The Poisson distribution

Counts with no upper limit

Some counts have no fixed number of trials behind them: customer arrivals in an hour, disputed invoices in a week, typos on a page. The Poisson distribution models the count of events in a fixed interval when events occur singly, at a constant average rate, independently of one another. Its one parameter is the average count per interval, $\lambda$:

$$P(X=x)=\frac{e^{-\lambda }{\lambda }^x}{x!},x=0,1,2,\dots$$

Write $X\sim \mathrm{Poisson}(\lambda )$. The mean and variance are the same number:

$$\mu =\lambda ,{\sigma }^2=\lambda ,\sigma =\sqrt{\lambda }$$

There is no $n$ and no upper bound on $x$ — the distinction from the binomial that Lesson 6.7 turns into a rule.

Worked example — arrivals at Cedar Valley's counter

Cedar Valley Credit Union's branch counter sees an average of 3.2 customers per hour on a weekday morning. Let $X$ = the number of arrivals in a given hour, $X\sim \mathrm{Poisson}(3.2)$.

Exactly two arrivals:

$$P(X=2)=\frac{e^{-3.2}3.2^2}{2!}=\frac{0.04076\times 10.24}{2}=0.2087$$

At most one:

$$P(X\le 1)=e^{-3.2}\left(1+3.2\right)=0.04076\times 4.2=0.1712$$

The $e^{-\lambda }$ factor is common to every term; compute it once and reuse it. On a non-programmable calculator: $e^x$ key, then the powers and factorials by hand.

Rescaling the rate

The question rarely asks about the interval the rate was quoted for. Probability of no arrivals in a 15-minute window? The rate per hour is 3.2, so per quarter-hour it is

$${\lambda }_{15}=3.2\times \frac{15}{60}=0.8$$

and $P(X=0)=e^{-0.8}=0.4493$. The rate scales proportionally with the interval — a two-hour window has $\lambda =6.4$, a day of seven counter hours $\lambda =22.4$. Rescale first, then compute; using 3.2 in a 15-minute question is the error the marker sees most.

The scaffold, again

Define $X$ in words with its interval ("$X$ = arrivals in a 15-minute window"), state the distribution with the rescaled rate ($X\sim \mathrm{Poisson}(0.8)$), translate the phrase, compute. "At least one arrival in 15 minutes" is $1-P(X=0)=0.5507$ — the complement, as always.

A note on symbols

IBS1 writes the Poisson parameter as $\mu$, and its glossary adds "$\mu =np$". The course writes $\lambda$ for the rate, and $\lambda =np$ appears only in one place — the approximation to the binomial in Lesson 6.7, where a binomial's $np$ is borrowed as a Poisson rate. Everywhere else, $\lambda$ is the average count in the interval, full stop.