Poisson Distribution with Examples in Statistics

 

The Poisson distribution is a discrete probability distribution that describes the number of events occurring in a fixed interval of time or space, assuming these events happen independently at a constant average rate $\lambda$ . It is widely used in various fields such as queuing theory, telecommunications, biology, and epidemiology to model rare events or phenomena where the occurrence of events is random and infrequent. Here, we'll explore the Poisson distribution in detail, including its probability mass function, mean, variance, examples, and numerical illustrations.

Definition and Notation:

For a random variable $X$ following a Poisson distribution $X \sim \text{Poisson}(\lambda)$ :

• $\lambda$ (lambda) is the average rate or mean number of events occurring in the given interval.
• The outcomes $X$ can take are $0, 1, 2, \ldots$ .

Probability Mass Function (PMF):

The probability mass function of $X$ is given by:

$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$

Where:

• $k$ is a non-negative integer representing the number of events.
• $e$ is the base of the natural logarithm (approximately 2.71828).
• $\lambda$ is the average number of events occurring in the interval.

Mean and Variance:

The mean (expected value) and variance of $X$ are both equal to $\lambda$ :

$E(X) = \lambda$

$\text{Var}(X) = \lambda$

These properties indicate that the Poisson distribution is characterized by its single parameter $\lambda$ , which determines both the average number of events and the spread or variability around this average.

Examples:

Example 1: Number of Calls Received

In a call center, calls arrive at an average rate of 2 calls per minute.

• Random Variable $X$ : Number of calls received in a minute.

• Distribution: $X \sim \text{Poisson}(2)$ .

• Probability of Specific Outcomes: $P(X = 3) = \frac{2^3 e^{-2}}{3!} = \frac{8 e^{-2}}{6} \approx 0.1804$

• Mean and Variance: $E(X) = 2$ $\text{Var}(X) = 2$

Example 2: Number of Emails Received

An employee receives emails at an average rate of 5 emails per hour.

• Random Variable $X$ : Number of emails received in an hour.

• Distribution: $X \sim \text{Poisson}(5)$ .

• Calculate Probabilities: $P(X = 4) = \frac{5^4 e^{-5}}{4!} = \frac{625 e^{-5}}{24} \approx 0.1755$

• Mean and Variance: $E(X) = 5$ $\text{Var}(X) = 5$

Numerical Illustration:

Let's calculate $P(X = 2)$ for $X \sim \text{Poisson}(3)$ :

$P(X = 2) = \frac{3^2 e^{-3}}{2!} = \frac{9 e^{-3}}{2} \approx 0.224$

Thus, the probability of observing exactly 2 events in a Poisson process with an average rate of 3 events is approximately $0.224$ .

Conclusion:

The Poisson distribution is a powerful tool for modeling the occurrence of rare events or phenomena where the events happen independently at a constant average rate $\lambda$ . Its simplicity and applicability make it suitable for various real-world applications, particularly in situations involving counts or occurrences of events over time or space intervals. Understanding the Poisson distribution, including its probability mass function, mean, variance, and practical examples, enables statisticians, analysts, and researchers to effectively model and analyze data in fields such as insurance risk assessment, traffic flow analysis, and public health epidemiology.