After-hours arrivals at a rural urgent-care clinic
A rural urgent-care clinic is modeling after-hours arrivals to decide how many exam rooms to staff. Historical logs show arrivals average 3.5 per hour, with no strong trend across the evening; the clinic treats arrivals as independent and occurring at a constant rate. The clinic can handle up to 7 arrivals in any one hour without diversion, but a diversion is triggered at 8 or more. The administrator estimates the diversion probability as "about 1 in 5" by treating the hourly count as approximately normal with mean 3.5. A nurse instead wants the probability of zero arrivals in a specific 20-minute window, because a single triage nurse takes a scheduled break then. The clinic also asks whether a Poisson model is justified, given that a nearby factory shift ends at 10 p.m. and sometimes sends several patients together.
1. Compute the exact probability that 8 or more patients arrive in one hour under the stated Poisson model. Compare it with the administrator's "1 in 5" and with a normal approximation using a continuity correction.
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Let X be hourly arrivals, X ~ Poisson(lambda = 3.5). Then P(X >= 8) = 1 - P(X <= 7). Using e^-3.5 = 0.03020: P(0) = 0.03020, P(1) = 0.10569, P(2) = 0.18496, P(3) = 0.21579, P(4) = 0.18881, P(5) = 0.13217, P(6) = 0.07710, P(7) = 0.03855. Their sum is 0.97326, so P(X >= 8) = 0.02674, about 2.7%, or roughly 1 in 37 hours, not 1 in 5.
A normal approximation with continuity correction uses mean 3.5 and SD sqrt(3.5) = 1.871: z = (7.5 - 3.5)/1.871 = 2.14, giving an upper tail of about 0.016. Without continuity correction, z = (8 - 3.5)/1.871 = 2.41 and the tail is about 0.008. The normal approximation understates the exact Poisson tail here because lambda is small and the distribution is discrete and right-skewed. The exact Poisson calculation is preferable.
2. Find the probability of zero arrivals in a specific 20-minute window. State the assumptions and explain how the nearby factory shift could change the answer.
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Over 20 minutes the mean is lambda = 3.5/3 = 1.1667. Then P(X = 0) = e^-1.1667 = 0.311, so there is about a 31% chance of no arrivals in that window. This calculation uses the same Poisson assumptions: independent arrivals and a constant rate within the window.
If the triage nurse's break is scheduled right after the nearby factory shift, the constant-rate and independence assumptions may fail. Arrivals may cluster, so the effective rate in that 20-minute period is higher than 1.1667, and the probability of zero arrivals is lower than 0.311. For break scheduling, use a time-specific rate estimated from that period rather than the evening average. Also ensure that the 20-minute and hourly calculations use a consistent rate; the 20-minute window is nested within the hourly count.
3. Is a Poisson model or a binomial model more appropriate for the number of arrivals in an hour? What evidence would make the Poisson model doubtful, and what alternative would you consider?
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Poisson is the natural first model for counts of arrivals in a fixed time interval: there is no fixed number of "trials," events occur independently, and the rate is constant. A binomial model would require a known number of potential patients n and a constant probability p per patient, which is artificial here. Poisson can be viewed as a limiting case of a binomial model when n is large and p is small.
The factory shift is a warning. If several patients arrive together after 10 p.m., arrivals are not independent and the rate is not constant, producing overdispersion, meaning the variance is greater than the mean. Check the variance-to-mean ratio across hours and use a dispersion or time-series test. If overdispersion is present, use a negative binomial model or a time-varying Poisson rate with hour-of-evening effects. Do not simply plug the average into a Poisson model and report a precise probability.