Choosing the Right Distribution
For each scenario, name the most appropriate distribution, state its parameter(s) from the information given, and compute the quantity asked.
- A checkout page has a 4% conversion rate. 50 independent visitors arrive. What is the expected number of conversions and the standard deviation of that count? Roughly what is the probability of exactly zero conversions?
- A support inbox receives on average 2 urgent tickets per hour, arriving independently. What is the probability of receiving no urgent tickets in a given hour? At least 3?
- Servers in a fleet fail at a constant rate with a mean time between failures of 500 hours. A server has already run for 200 hours without failing. What is the probability it survives another 100 hours?
- Which of the above would you not model with a Normal distribution, and why?
1. Conversions — Binomial(n = 50, p = 0.04)
Each visitor is a Bernoulli(0.04) trial; the count of successes across 50 independent trials is Binomial. Mean np = 50 \times 0.04 = 2. Variance np(1-p) = 50 \times 0.04 \times 0.96 = 1.92, so the standard deviation is \sqrt{1.92} \approx 1.39. P(X = 0) = 0.96^{50} \approx 0.13 — about a 13% chance of seeing no conversions at all, which is why small samples produce alarming-looking zero-conversion days.
2. Urgent tickets — Poisson(λ = 2)
Independent arrivals at a constant average rate over a fixed window is the Poisson story. P(X = 0) = e^{-2} \approx 0.135. P(X \ge 3) = 1 - [P(0) + P(1) + P(2)] = 1 - e^{-2}(1 + 2 + 2) = 1 - 5e^{-2} \approx 1 - 0.677 = 0.323. About a third of hours will see three or more urgent tickets even though the average is two.
3. Server survival — Exponential(λ = 1/500 per hour)
"Constant failure rate" is the signature of the Exponential distribution, with mean 1/\lambda = 500 hours. By memorylessness, the 200 hours already survived are irrelevant: P(T > 300 \mid T > 200) = P(T > 100) = e^{-100/500} = e^{-0.2} \approx 0.82. If failures instead showed wear-out (hazard rising with age), the Exponential would be the wrong model and the prior 200 hours would matter.
4. Where the Normal is inappropriate
Scenarios 1 and 2 are small counts (means of 2): they are bounded below by zero, discrete, and noticeably right-skewed, so a Normal approximation would assign probability to negative counts and misstate tail probabilities. Scenario 3 is a strictly positive, heavily right-skewed waiting time — also not Normal. The Normal becomes appropriate for the sample mean of many such observations (CLT), or for a Binomial/Poisson whose mean is large (roughly np and n(1-p) both ≥ 10, or λ ≥ 10).
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