Source: Quiz 3 Solutions, Principles of Statistics I (Texas A&M University)
Tags: Poisson distribution, rare events, defective parts, lambda, rate parameter, Poisson probability, STAT 211, probability distributions
The Poisson distribution models the number of events occurring in a fixed interval of time or space, where events happen independently at a constant average rate. It is the go-to distribution for counting rare, independent events, and only requires one parameter: λ (the average rate).
Poisson distribution
A discrete probability distribution that gives the probability of a certain number of events occurring in a fixed interval, given a known constant mean rate (λ). Written X ~ Poisson(λ).
Lambda (λ)
The rate parameter of the Poisson distribution. It equals both the mean and the variance: E(X) = Var(X) = λ. For a different-sized interval, scale λ proportionally.
Rate scaling
If the rate is given per one unit of time (e.g. 0.01 per hour), then for a different interval length t, the effective rate is λ_new = λ · t. For example, 0.01 defective parts per hour over 4 hours gives λ = 0.01 × 4 = 0.04.
The Poisson distribution applies when:
You are counting the number of events in a fixed interval (time, area, volume, etc.)
Events occur independently of one another
The average rate of occurrence (λ) is constant
Two events cannot occur at exactly the same instant
Classic scenarios include defective parts arriving at a station, phone calls received per hour, typos per page, and accidents per month.
The quiz gives this setup: defective parts arrive at a rate of 0.01 per hour. You are asked about the number of defective parts D in a given time span.
This is Poisson because:
Events (defective parts) arrive independently
There is a known average rate (0.01 per hour)
You are counting occurrences in a fixed interval
The quiz offered three choices: Binomial, Discrete Uniform, and Poisson. The correct answer is Poisson.
A binomial would require a fixed number of trials with a success/failure outcome. A discrete uniform would mean every count is equally likely. Neither fits here.
The probability of observing exactly k events is:
P(X = k) = (e^(−λ) · λ^k) / k!
where e ≈ 2.71828, λ is the rate parameter, and k! is k factorial.
The quiz gives λ = 0.01 per hour, then asks about a 4-hour span.
Scaled λ = 0.01 × 4 = 0.04
This new λ = 0.04 is what you plug into the PMF for any probability calculation over the 4-hour window.
P(X = 1 | λ = 0.04) = (e^(−0.04) · 0.04^1) / 1!
P(X = 1) = (0.9608)(0.04) / 1 = 0.0384
Use the complement rule. "One or more" means P(X ≥ 1), which equals 1 − P(X = 0).
P(X = 0 | λ = 0.04) = (e^(−0.04) · 0.04^0) / 0! = e^(−0.04) = 0.9608
P(X ≥ 1) = 1 − 0.9608 = 0.0392
The complement approach is almost always the fastest way to handle "at least one" Poisson questions.
Poisson PMF: P(X = k) = (e^(−λ) · λ^k) / k!
Mean: E(X) = λ
Variance: Var(X) = λ
Standard deviation: σ = √λ
Rate scaling: λ_new = λ_original × t (where t is the new interval length in the same units)
At least one event: P(X ≥ 1) = 1 − e^(−λ)
⚠️ The Poisson distribution has the special property that its mean and variance are equal (both λ). This is a classic exam fact and a quick way to identify or verify a Poisson model.
⚠️ Always check whether λ needs to be scaled. If the rate is given per hour but the question asks about 4 hours, multiply λ by 4 before computing.
⚠️ For "at least one" questions, use the complement: P(X ≥ 1) = 1 − P(X = 0). Computing P(X = 1) + P(X = 2) + ... directly is an infinite sum and not feasible.
⚠️ Remember that 0! = 1 and anything raised to the power 0 equals 1. These come up when computing P(X = 0).
⚠️ On a multiple-choice question asking which distribution applies, the Poisson signal words are: rate, per unit time, arrivals, occurrences, average number per interval.
Q: Emails arrive at a server at a rate of 5 per minute. What distribution models the number of emails in a 2-minute window, and what is λ?
A: Poisson distribution with λ = 5 × 2 = 10.
Q: Using the quiz setup (λ = 0.01 per hour), what is the probability of zero defective parts in a 4-hour span?
A: P(X = 0) = e^(−0.04) ≈ 0.9608.
Q: Why is the complement rule used for P(X ≥ 1) in a Poisson setting?
A: Because summing P(X = 1) + P(X = 2) + P(X = 3) + ... would require infinitely many terms. Computing 1 − P(X = 0) gives the same result in one step.
Q: A Poisson random variable has variance 3. What is its mean?
A: The mean is also 3. For a Poisson distribution, E(X) = Var(X) = λ.
Q: If the mean number of accidents at an intersection is 2 per week, what is the probability of exactly 3 accidents next week?
A: P(X = 3) = (e^(−2) · 2³) / 3! = (0.1353)(8) / 6 ≈ 0.1804.
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