Poisson model and probabilitiesAQA A-Level Further Maths: Revision notes

A Poisson distribution models the number of times an event occurs in a fixed interval of time, length, area or volume. We write $X\sim\mathrm{Po}(\lambda)$, where the parameter $\lambda>0$ is the average number of occurrences in the interval.

The model is valid when events:

• occur singly (one at a time) and at random,
• occur independently of one another,
• occur at a constant average rate (so the mean is proportional to the length of the interval).

Examples: faults in a roll of fabric, calls to a call centre, flaws per square metre of glass. There is no upper limit on $X$: it can take any value $0,1,2,3,\dots$

If $X\sim\mathrm{Po}(\lambda)$ then $\mathrm{P}(X=r)=\frac{e^{-\lambda}\lambda^{r}}{r!},\qquad r=0,1,2,\dots$ Example: $X\sim\mathrm{Po}(2.5)$. Then $\mathrm{P}(X=2)=\frac{e^{-2.5}\times2.5^2}{2!}=0.0821\times3.125=0.2565$.

On a calculator use the Poisson probability function for $\mathrm{P}(X=r)$ and the cumulative function for $\mathrm{P}(X\le r)$. Check your calculator output against the formula for one value.

Because $X$ has no upper limit, 'at least' probabilities use the complement. $\mathrm{P}(X\le r)=\sum_{k=0}^{r}\mathrm{P}(X=k),\qquad \mathrm{P}(X\ge r)=1-\mathrm{P}(X\le r-1).$ Example: $X\sim\mathrm{Po}(1.8)$. $\mathrm{P}(X\ge3)=1-\mathrm{P}(X\le2)=1-e^{-1.8}\left(1+1.8+\frac{1.8^2}{2}\right)=0.269$.

Also $\mathrm{P}(X>r)=\mathrm{P}(X\ge r+1)$ and $\mathrm{P}(a\le X\le b)=\mathrm{P}(X\le b)-\mathrm{P}(X\le a-1)$.

If $X\sim\mathrm{Po}(\lambda)$ then $\mathrm{E}(X)=\lambda,\qquad \mathrm{Var}(X)=\lambda,\qquad \text{standard deviation}=\sqrt{\lambda}.$ The mean and variance are equal. This is a useful check: if data have a sample mean far from the sample variance, a Poisson model is probably unsuitable.

Example: $Y\sim\mathrm{Po}(6)$ has $\mathrm{Var}(Y)=6$ and standard deviation $\sqrt6=2.45$.

Because the rate is constant, $\lambda$ scales with the size of the interval. If faults occur at 2.5 per metre, a 2 m length has $\lambda=5$ and a 10 cm length has $\lambda=0.25$.

Example: $\mathrm{P}(\text{no faults in 2 m})=e^{-5}=0.00674$.

Counts in separate, independent intervals can be combined with the binomial distribution. If a page has no errors with probability $p=e^{-1.8}=0.1653$, the number of error-free pages out of 5 is $\mathrm{B}(5,0.1653)$.

Sometimes $\lambda$ must be found from given information. If $\mathrm{P}(X=0)=0.2$ then $e^{-\lambda}=0.2$, so $\lambda=-\ln0.2=\ln5=1.61$.

Once $\lambda$ is known, probabilities, the mean $\lambda$ and the standard deviation $\sqrt\lambda=1.27$ follow. 'Within one standard deviation of the mean' means $\lambda-\sqrt\lambda<X<\lambda+\sqrt\lambda$; here $0.34<X<2.88$, so $X=1$ or $2$ because $X$ is a whole number.