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How do the Poisson and binomial distributions model counts, and when does one approximate the other?

The Poisson distribution as a model for random events, its mean and variance, the binomial distribution, the additive property of Poisson variables, and the Poisson approximation to the binomial.

A focused answer to the Edexcel A-Level Further Mathematics Further Statistics content on the Poisson and binomial distributions, covering the Poisson model and its mean and variance, the binomial distribution, the additive property of independent Poisson variables, and the Poisson approximation to the binomial.

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  1. What this dot point is asking
  2. The Poisson distribution
  3. The binomial and the additive property
  4. The Poisson approximation to the binomial
  5. Examples in context
  6. Try this

What this dot point is asking

Edexcel Further Statistics wants you to use the Poisson distribution to model random events over a fixed interval, state and apply its mean and variance, use the binomial distribution, add independent Poisson variables, and apply the Poisson approximation to the binomial when nn is large and pp small. Justifying the choice of model and approximation, with clear method, is what earns the marks.

The Poisson distribution

The Poisson models the number of random events occurring in a fixed interval of time or space, when events happen independently at a constant average rate λ\lambda. The distinctive feature is that its mean and variance are equal, both λ\lambda, which is also the practical test of whether a Poisson model is plausible for a data set: if the sample mean and variance are close, a Poisson fit is reasonable.

The binomial and the additive property

The binomial distribution counts the number of successes in a fixed number nn of independent trials, each with the same success probability pp. Independent Poisson variables have a clean additive property: their sum is again Poisson, with the rates added, which lets you combine, for example, the number of events over two separate intervals into a single Poisson model.

The Poisson approximation to the binomial

When the number of trials is large and the success probability is small, the binomial probabilities are awkward to compute but the Poisson distribution with λ=np\lambda = np gives a very close approximation. The rule of thumb is nn large (commonly n>50n > 50) and pp small (commonly p<0.1p < 0.1), so that npnp is moderate.

Examples in context

The Poisson and binomial are the two most heavily examined discrete models. Their means and variances are derived using the expectation machinery of the discrete-distributions dot point. They contrast with the geometric and negative binomial "waiting time" models, which count trials rather than successes. Both distributions are the usual subjects of chi-squared goodness of fit tests, where the mean is often estimated from the data (costing a degree of freedom). The additive property of the Poisson reflects the way independent rates combine, and the approximation to the binomial is the historical origin of the Poisson distribution as the "law of rare events".

Try this

Q1. For XPo(3)X \sim \operatorname{Po}(3), find P(X=0)P(X = 0). [2 marks]

  • Cue. P(X=0)=e30.0498P(X = 0) = e^{-3} \approx 0.0498.

Q2. State the mean and variance of B(50,0.2)B(50, 0.2). [2 marks]

  • Cue. Mean =10= 10, variance =50(0.2)(0.8)=8= 50(0.2)(0.8) = 8.

Q3. Independent XPo(2)X \sim \operatorname{Po}(2) and YPo(6)Y \sim \operatorname{Po}(6). State the distribution of X+YX + Y. [1 mark]

  • Cue. X+YPo(8)X + Y \sim \operatorname{Po}(8).

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20196 marksThe number of emails arriving at an inbox follows a Poisson distribution with mean 44 per hour. Find the probability that exactly 33 emails arrive in a given hour, and the probability that more than 22 arrive.
Show worked answer →

Apply the Poisson pmf, then use the complement for the "more than" probability.

XPo(4)X \sim \operatorname{Po}(4). P(X=3)=e4433!=e46460.1954P(X = 3) = e^{-4}\frac{4^3}{3!} = e^{-4}\frac{64}{6} \approx 0.1954 (M1 A1).

P(X>2)=1P(X2)=1[P(0)+P(1)+P(2)]P(X > 2) = 1 - P(X \le 2) = 1 - [P(0) + P(1) + P(2)] (M1). P(0)=e40.0183P(0) = e^{-4} \approx 0.0183, P(1)=4e40.0733P(1) = 4e^{-4} \approx 0.0733, P(2)=8e40.1465P(2) = 8e^{-4} \approx 0.1465 (A1).

P(X2)0.2381P(X \le 2) \approx 0.2381, so P(X>2)10.2381=0.7619P(X > 2) \approx 1 - 0.2381 = 0.7619 (M1 A1).

Edexcel 20225 marksA machine produces components, each defective independently with probability 0.0050.005. In a batch of 400400 components, use a suitable approximation to estimate the probability that exactly 33 are defective. Justify your choice of approximation.
Show worked answer →

Recognise large nn and small pp, justify the Poisson approximation, then apply it.

The exact model is XB(400,0.005)X \sim B(400, 0.005). Since n=400n = 400 is large and p=0.005p = 0.005 is small, the Poisson approximation Po(λ)\operatorname{Po}(\lambda) with λ=np\lambda = np is appropriate (B1 for the justification).

λ=np=400×0.005=2\lambda = np = 400 \times 0.005 = 2 (M1 A1).

P(X=3)=e2233!=e286=43e20.1804P(X = 3) = e^{-2}\frac{2^3}{3!} = e^{-2}\frac{8}{6} = \frac{4}{3}e^{-2} \approx 0.1804 (M1 A1).

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