This note covers the Poisson distribution: when it is a suitable model, how to calculate probabilities, its mean and variance, how to change the time or space interval, and how to add independent Poisson variables. It is used for counting rare, random events such as emails arriving, faults in cable or radioactive decays.
Key ideas
A Poisson random variable counts how many times an event happens in a fixed interval of time, length, area or volume. We write \(X\sim\text{Po}(\lambda)\), where \(\lambda\gt 0\) is the mean number of events in that interval.
Poisson is a good model when:
- events happen one at a time (not in clusters),
- events happen independently of each other,
- events happen at a constant average rate, so the mean number depends only on the size of the interval.
Important properties:
- \(X\) can take the values \(0,1,2,3,\dots\) with no upper limit.
- The mean and the variance are equal: \(\mathrm{E}(X)=\mathrm{Var}(X)=\lambda\). If a data set has a mean and variance that are very different, Poisson is probably not a good model.
- Changing the interval. If the rate is \(r\) per unit, then in an interval of length \(k\) units the mean is \(\lambda=kr\). Always work out the mean for the interval in the question before calculating.
- Sums. If \(X\sim\text{Po}(\lambda_1)\) and \(Y\sim\text{Po}(\lambda_2)\) are independent, then \(X+Y\sim\text{Po}(\lambda_1+\lambda_2)\).
On the GDC, use the Poisson probability distribution function (pdf) for \(P(X=r)\) and the Poisson cumulative distribution function (cdf) for \(P(X\le r)\). The cdf adds probabilities from 0 up to and including \(r\).
For "more than" and "at least" use complements. Since the possible values have no upper limit, you cannot add them all up directly:
- \(P(X\ge r)=1-P(X\le r-1)\)
- \(P(X\gt r)=1-P(X\le r)\)
Checking whether Poisson fits. Suppose a shop records the number of customers entering each minute over many minutes and finds a mean of 5.1 and a variance of 5.0. These are close, which supports a Poisson model. If the variance were much larger than the mean, events would be bunching together and Poisson would be a poor choice. Also think about the context. Goals in a football match or breakdowns of a machine can often be modelled this way, but the number of people in a lift cannot, because the events are not independent and there is a fixed upper limit.
Shape. For a small \(\lambda\) the distribution is strongly skewed to the right, (when \(\lambda\lt 1\) the most likely value is 0). As \(\lambda\) increases, the distribution becomes more symmetrical and its peak moves to the right. The probabilities for all values \(0,1,2,\dots\) sum to 1.
Reading wording. "Fewer than 3" means \(X\le 2\). "At most 3" means \(X\le 3\). "More than 3" means \(X\ge 4\). "At least 3" means \(X\ge 3\). Write the inequality in symbols before using the GDC. "Between 2 and 5 inclusive" means \(P(X\le 5)-P(X\le 1)\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(X\sim\text{Po}(\lambda)\) | \(X\) is the number of events in an interval with mean \(\lambda\). |
| \(P(X=r)=\dfrac{e^{-\lambda}\lambda^{r}}{r!}\) | Probability of exactly \(r\) events, \(r=0,1,2,\dots\) |
| \(\mathrm{E}(X)=\lambda\) | Mean number of events. |
| \(\mathrm{Var}(X)=\lambda\) | Variance equals the mean, so the standard deviation is \(\sqrt{\lambda}\). |
| \(\lambda=k\times\text{rate}\) | Scaling the mean to an interval \(k\) times as long. |
| \(X+Y\sim\text{Po}(\lambda_1+\lambda_2)\) | Sum of independent Poisson variables. |
| \(P(X\ge r)=1-P(X\le r-1)\) | Complement rule for "at least". |
Worked examples
Example 1 A help desk receives calls at a mean rate of 3.2 per hour. Assume the number of calls in an hour is Poisson. Find (a) the probability of exactly 5 calls in an hour, (b) the probability of at least 2 calls in an hour.
Let \(X\sim\text{Po}(3.2)\).
(a) \(P(X=5)=\frac{e^{-3.2}\times 3.2^{5}}{5!}=\frac{0.040762\times 335.54}{120}\approx 0.114\). A Poisson pdf on the GDC with \(\lambda=3.2\) and \(r=5\) gives the same value.
(b) \(P(X\ge 2)=1-P(X\le 1)=1-\big(P(X=0)+P(X=1)\big)\).
\(P(X=0)=e^{-3.2}\approx 0.04076\) and \(P(X=1)=3.2e^{-3.2}\approx 0.13044\).
\(P(X\ge 2)=1-0.04076-0.13044\approx 0.829\).
Answer: (a) 0.114 (b) 0.829
Example 2 Using the same help desk (3.2 calls per hour), find the probability that in a two-hour period there are at most 4 calls, and the probability that there are no calls.
The interval is twice as long, so the mean doubles: \(\lambda=2\times 3.2=6.4\). Let \(Y\sim\text{Po}(6.4)\).
At most 4 calls: use the Poisson cdf with \(\lambda=6.4\) and upper value 4. This gives \(P(Y\le 4)\approx 0.235\).
No calls: \(P(Y=0)=e^{-6.4}\approx 0.00166\).
Answer: \(P(Y\le 4)\approx 0.235\), \(P(Y=0)\approx 0.00166\).
Example 3 Two independent machines make surface flaws on a roll of fabric. Machine P makes flaws at a mean of 2.5 per roll and machine Q at a mean of 1.8 per roll, each Poisson. Find the probability that exactly 3 flaws are made by the two machines on one roll, and state the variance of the total.
The total \(T=X+Y\) is the sum of independent Poisson variables, so \(T\sim\text{Po}(2.5+1.8)=\text{Po}(4.3)\).
\(P(T=3)=\frac{e^{-4.3}\times 4.3^{3}}{3!}=\frac{0.013569\times 79.507}{6}\approx 0.180\).
The variance equals the mean, so \(\mathrm{Var}(T)=4.3\).
Answer: \(P(T=3)\approx 0.180\); variance 4.3.
Common errors
- Forgetting to rescale \(\lambda\) when the question changes the interval, such as moving from one hour to two hours or one metre to three metres.
- Using the cdf when the question asks for an exact value, or using the pdf when it asks for "at most". The cdf includes the upper value.
- Writing \(P(X\ge 3)=1-P(X\le 3)\). The correct statement is \(1-P(X\le 2)\).
- Adding the variances incorrectly for a sum. For independent Poisson variables the means add, and the variance of the sum is the new mean.
- Applying Poisson when events are not independent or the rate is not constant, such as arrivals that clearly cluster at rush hour. State the assumptions.
Exam tip: Write the distribution with its new mean first, for example \(X\sim\text{Po}(6.4)\), then the probability statement such as \(P(X\le 4)\), then the value. This shows the interval was rescaled.
Check yourself
- \(X\sim\text{Po}(1.5)\). Find \(P(X=2)\).
Answer
\(P(X=2)=\frac{e^{-1.5}\times 1.5^2}{2!}=\frac{0.22313\times 2.25}{2}\approx 0.251\). - \(X\sim\text{Po}(4)\). Find \(P(X\gt 5)\).
Answer
\(P(X\gt 5)=1-P(X\le 5)=1-0.7851\approx 0.215\). - Cars pass a sensor at a mean of 4.5 per 10 minutes. Find the probability that exactly 10 cars pass in 25 minutes.
Answer
The mean for 25 minutes is \(4.5\times 2.5=11.25\). \(P(X=10)=\frac{e^{-11.25}\times 11.25^{10}}{10!}\approx 0.116\). - A cable has flaws at a mean of 0.8 per metre. Find the probability that a 3 m length has no flaws.
Answer
\(\lambda=0.8\times 3=2.4\). \(P(X=0)=e^{-2.4}\approx 0.0907\).