The Poisson Distribution (AQA A-Level Further Maths): Revision Notes

Worked Example 1: Radioactive Disintegrations

Problem: A Geiger counter detects radioactive disintegrations at a mean rate of 23 per minute. Find the probability that in a randomly chosen minute there will be exactly 20 disintegrations.

Solution:

Let $X$ be the number of disintegrations in a randomly chosen minute.

Step 1: Check conditions. Radioactive disintegrations occur randomly and independently, so we can use a Poisson distribution with $\lambda = 23$.

Therefore: $X \sim \text{Po}(23)$

Step 2: We need to find $P(X = 20)$.

Using the probability distribution function:

$P(X = 20) = \frac{e^{-23} \times 23^{20}}{20!} = 0.072 \text{ (3 dp)}$

Answer: The probability is 0.072 or 7.2%

Worked Example 2: Between 21 and 23 Disintegrations

Problem: Using the same scenario (mean rate = 23 per minute), find the probability that there will be between 21 and 23 inclusive disintegrations.

Solution:

We need: $P(21 \leq X \leq 23) = P(X = 21) + P(X = 22) + P(X = 23)$

Substitute $x = 21$, $x = 22$, and $x = 23$ into the formula $P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}$ with $\lambda = 23$:

$P(21 \leq X \leq 23) = \frac{e^{-23} \times 23^{21}}{21!} + \frac{e^{-23} \times 23^{22}}{22!} + \frac{e^{-23} \times 23^{23}}{23!}$

$= 0.079 + 0.083 + 0.083 = 0.245 \text{ (3 dp)}$

Answer: The probability is 0.245 or 24.5%

Calculator tip: Always verify your answers using your calculator's Poisson function to avoid arithmetic errors.

Worked Example 3: Completing a Probability Table

Problem: A random variable $X$ has a Poisson distribution with parameter $\lambda = 2$. Complete the following probability table:

Solution:

Since $X \sim \text{Po}(2)$, we use $P(X = x) = \frac{e^{-2} \times 2^x}{x!}$ to find the missing probabilities:

For $x = 2$: $P(X = 2) = \frac{e^{-2} \times 2^2}{2!} = \frac{e^{-2} \times 4}{2} = 0.271$

For $x = 3$: $P(X = 3) = \frac{e^{-2} \times 2^3}{3!} = \frac{e^{-2} \times 8}{6} = 0.180$

For $x = 4$: $P(X = 4) = \frac{e^{-2} \times 2^4}{4!} = \frac{e^{-2} \times 16}{24} = 0.090$

For $x = 5$: $P(X = 5) = \frac{e^{-2} \times 2^5}{5!} = \frac{e^{-2} \times 32}{120} = 0.036$

The completed table shows:

Worked Example 4: Probability Within One Standard Deviation

Problem: For a random variable $X$ with Poisson distribution $\lambda = 2$:

• Write down the mean and variance
• Find the probability that a randomly chosen value will be within 1 standard deviation of the mean

Solution:

Part a: Since $X \sim \text{Po}(2)$:

• Mean $= 2$
• Variance $= 2$

Part b: The standard deviation is the square root of the variance:

$\text{Standard deviation} = \sqrt{2} = 1.414 \text{ (3 dp)}$

We need: $P(\mu - \sigma < X < \mu + \sigma) = P(2 - 1.414 < X < 2 + 1.414)$

$= P(0.586 < X < 3.414)$

Since $X$ is discrete, this is equivalent to:

$P(X = 1 \text{ or } X = 2 \text{ or } X = 3)$

From the table in Example 3:

$= 0.271 + 0.271 + 0.180 = 0.722$

Answer: The probability is 0.722 or 72.2%

Exam tip: Remember that for discrete distributions, you must work with integer values only. The inequality $0.586 < X < 3.414$ translates to $X \in \{1, 2, 3\}$.

Worked Example 5: Deaths from Horse Kicks

Problem: The table shows data about deaths from horse kicks in 10 units of the Prussian Army over 20 years (200 observations in total). Decide whether this data is likely to follow a Poisson distribution.

Solution:

First, calculate the mean from the frequency table:

$\text{Mean} = \frac{0(109) + 1(65) + 2(22) + 3(3) + 4(1)}{200} = \frac{0 + 65 + 44 + 9 + 4}{200} = \frac{122}{200} = 0.610$

Next, calculate the variance:

$\text{Variance} = \frac{0^2(109) + 1^2(65) + 2^2(22) + 3^2(3) + 4^2(1)}{200} - (0.610)^2$

$= \frac{0 + 65 + 88 + 27 + 16}{200} - 0.372 = 0.980 - 0.372 = 0.608$

Conclusion: Mean $\approx$ variance (both approximately 0.61), so the data is likely to be Poisson distributed.

Calculator tip: Use your calculator's statistical functions to find these values quickly and accurately.

Worked Example 6: Events in Multiple Time Intervals

Problem: Events occur randomly at a mean rate of $\mu$ per hour. Find the probability that exactly 3 events occur in a randomly chosen two-hour period.

Solution:

Let $X_1$ be the number of events in the first hour, and $X_2$ be the number in the second hour.

Then: $X_1 \sim \text{Po}(\mu)$ and $X_2 \sim \text{Po}(\mu)$

The total number of events in 2 hours is: $X_T = X_1 + X_2$

By the addition property: $X_T \sim \text{Po}(2\mu)$

We need to find $P(X_T = 3)$.

There are multiple ways this could happen:

• 3 events in first hour, 0 in second: $P(X_1 = 3) \cdot P(X_2 = 0)$
• 2 events in first hour, 1 in second: $P(X_1 = 2) \cdot P(X_2 = 1)$
• 1 event in first hour, 2 in second: $P(X_1 = 1) \cdot P(X_2 = 2)$
• 0 events in first hour, 3 in second: $P(X_1 = 0) \cdot P(X_2 = 3)$

However, it's much simpler to use the addition property that $X_T \sim \text{Po}(2\mu)$:

$P(X_T = 3) = \frac{e^{-2\mu} (2\mu)^3}{3!}$

This is the probability distribution function of a Poisson distribution with mean $2\mu$.

Worked Example 7: Cars Passing a Checkpoint

Problem: Cars pass a traffic checkpoint at a rate of 2 per hour. Find the probability that in a two-hour period, more than 2 cars pass, assuming the conditions for a Poisson distribution are met.

Solution:

Step 1: Check conditions. A Poisson distribution requires random arrivals with free-flowing traffic (no queues).

Step 2: Let $X$ be the number of cars passing in 2 hours.

A mean of 2 per hour is equivalent to $\lambda = 4$ per 2 hours.

Therefore: $X \sim \text{Po}(4)$

Step 3: We need $P(X > 2)$.

Using the complement rule:

$P(X > 2) = 1 - P(X \leq 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]$

$= 1 - \left[\frac{e^{-4} \times 4^0}{0!} + \frac{e^{-4} \times 4^1}{1!} + \frac{e^{-4} \times 4^2}{2!}\right]$

$= 1 - [e^{-4} + 4e^{-4} + 8e^{-4}]$

$= 1 - 13e^{-4} = 0.762$

Answer: The probability is 0.762 or 76.2%

Exam tip: When finding $P(X > k)$, it's often easier to calculate $1 - P(X \leq k)$, especially for small values of $k$.

Worked Example 8: Two Independent Time Periods

Problem: The number of cars passing a point on a quiet rural road between 7am and 8am is given by random variable $X$. The same statistic for the interval 8am to 9am is given by random variable $Y$.

Given that $X \sim \text{Po}(21)$ and $Y \sim \text{Po}(34)$, find the probability that the total number of cars between 7am and 9am is between 50 and 60 inclusive.

Solution:

Part a: On a quiet rural road, cars are likely to arrive randomly and independently with no queuing, so the Poisson distribution is appropriate.

Part b: Let $T$ be the total number of cars passing between 7am and 9am.

Then: $T = X + Y$

Since $X$ and $Y$ are independent Poisson variables:

$T \sim \text{Po}(21 + 34) = \text{Po}(55)$

We need: $P(50 \leq T \leq 60)$

$= P(T \leq 60) - P(T < 50)$

$= P(T \leq 60) - P(T \leq 49)$

Using a calculator or tables:

$= 0.848 - 0.322 = 0.53 \text{ (2 dp)}$

Answer: The probability is 0.53 or 53%