(N=0)=e^{-\lambda x}
    Poisson probability of zero.
  4. f(x)=ddx(1−e−λx)=λe−λxf(x)=\frac{d}{dx}(1-e^{-\lambda x})=\lambda e^{-\lambda x}
    Differentiate the cdf.
Worked exampleWaiting more than 30 minutes

3 calls per hour. Find P(X>0.5)P(X>0.5).

Goal
Probability.
Strategy
Hours throughout.
  1. λ=3, x=0.5\lambda=3,\ x=0.5
    Same units.
  2. P=e−1.5P=e^{-1.5}
    Survival function.
Answer≈0.223\approx0.223.
✓ CheckMean wait 2020 min, so waiting 3030+ should be fairly unlikely ✓.

2. Mean, median, memory

Mean by integration by parts
  1. E(X)=∫0∞xλe−λx dxE(X)=\int_0^\infty x\lambda e^{-\lambda x}\,dx
    Definition.
  2. =[−xe−λx]0∞+∫0∞e−λx dx=\left[-xe^{-\lambda x}\right]_0^\infty+\int_0^\infty e^{-\lambda x}\,dx
    Parts with u=xu=x.
  3. =0+1λ=0+\frac1\lambda
    Exponential beats xx at infinity.
Worked exampleMemoryless property

Lifetime ∼Exp(0.1)\sim\text{Exp}(0.1) years. Given 5 years' survival, find P(survives to 10)P(\text{survives to }10).

Goal
Conditional probability.
Strategy
Use the definition.
  1. P(X>10∣X>5)=P(X>10)P(X>5)P(X>10\mid X>5)=\frac{P(X>10)}{P(X>5)}
    Conditional probability.
  2. =e−1e−0.5=e−0.5=\frac{e^{-1}}{e^{-0.5}}=e^{-0.5}
    Survival function.
Answere−0.5≈0.607e^{-0.5}\approx0.607, the same as a new component lasting 5 years.
✓ CheckP(X>5)=e−0.5P(X>5)=e^{-0.5} ✓: the past doesn't matter.
Two sides of a Poisson process

Counting events and timing them are two views of one process.

Common misconceptionRate vs mean λ\lambda is events per unit time; the mean wait is 1λ\frac1\lambda.
Key takeawayMedian e−λm=12⇒m=ln⁡2λe^{-\lambda m}=\frac12\Rightarrow m=\frac{\ln2}\lambda, less than the mean: the distribution is right-skewed.
Pause and think

Why isn't the exponential a good model for the lifetime of a human?

Summary

  • P(X>x)=e−λxP(X>x)=e^{-\lambda x}.
  • Mean 1λ\frac1\lambda, variance 1λ2\frac1{\lambda^2}.
  • Median ln⁡2λ\frac{\ln2}\lambda.
  • Memoryless.
Key words
Poisson process
Events occurring singly, independently, at constant rate.
Memoryless
Future waiting time independent of time already waited.
HistoryRadioactive decay. Rutherford and Soddy (1902) found atoms decay at random: the lifetime of each atom follows an exponential distribution, giving the half-life law.
Fun factHalf-life. The median of Exp(λ)\text{Exp}(\lambda) is ln⁡2λ\frac{\ln2}\lambda: exactly the half-life of a radioactive substance.
Section 3 of 14

Key formulae

pdf and cdff(x)=λe−λx,F(x)=1−e−λx(x≥0)f(x)=\lambda e^{-\lambda x},\quad F(x)=1-e^{-\lambda x}\quad(x\ge0)
MomentsE(X)=1λ,Var(X)=1λ2E(X)=\frac1\lambda,\quad\text{Var}(X)=\frac1{\lambda^2}
MemorylessP(X>a+b∣X>a)=P(X>b)P(X>a+b\mid X>a)=P(X>b)
TrickUnits of λ. Keep time units consistent: 3 calls per hour means λ=3\lambda=3 with xx in hours, so 30 minutes is x=0.5x=0.5.
Section 4 of 14

Worked examples

Example 1 · Waiting0 / 2 steps
Calls arrive at 3 per hour. Find P(wait>30 min)P(\text{wait}>30\text{ min}).
    Example 2 · Median0 / 2 steps
    Find the median of Exp(3)\text{Exp}(3).
      Section 5 of 14

      Explore the graph

      Exponential pdf
      Loading the interactive graph…
      Try this
      1. The vertical line is the mean 1λ\frac1\lambda.
      2. Bigger λ\lambda: events come faster, waits are shorter.

      f(x)=λe−λxf(x)=\lambda e^{-\lambda x}.

      In the real worldReliability. Electronic components with constant failure rate have exponential lifetimes; engineers quote the mean time between failures, 1λ\frac1\lambda.
      Section 6 of 14

      Common mistakes

      IncorrectMean =λ=\lambda
      Correct1λ\frac1\lambda
      λ\lambda is a rate; the mean is a time.
      IncorrectMixed units
      CorrectSame time unit throughout
      Convert minutes to hours or vice versa.
      IncorrectP(X>x)=1−e−λxP(X>x)=1-e^{-\lambda x}
      Correcte−λxe^{-\lambda x}
      That's P(X≤x)P(X\le x).

      Bus is overdue, must be soon

      Exponential distribution: I have no memory.

      Section 7 of 14

      Exam tip

      Exam tipStandard deviation equals mean. For the exponential, σ=μ=1λ\sigma=\mu=\frac1\lambda: a quick check on modelling questions.
      Section 8 of 14

      Guided practice

      SA-ST-EX-G012 marksL2 StandardCalculatorProbabilities
      Calls arrive at 3 per hour. Find the probability of waiting more than 30 minutes for the next call (3 s.f.).

      Sign in or create a free account to check answers, get hints and track your progress.

      Salam Academy Original Practice · Proposed mark scheme
      Section 9 of 14

      Practice questions

      SA-ST-EX-0011 markL1 FoundationNon-calculatorMean and memory
      Find the variance of X∼Exp(4)X\sim\text{Exp}(4).

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      Salam Academy Original Practice · Proposed mark scheme
      SA-ST-EX-0022 marksL2 StandardCalculatorMean and memory
      Find the median of Exp(3)\text{Exp}(3) (3 s.f.).

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      Salam Academy Original Practice · Proposed mark scheme
      SA-ST-EX-0032 marksL2 StandardCalculatorProbabilities
      X∼Exp(0.5)X\sim\text{Exp}(0.5). Find P(1<X<3)P(1<X<3) (3 s.f.).

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      Salam Academy Original Practice · Proposed mark scheme

      More questions on every skill in this subtopic are in the question bank under Practise.

      Section 10 of 14

      Challenge

      SA-ST-EX-0042 marksL3 ExamCalculatorMean and memory
      A component's lifetime is Exp(0.1)\text{Exp}(0.1) years. Given it has lasted 5 years, find the probability it lasts at least 10 years in total (3 s.f.).

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      Salam Academy Original Practice · Proposed mark scheme
      Section 11 of 14

      Summary

      • P(X>x)=e−λxP(X>x)=e^{-\lambda x}.
      • Mean 1λ\frac1\lambda, variance 1λ2\frac1{\lambda^2}.
      • Median ln⁡2λ\frac{\ln2}\lambda.
      • Memoryless.
      Section 12 of 14

      Quick quiz

      Q1 · SA-ST-EX-Q011 markL1 FoundationNon-calculatorProbabilities
      If events follow a Poisson process with rate λ\lambda, the time between events is
      Choose one option

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      Salam Academy Original Practice · Proposed mark scheme
      Section 13 of 14

      Same topic, different board

      Specification
      7367 A-level Further Mathematics
      Where
      Statistics · Exponential distribution
      Expected
      pdf λe^{−λx}, cdf, mean 1/λ, variance 1/λ², link with the Poisson process, the memoryless property.
      Calculator
      Permitted on all papers.
      Typical style
      AQA mixes short multiple-choice questions with longer structured ones.
      Open in
      7367 · Paper 3
      Section 14 of 14

      Past paper questions

      No past paper questions are tagged to this subtopic yet. We add papers every week.

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      Salam Academy original content · Content status: reviewed and published. Syllabus mapping: AQA 7367, First teaching 2017 (official specification).