Some previous studies have shown a relationship between emergency room admission

Question

Some previous studies have shown a relationship between emergency room admissions per day and level of pollution on a given day. A small local hospital finds that the number of admissions to the emergency ward on a single day ordinarily (unless there is unusually high pollution) follows a Poisson distribution with mean=2.0 admissions per day. Suppose each admitted person to the emergency ward stays there for exactly 1 day and is then discharged. The hospital is planning a new emergency-room facility. It wants enough beds in the emergency ward so that for at least 95% of normal-pollution days it will not need to turn anyone away. What is the smallest number of beds it should have to satisfy this criterion?

Explanation / Answer

The way to solve this problem is to find the first x such that P(X <= x) is greater than .95.
Since this is a Poisson distribution with parameter 2.0,
P(x) =e -x/x! = e-22x /x!

P(0) = e-220 /0! = e-2 * 1/1 = e-2 = 0.1353

P(1) = e-221 /1! = e-2 * 2/1 = 2e-2 = 0.2707

P(2) = e-222 /2! = e-2 * 4/2 = 2e-3 = 0.2707

P(3) = e-223 /3! = e-2 * 8/6 = 4/3e-3 = 0.1804

P(4) = e-224 /4! = e-2 * 16/24 = 2/3e-3 = 0.0902

P(5) = e-225 /5! = e-2 * 32/120 = 4/15e-3 = 0.0361

P(X<= 4) = 0.1353 + 0.2707 + 0.2707 + 0.1804 + 0.0902 = .9473

P(X <=5) = .9473 + 0.0361 = .9834

If we consider "at least 95%" to not allow rounding, then it will require 5 beds.

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