Understanding the Poisson Distribution
The Poisson Distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space. These events must occur with a known constant mean rate and independently of the time since the last event.
Conditions for a Poisson Experiment
- Events are independent of each other. The occurrence of one event does not affect the probability of another event occurring.
- The average rate of occurrence (λ, lambda) is constant for the interval.
- Two events cannot occur at the exact same instant.
- The probability of an event in a small sub-interval is proportional to the length of that sub-interval.
The Poisson Formula
The probability of observing exactly \(k\) events in an interval is given by the Probability Mass Function (PMF):
$$P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$$
Where:
- \(k\) is the number of occurrences (k = 0, 1, 2, ...).
- \(\lambda\) (lambda) is the average number of events per interval.
- \(e\) is Euler's number (approximately 2.71828).
A key property of the Poisson distribution is that its mean and variance are equal:
Mean / Expected Value: \(E[X] = \lambda\)
Variance: \(Var[X] = \lambda\)
Examples with Step-by-Step Answers
Example 1: Call Center
Problem: A call center receives an average of 4 calls per minute. What is the probability of receiving exactly 2 calls in a given minute?
1. Identify the parameter:
- The average rate, \(\lambda = 4\) calls per minute.
- The number of occurrences we are interested in, \(k = 2\).
2. Apply the formula:
$$P(X=2) = \frac{4^2 e^{-4}}{2!}$$
3. Calculate the parts: \(4^2 = 16\), \(e^{-4} \approx 0.0183\), \(2! = 2\).
4. Combine the results: \(P(X=2) = \frac{16 \times 0.0183}{2} = \frac{0.2928}{2} \approx 0.1464\)
Answer: There is approximately a 14.64% chance of receiving exactly 2 calls in a minute.
Example 2: Website Errors
Problem: A website experiences an average of 1.5 server errors per day. What is the probability of experiencing at most 1 error tomorrow?
1. Identify the parameter:
- The average rate, \(\lambda = 1.5\) errors per day.
- "At most 1" means we need to find \(P(X \le 1)\), which is \(P(X=0) + P(X=1)\).
2. Calculate the probability for each k:
\(P(X=0) = \frac{1.5^0 e^{-1.5}}{0!} = \frac{1 \times e^{-1.5}}{1} \approx 0.2231\)
\(P(X=1) = \frac{1.5^1 e^{-1.5}}{1!} = \frac{1.5 \times e^{-1.5}}{1} \approx 0.3347\)
3. Sum the probabilities: \(P(X \le 1) = 0.2231 + 0.3347 = 0.5578\)
Answer: The probability of experiencing 0 or 1 server error tomorrow is about 55.78%.
How to Do This in Statistical Software
The default example used in this calculator (λ = 5, point a = 3, point b = 7) can be reproduced directly in the following tools.
RStudio
R's built-in dpois() and ppois() functions give the PMF and CDF directly:
lambda <- 5
a <- 3
b <- 7
P_X_eq_a <- dpois(a, lambda)
P_X_lt_a <- ppois(a - 1, lambda)
P_X_gt_b <- 1 - ppois(b, lambda)
P_a_le_X_le_b <- ppois(b, lambda) - ppois(a - 1, lambda)
c(P_X_eq_a, P_X_lt_a, P_X_gt_b, P_a_le_X_le_b)
Python
scipy.stats.poisson provides matching pmf and cdf methods:
from scipy.stats import poisson
lam, a, b = 5, 3, 7
p_exact_a = poisson.pmf(a, lam)
p_less_a = poisson.cdf(a - 1, lam)
p_greater_b = 1 - poisson.cdf(b, lam)
p_between_ab = poisson.cdf(b, lam) - poisson.cdf(a - 1, lam)
Excel / Google Sheets
=POISSON.DIST(3, 5, FALSE) // P(X = a)
=POISSON.DIST(2, 5, TRUE) // P(X < a)
=1 - POISSON.DIST(7, 5, TRUE) // P(X > b)
=POISSON.DIST(7, 5, TRUE) - POISSON.DIST(2, 5, TRUE) // P(a ≤ X ≤ b)
SPSS
Via Transform → Compute Variable, using the PDF/CDF function groups:
COMPUTE P_exact_a = PDF.POISSON(3,5).
COMPUTE P_less_a = CDF.POISSON(2,5).
COMPUTE P_greater_b = 1 - CDF.POISSON(7,5).
COMPUTE P_between_ab = CDF.POISSON(7,5) - CDF.POISSON(2,5).
EXECUTE.
JASP
Enable the Distributions module, choose Poisson, set Lambda to 5, and read P(X ≤ k) from the interactive panel — k = 2 for P(X < 3), k = 7 for P(X ≤ 7).