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I'm having some trouble with a basic question regarding Poisson distributions.

Question: the average rate of admission to the ICU is 48 patients per day, what is the probability of less than two patients being admitted in an hour?

With the daily $\lambda = 48$ and hourly $X \leq 1$ what would be the chosen distribution?

I don't think it would be right to assume an equal rate for hours and choose $\lambda = 2, X \leq 1$, then again even $\lambda = 48, X \leq 24$ seems too optimistic since it does not take into account hours where $X = 0$ (assumes an $X = 1$ for each hour).

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  • $\begingroup$ Welcome to MSE. Can you post the full question? $\endgroup$
    – Sergio
    Dec 23, 2020 at 6:47
  • $\begingroup$ @Sergio thank you for the suggestions. $\endgroup$ Dec 23, 2020 at 6:59

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With Poisson processes, you can scale up or down the time intervals as needed to get to the interval asked about in the problem. So you can convert your average 48 events per day to an average of 2 events per hour and use $ \lambda = 2 $.

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    $\begingroup$ Then with $X \sim \mathsf{Pois}(\lambda=2),$ you get $P(X \le 1) = P(X=0) + P(X=1) = 0.04060.$ Or in R, where ppois is a Poisson CDF, code ppois(1,2) returns $0.4060058.$ $\endgroup$
    – BruceET
    Dec 24, 2020 at 2:57

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