The Poisson Distribution and Its Uses

The Poisson distribution models how many times an event happens in a fixed interval of time or space, such as calls to a helpline per hour or typing errors per page. It is one of the most useful distributions in A Level and university statistics because it needs only a single number, the average rate, to describe it. This guide explains the formula, its conditions and its main uses, with worked examples.

The Poisson Formula

A random variable \( X \) follows a Poisson distribution with parameter \( \lambda \), written \( X \sim \text{Po}(\lambda) \), if

$$ P(X = r) = \frac{e^{-\lambda} \lambda^r}{r!}, \qquad r = 0, 1, 2, \ldots $$

Here \( \lambda \) is the mean number of events in the interval, and \( e \approx 2.71828 \).

Mean and Variance

A remarkable property of the Poisson distribution is that

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

So if a dataset's mean and variance are roughly equal, a Poisson model is worth considering.

When Can You Use It?

A Poisson model is suitable when events:

1. Occur singly, meaning two events cannot happen at exactly the same instant.
2. Occur independently of one another.
3. Occur at a constant average rate across the interval.
4. Are counted over a fixed interval of time, length, area or volume.

Typical Uses

• Number of customers arriving at a bank counter per 10 minutes.
• Number of defects per metre of cloth.
• Number of accidents at a junction per month.
• Number of radioactive decays per second.
• Rare events in a large population (see the binomial approximation below).

Scaling the Rate

If the interval changes, scale \( \lambda \) in proportion. If calls arrive at a rate of 3 per hour, then over 2 hours \( \lambda = 6 \), and over 20 minutes \( \lambda = 1 \).

Worked Example 1: Calls to a Helpline

A helpline receives an average of 3 calls per hour. Find the probability that in a given hour it receives (a) exactly 2 calls, (b) at least 1 call.

Let \( X \sim \text{Po}(3) \).

(a) Exactly 2 calls.

$$ P(X = 2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{9}{2} e^{-3} = 4.5 e^{-3} $$

Using \( e^{-3} \approx 0.049787 \):

$$ P(X = 2) \approx 4.5 \times 0.049787 \approx 0.2240 $$

(b) At least 1 call. Use the complement, since "at least 1" includes infinitely many values:

$$ P(X \geq 1) = 1 - P(X = 0) = 1 - \frac{e^{-3} \cdot 3^0}{0!} = 1 - e^{-3} $$

$$ P(X \geq 1) \approx 1 - 0.049787 = 0.9502 $$

Worked Example 2: Typing Errors

A typist makes on average 0.4 errors per page. Find the probability that a 5-page document contains at most 1 error.

Step 1: Scale the rate. Over 5 pages,

$$ \lambda = 0.4 \times 5 = 2 $$

So \( X \sim \text{Po}(2) \).

Step 2: Write out "at most 1".

$$ P(X \leq 1) = P(X = 0) + P(X = 1) $$

Step 3: Calculate each term.

$$ P(X = 0) = \frac{e^{-2} \cdot 2^0}{0!} = e^{-2} $$

$$ P(X = 1) = \frac{e^{-2} \cdot 2^1}{1!} = 2e^{-2} $$

Step 4: Add.

$$ P(X \leq 1) = e^{-2} + 2e^{-2} = 3e^{-2} \approx 3 \times 0.135335 \approx 0.4060 $$

So there is roughly a 40.6% chance the document has no more than one error.

The Poisson Approximation to the Binomial

When \( X \sim B(n, p) \) with \( n \) large and \( p \) small, calculating binomial probabilities is tedious. A common rule of thumb: if \( n > 50 \) and \( np < 5 \), use

$$ X \approx \text{Po}(np) $$

The idea is that many trials, each with a tiny chance of "success", behave like a stream of rare, independent events.

Worked Example 3: Faulty Components

A machine produces components, each with probability 0.004 of being faulty. In a batch of 500, find the probability that none are faulty.

Step 1: Check conditions. \( n = 500 \) is large and \( p = 0.004 \) is small. \( np = 500 \times 0.004 = 2 < 5 \).

Step 2: Approximate. \( X \approx \text{Po}(2) \).

$$ P(X = 0) \approx e^{-2} \approx 0.1353 $$

Step 3: Compare with the exact answer. The exact binomial value is

$$ P(X = 0) = (0.996)^{500} \approx 0.1348 $$

The approximation is accurate to about three decimal places, with far less work.

Common Mistakes

• Forgetting to scale \( \lambda \). Always match \( \lambda \) to the interval in the question.
• Mishandling "at least" and "more than". \( P(X > 2) = 1 - P(X \leq 2) \), which includes \( r = 0, 1, 2 \) in the subtraction.
• Forgetting \( 0! = 1 \). This makes \( P(X = 0) = e^{-\lambda} \).
• Using Poisson when the rate is not constant. Rush-hour traffic and quiet night traffic should not share one \( \lambda \).
• Applying the binomial approximation when \( p \) is large. If \( np \) is large, a normal approximation is usually better.

Key Takeaways

• \( P(X = r) = e^{-\lambda}\lambda^r / r! \) models counts of independent events at a constant average rate.
• The mean and variance are both equal to \( \lambda \).
• Scale \( \lambda \) whenever the length of the interval changes.
• Use complements for "at least" questions.
• For large \( n \) and small \( p \), \( B(n, p) \) is well approximated by \( \text{Po}(np) \).

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