The Poisson Distribution

How many calls will a help line get in the next hour? How many typos are on a page, or how many cars pass a checkpoint in a minute? Each of these counts events that happen at random over time or space, with no fixed number of trials. The Poisson distribution is the standard model for counts like these, and you only need one number, the average rate, to use it.

Key ideas

When to use a Poisson model

Let $X$ be the number of times an event happens in a fixed interval (of time, length, area, or volume). A Poisson model is appropriate when:

1. events occur independently of each other, and
2. events occur at a uniform average rate during the period of interest (the rate doesn’t change).

If the mean number of events in the interval is $m$, write

$$X \sim \mathrm{Po}(m)$$

Unlike a binomial distribution, there’s no fixed number of trials and no upper limit: $X$ can be $0, 1, 2, 3, \dots$

The probability formula

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

In particular, $P(X = 0) = e^{-m}$. On a GDC, use the Poisson pdf function for $P(X = x)$ and the Poisson cdf function for $P(X \le x)$.

Mean and variance

For $X \sim \mathrm{Po}(m)$,

$$E(X) = m \qquad\text{and}\qquad \mathrm{Var}(X) = m$$

The mean and variance are equal, so the standard deviation is $\sqrt{m}$. This gives a quick check on real data: if the sample mean and variance of a set of counts are close, a Poisson model may fit; if the variance is much bigger or smaller than the mean, it probably won’t. (Proving these results is not required.)

$\mathrm{Po}(1.5)$ is skewed to the right; $\mathrm{Po}(6)$ is wider and more symmetric. Larger $m$ shifts the distribution right and spreads it out.

Changing the interval

Because the rate is uniform, the mean scales with the length of the interval. If calls arrive at $4$ per hour, then in $30$ minutes the number of calls is $\mathrm{Po}(2)$, and in $3$ hours it’s $\mathrm{Po}(12)$. Always find the right $m$ for the interval in the question before calculating.

Sums of independent Poisson variables

If $X \sim \mathrm{Po}(m_1)$ and $Y \sim \mathrm{Po}(m_2)$ are independent, then

$$X + Y \sim \mathrm{Po}(m_1 + m_2)$$

For example, if one road has $3$ cars per minute and another has $2$, the total on both roads is $\mathrm{Po}(5)$ per minute. (This doesn’t work for differences: $X - Y$ can be negative, so it isn’t Poisson.)

Cumulative probabilities

Most questions need more than one value of $x$. Translate the words into $P(X \le \dots)$, which the cdf gives directly:

Words	Probability	Using the cdf
at most $3$	$P(X \le 3)$	$P(X \le 3)$
fewer than $3$	$P(X \le 2)$	$P(X \le 2)$
at least $3$	$P(X \ge 3)$	$1 - P(X \le 2)$
more than $3$	$P(X \ge 4)$	$1 - P(X \le 3)$
between $2$ and $5$ inclusive	$P(2 \le X \le 5)$	$P(X \le 5) - P(X \le 1)$

Many GDCs let you enter a lower and upper bound directly, which avoids these conversions.

Choosing a model

Model	Use it for
Binomial $B(n, p)$	the number of successes in a fixed number $n$ of independent trials, each with the same probability $p$
Poisson $\mathrm{Po}(m)$	the number of events in a fixed interval of time or space, occurring independently at a constant average rate
Normal $N(\mu, \sigma^2)$	a continuous measurement (length, mass, time) that is symmetric and bell-shaped

Worked examples

Example 1: Using the formula

Calls to a help line arrive independently at an average rate of $4$ per hour. Find the probability that in a given hour there are:

• (a) exactly $2$ calls
• (b) no calls

Solution. Let $X$ be the number of calls in an hour, so $X \sim \mathrm{Po}(4)$.

(a)

$$P(X = 2) = \frac{4^2 e^{-4}}{2!} = 8e^{-4} = 0.147 \text{ (3 s.f.)}$$

(b)

$$P(X = 0) = \frac{4^0 e^{-4}}{0!} = e^{-4} = 0.0183 \text{ (3 s.f.)}$$

Example 2: Changing the interval and using the cdf

A newspaper has typos at an average rate of $1.2$ per page, independently. A $5$-page article is checked. Find the probability that it has:

• (a) at most $3$ typos
• (b) at least $3$ typos
• (c) between $2$ and $5$ typos inclusive

Solution. For $5$ pages, $m = 5 \times 1.2 = 6$, so $T \sim \mathrm{Po}(6)$.

(a) $P(T \le 3) = 0.151$ (3 s.f.), from the Poisson cdf.

(b) $P(T \ge 3) = 1 - P(T \le 2) = 0.938$ (3 s.f.)

(c) $P(2 \le T \le 5) = P(T \le 5) - P(T \le 1) = 0.428$ (3 s.f.)

Example 3: Two roads

Cars arrive at a junction from road A at an average rate of $3$ per minute and from road B at $2$ per minute, independently and at random. Find the probability that more than $12$ cars arrive in a $2$-minute period.

Solution. In one minute the total is $\mathrm{Po}(3 + 2) = \mathrm{Po}(5)$, so in $2$ minutes it’s $N \sim \mathrm{Po}(10)$.

$$P(N \gt 12) = 1 - P(N \le 12) = 1 - 0.79155\ldots = 0.208 \text{ (3 s.f.)}$$

Example 4: Does a Poisson model fit?

The number of customers arriving at a coffee kiosk was counted in $50$ randomly chosen $5$-minute intervals.

Customers, $x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
Frequency	$7$	$14$	$14$	$8$	$4$	$2$	$1$

• (a) Find the mean and variance of the data. Explain why a Poisson model may be suitable.
• (b) Using a Poisson model with the same mean, find the expected number of intervals with no customers.

Solution.

(a) From one-variable statistics on the GDC: $\bar{x} = 1.96$ and $\sigma x = 1.4136\ldots$, so the variance is $1.9984 \approx 2.00$. The mean and variance are very close, which is what a Poisson model predicts. It’s also reasonable that customers arrive independently at a fairly constant rate over short periods.

(b) Using $X \sim \mathrm{Po}(1.96)$:

$$P(X = 0) = e^{-1.96} = 0.14086\ldots$$

Expected number of intervals $= 50 \times 0.14086\ldots = 7.04$ (3 s.f.), which is very close to the $7$ observed.

Common mistakes

Forgetting to rescale m. If the rate is $1.2$ per page and the question asks about $5$ pages, use $m = 6$. Using $1.2$ answers the wrong question.

Mixing up “more than” and “at least”. “More than $3$” is $P(X \ge 4) = 1 - P(X \le 3)$. “At least $3$” is $1 - P(X \le 2)$. Write the inequality before using the GDC.

Using pdf when you need cdf. “At most $3$” needs $P(X \le 3)$ (cdf), not $P(X = 3)$ (pdf).

Taking the standard deviation as m. For $\mathrm{Po}(9)$, the variance is $9$ and the standard deviation is $3$.

Using Poisson when there’s a fixed number of trials. “The number of defective phones in a box of $20$” has a maximum of $20$ and a fixed $n$: that’s binomial. Poisson counts events in an interval with no fixed number of trials.

Ignoring the conditions. If the rate changes (rush hour versus midnight) or events come in clusters (people arriving in groups), the Poisson model isn’t appropriate. Say which condition fails.

Practice

1. (Warm-up) $X \sim \mathrm{Po}(3)$. Find $P(X = 4)$, $E(X)$, and $\mathrm{Var}(X)$.

Solution

$$P(X = 4) = \frac{3^4 e^{-3}}{4!} = \frac{81e^{-3}}{24} = 0.168 \text{ (3 s.f.)}$$

$E(X) = 3$ and $\mathrm{Var}(X) = 3$.

2. (Warm-up) Choose the most suitable model (binomial, Poisson, or normal) for each variable.

• (a) The number of heads in $20$ tosses of a coin.
• (b) The number of meteors seen in an hour on a clear night.
• (c) The heights of students at a school.
• (d) The number of patients arriving at an emergency department between 2 a.m. and 3 a.m.

Solution

(a) Binomial: a fixed number ($20$) of independent trials with the same probability.

(b) Poisson: random, independent events in a fixed time interval.

(c) Normal: a continuous, roughly symmetric measurement.

(d) Poisson: arrivals in a fixed time interval, at a roughly constant rate over that hour.

3. (Warm-up) $X \sim \mathrm{Po}(m)$ and $P(X = 0) = 0.2$. Find $m$.

Solution

$$e^{-m} = 0.2 \quad\Rightarrow\quad m = -\ln 0.2 = \ln 5 = 1.61 \text{ (3 s.f.)}$$

4. (Core) A machine breaks down at an average rate of $0.6$ times per week, at random. Find the probability that it breaks down at least once in a $4$-week period.

Solution

In $4$ weeks, $m = 4 \times 0.6 = 2.4$, so $B \sim \mathrm{Po}(2.4)$.

$$P(B \ge 1) = 1 - P(B = 0) = 1 - e^{-2.4} = 0.909 \text{ (3 s.f.)}$$

5. (Core) A bakery sells custom cakes from two shops. Shop A gets orders at an average of $3.2$ per day and Shop B at $1.8$ per day, independently and at random. Find the probability that on a given day the bakery gets at most $4$ orders in total.

Solution

The total is $\mathrm{Po}(3.2 + 1.8) = \mathrm{Po}(5)$.

$$P(\text{total} \le 4) = 0.440 \text{ (3 s.f.)}$$

6. (Core) $X \sim \mathrm{Po}(9)$. Find the standard deviation of $X$, and the probability that $X$ is within one standard deviation of its mean.

Solution

$\mathrm{Var}(X) = 9$, so the standard deviation is $3$. Within one standard deviation means $6 \le X \le 12$:

$$P(6 \le X \le 12) = P(X \le 12) - P(X \le 5) = 0.760 \text{ (3 s.f.)}$$

7. (Core) Accidents at a busy intersection happen at random at an average rate of $0.5$ per day.

• (a) Find the probability that there are no accidents on a given day.
• (b) Find the probability that, in a week of $7$ days, there are exactly $2$ days with no accidents.

Solution

(a) $P(X = 0) = e^{-0.5} = 0.60653\ldots = 0.607$ (3 s.f.)

(b) Each day is independent and has probability $q = 0.60653\ldots$ of no accidents, so the number of accident-free days is $Y \sim B(7, 0.60653\ldots)$.

$$P(Y = 2) = 0.0729 \text{ (3 s.f.)}$$

Notice how the Poisson model gives the probability for one day, and then a binomial model counts the days.

8. (Challenge) $X \sim \mathrm{Po}(m)$ and $P(X = 2) = P(X = 3)$. Find $m$ and $P(X \ge 2)$.

Solution

$$\frac{m^2 e^{-m}}{2!} = \frac{m^3 e^{-m}}{3!} \quad\Rightarrow\quad \frac{m^2}{2} = \frac{m^3}{6} \quad\Rightarrow\quad m = 3$$

(dividing by $m^2 e^{-m}$, which isn’t $0$). Then

$$P(X \ge 2) = 1 - P(X \le 1) = 1 - 4e^{-3} = 0.801 \text{ (3 s.f.)}$$

9. (Challenge) The number of emails a manager receives in a $10$-minute period is $X \sim \mathrm{Po}(m)$. The probability of receiving at most one email in a $10$-minute period is $0.25$.

• (a) Use your GDC to find $m$.
• (b) Given that at least one email arrives in a $10$-minute period, find the probability that at least $3$ arrive.

Solution

(a) $P(X \le 1) = e^{-m} + me^{-m} = (1 + m)e^{-m}$. Solve $(1 + m)e^{-m} = 0.25$ with the GDC (graph both sides and find the intersection, or use the solver):

$$m = 2.6926\ldots = 2.69 \text{ (3 s.f.)}$$

(b) Since “at least $3$” is inside “at least $1$”,

$$P(X \ge 3 \mid X \ge 1) = \frac{P(X \ge 3)}{P(X \ge 1)} = \frac{1 - P(X \le 2)}{1 - e^{-m}} = \frac{0.50456\ldots}{0.93229\ldots} = 0.541 \text{ (3 s.f.)}$$

using the unrounded value of $m$.