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4.17 Poisson distributionIB Maths: Applications and Interpretation HL: Revision notes

Section 1

When a Poisson model is appropriate

The Poisson distribution models the number of events XX occurring in a fixed interval of time or space. It is appropriate when:

  • events are independent of one another, and
  • events occur at a uniform average rate (constant over the period of interest), and can occur at any point in the interval. We write X∼Po(λ)X\sim\mathrm{Po}(\lambda), where λ>0\lambda>0 is the mean number of events in the interval. Examples: calls to an office per hour, flaws per square metre of fabric, typing errors per page. If events come in clusters (for example several patients from one accident), or the rate changes through the day, the model is not appropriate.
Key termsPoisson distributionuniform average rate
Common mistake

Using a Poisson model when the rate varies (for example rush hour) or when events trigger each other.

Section 2

Probabilities, mean and variance

For X∼Po(λ)X\sim\mathrm{Po}(\lambda): P(X=r)=e−λλrr!,r=0,1,2,…\mathrm{P}(X=r)=\frac{e^{-\lambda}\lambda^r}{r!},\quad r=0,1,2,\dots E(X)=λ,Var(X)=λ.\mathrm{E}(X)=\lambda,\qquad\mathrm{Var}(X)=\lambda. The mean and variance are equal. Use the GDC (Poisson pdf for P(X=r)\mathrm{P}(X=r), Poisson cdf for P(X≤r)\mathrm{P}(X\le r)). Example: λ=4\lambda=4: P(X=3)=0.195\mathrm{P}(X=3)=0.195, P(X≤3)=0.433\mathrm{P}(X\le3)=0.433. For P(X≥r)\mathrm{P}(X\ge r) use 1−P(X≤r−1)1-\mathrm{P}(X\le r-1); so P(X>3)=1−P(X≤3)\mathrm{P}(X>3)=1-\mathrm{P}(X\le3). Formal proofs of the mean and variance are not required.

Key termsmeanvariance
Common mistake

Getting the boundary wrong: 'more than 33' is X≥4X\ge4, so 1−P(X≤3)1-\mathrm{P}(X\le3).

Section 3

Changing the interval

The parameter λ\lambda is the mean for the interval in the question. If the rate is 44 per hour then for 3030 minutes λ=2\lambda=2, and for three hours λ=12\lambda=12 (so the variance is also 1212). Always rescale λ\lambda first, then use the distribution for the new interval. Example: patients arrive at 55 per hour. In two hours Y∼Po(10)Y\sim\mathrm{Po}(10) and P(Y>12)=1−P(Y≤12)=0.208\mathrm{P}(Y>12)=1-\mathrm{P}(Y\le12)=0.208.

Key termsrescale
Exam tip

Write down the interval and its mean before touching the calculator.

Section 4

Sum of independent Poisson distributions

If X∼Po(λ)X\sim\mathrm{Po}(\lambda) and Y∼Po(μ)Y\sim\mathrm{Po}(\mu) are independent, then X+Y∼Po(λ+μ).X+Y\sim\mathrm{Po}(\lambda+\mu). Example: flaws at 0.80.8 and 1.51.5 per square metre; one square metre of each gives Po(2.3)\mathrm{Po}(2.3) and P(none)=e−2.3=0.100\mathrm{P}(\text{none})=e^{-2.3}=0.100. For joint events use independence: P(X=0 and Y=2)=P(X=0) P(Y=2)\mathrm{P}(X=0\text{ and }Y=2)=\mathrm{P}(X=0)\,\mathrm{P}(Y=2).

Key termssum of Poisson variables
Common mistake

Using the average of the two means, or subtracting. X−YX-Y is not Poisson.

Section 5

Choosing between normal, binomial and Poisson

  • Binomial B(n,p)\mathrm{B}(n,p): a fixed number nn of independent trials, each success or failure with constant probability pp (a count out of nn).
  • Poisson Po(λ)\mathrm{Po}(\lambda): events occurring at random in a continuous interval of time or space, at a constant average rate, with no upper limit on the count.
  • Normal N(μ,σ2)N(\mu,\sigma^2): a continuous measurement (mass, time, height) that is symmetric about the mean. Check the context: 'out of 8080 bottles, each cracked with probability 0.030.03' is binomial; '0.50.5 bubbles per bottle on average' is Poisson; the mass of a loaf is normal.
Key termsbinomialnormal
Exam tip

Ask: is it a count out of a fixed number of trials (binomial), a count at a random rate (Poisson), or a measurement (normal)?

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Exam questions on 4.17 Poisson distribution

  1. Calls to a school office arrive independently at a uniform average rate of 44 calls per hour. Let NN be the number of calls in a one-hour period.
    Find the probability that at least 22 calls arrive in a 3030-minute period.2 marks
  2. Flaws occur in rolls of fabric independently of each other, at a uniform average rate of 0.80.8 per square metre in fabric X and 1.51.5 per square metre in fabric Y.
    Two square metres of fabric X and one square metre of fabric Y are inspected. Use your GDC to find the probability that more than 33 flaws are found in total.2 marks
  3. Patients arrive at the emergency department of a hospital during the night at an average rate of 55 per hour. Let XX be the number of patients who arrive in a one-hour period during the night.
    State two conditions needed for XX to be modelled by a Poisson distribution, and give one reason why arrivals at an emergency department might not satisfy a condition.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).