Statistics From Zero · Lesson 24 of 25
The Poisson Distribution: How Many Customers in the Next Hour?
In one sentence: When events happen independently at a steady average rate, the Poisson distribution tells you how likely each count is in a fixed stretch of time or space, and how much random clumping to expect.
The idea
The binomial distribution (Lesson 20) counts successes out of a fixed number of tries. But many counts have no "tries" at all: customers walking up in an hour, emails arriving in a day, goals in a football match, typos on a page. Things just happen, now and then, at some average rate. The distribution for counts like these is the Poisson distribution, named after a French mathematician and pronounced roughly "pwa-SON".
It needs only one number: the average count, usually written λ (lambda).
The lemonade stand
Cousin Kevin runs a lemonade stand on Saturday afternoons. Over the summer he has averaged 4 customers an hour.
One Saturday, between 2 and 3 o'clock, he gets 9. He is now planning to hire his younger sister, open a second stand and possibly franchise. The following hour he gets 2 and blames the second stand, which doesn't exist yet.
The chances for each count
If customers arrive independently at an average of 4 an hour, the Poisson distribution gives:
| Customers in an hour | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 or more |
|---|---|---|---|---|---|---|---|---|---|
| Chance | 1.8% | 7.3% | 14.7% | 19.5% | 19.5% | 15.6% | 10.4% | 6.0% | 5.1% |
The single most likely count is 4, but only about one hour in five has exactly 4. Hours with 2 or 6 are completely normal. Nine or more happens about 2.1% of the time: unusual, roughly one hour in fifty, but over a summer of Saturday afternoons Kevin should expect a few such hours even if nothing has changed. One busy hour is not a business plan.
For the curious, the formula behind the table: the chance of exactly k events is λᵏ × e^-λ ÷ k!. You'll rarely calculate it by hand; any calculator or spreadsheet does it.
The neat trick: average and spread are the same number
For a Poisson count, the variance equals the average. With λ = 4, the variance is 4 and the standard deviation is √4 = 2. So a typical hour is 4 customers, give or take 2. That's a quick check on real data: if the spread is much bigger than the average suggests, something other than independent arrivals is going on.
Randomness clumps
People expect random events to be spread out evenly. They aren't. With an average of 4 an hour, quiet hours and rushes are both built in. The same goes for days with three bad events in a row, or a run of goals in five minutes: independent events at a steady rate produce clusters by themselves. Before looking for a reason behind a cluster, ask whether a Poisson count would produce it anyway.
Where you meet it
- Queues and staffing: calls to a help line, customers at a till, cars at a junction.
- Sport: goals in a football match are often modelled as Poisson for each team, which is why a match with 7 goals is rare but not suspicious.
- Rare events over many chances: accidents, defects, typos. The Poisson is what the binomial becomes when there are many tries and a small chance each time. 1,000 passers-by, each with a 0.4% chance of buying, give a binomial chance of exactly 4 sales of 19.58%; the Poisson with λ = 4 gives 19.54%.
Where it fails
- Events that come in groups. A family of five buying five lemonades isn't five independent customers. The counts get lumpier than Poisson: the spread is bigger than the average.
- A rate that changes. A hot afternoon and a rainy one don't share one λ. Mix them and you get more extreme hours than a single Poisson predicts. Model each kind of hour separately.
- Events that affect each other. A queue that puts people off, or one goal that changes how both teams play, breaks the independence.
Try it yourself
Aunt Linda gets on average 1.5 nuisance phone calls a day, arriving independently.
1. How often does she get a day with no nuisance calls at all? 2. How often does she get 3 or more in one day? 3. She says that on days with 3 calls "they must be targeting me". What would you tell her?
Show the answer
1. e^-1.5 = about 22% of days, so roughly one or two quiet days a week.
2. The chances of 0, 1 and 2 calls are about 22.3%, 33.5% and 25.1%, which add up to 80.9%. So 3 or more happens on about 19% of days: around 1.3 days every week.
3. A 3-call day is exactly what a steady average of 1.5 produces about once a week. It feels like targeting because we notice the clumps and forget the quiet days. If 3-call days started happening every other day, the rate itself would have gone up, and that would be worth looking into.
The words your lecturer uses
| Lecturer says | It means |
|---|---|
| Poisson distribution, Po(λ) | The chances of each count of independent events in a fixed interval, given the average rate λ |
| Rate parameter λ (lambda) | The average number of events per interval (per hour, per page, per match) |
| Poisson process | Events arriving independently, one at a time, at a constant average rate |
| Mean = variance | The defining property: the spread of the count matches its average |
| Overdispersion | More spread than Poisson predicts, a sign of grouping or a changing rate |
| Poisson approximation to the binomial | Using Poisson with λ = n × p when n is large and p is small |
Three things to remember
- Counts of independent events at a steady rate follow a Poisson distribution, set by one number: the average.
- Its variance equals its average, so the typical swing is the square root of the average.
- Random events clump. A busy hour or a bad day is not, on its own, evidence that anything changed.
Part of Statistics From Zero, a free beginner course by WrongStat. Not a substitute for your lecturer, but hopefully a good translator.