Further Mathematics · AQA 7367 · Option S · Topic SF

Exponential distribution

The exponential distribution for waiting times between random events.

  • 4 sub-topics
  • Taught in Year 13
  • 48 practice questions
  • 4 animated lessons
Practice questions (48) Lessons Worksheets All AQA Further Mathematics topics

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. SF1The exponential distribution and when to use itY13
  2. SF2Probabilities from the exponential distributionY13
  3. SF3Proofs of the mean, variance and standard deviationY13
  4. SF4Time between Poisson events is exponentially distributedY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 48 questions in this topic, free, no account needed.

Exponential distribution

Not needed for Year 12Easy

The variance of the exponential distribution · Year 13 · 1 mark

  1. XX has an exponential distribution with parameter λ\lambda. What is Var⁡(X)\operatorname{Var}(X)?
    (1)

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  1. The mean is 1λ\dfrac{1}{\lambda}, and so is the standard deviation.BB: 1λ2\dfrac{1}{\lambda^2}

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Exponential distribution

Not needed for Year 12Medium

The exponential distribution · Year 13 · 6 marks

The time TT hours between breakdowns of a machine is modelled by an exponential distribution with mean 0.5 hours.
  1. (a)
    Give a reason why an exponential distribution may be suitable.
    (1)
  2. (b)
    Find P(1<T<3)P(1 < T < 3).
    (3)
  3. (c)
    Find the median of TT.
    (2)
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Part (a)1 mark

  1. Breakdowns occur at random, at a constant average rate (a Poisson process).

Part (b)3 marks

  1. λ=2\lambda = 2
  2. P(T>t)\mathrm{P}(T > t)=e−λt{}= \mathrm{e}^{-\lambda t}.e−2−e−6\mathrm{e}^{-2} - \mathrm{e}^{-6}
  3. =0.1329= 0.1329

Part (c)2 marks

  1. 1−e−λm=0.51 - \mathrm{e}^{-\lambda m} = 0.5
  2. m=ln⁡22=0.347m = \frac{\ln 2}{2} = 0.347

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Exponential distribution

Not needed for Year 12Hard

Mean of the exponential distribution · Year 13 · 5 marks

The continuous random variable XX has probability density function f(x)=4e−4xf(x) = 4\mathrm{e}^{-4x} for x≥0x \geq 0.
  1. (a)
    Prove that E(X)=14\mathrm{E}(X) = \frac{1}{4}.
    (4)
  2. (b)
    Write down the standard deviation of XX.
    (1)

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Part (a)4 marks

  1. E(X)\mathrm{E}(X)=∫0∞4xe−4x dx{}= \int_0^{\infty} 4x\mathrm{e}^{-4x}\,\mathrm{d}x
  2. by parts: [−xe−λx]0∞\text{by parts: } \left[-x\mathrm{e}^{-\lambda x}\right]_0^{\infty}+∫0∞e−λx dx{}+ \int_0^{\infty}\mathrm{e}^{-\lambda x}\,\mathrm{d}x
  3. x e^{−λx} → 0 as x → ∞.=0+1λ= 0 + \frac{1}{\lambda}=14{}= \frac{1}{4}

Part (b)1 mark

  1. For an exponential distribution σ=μ=1/λ\sigma = \mu = 1/\lambda.14\frac{1}{4}

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