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Some previous studies have shown a relationship between emergency-room admission

ID: 3023799 • Letter: S

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.

a) 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?

b) The hospital also finds that on high-pollution days the number of admissions is Poisson-distributed with mean = 4.0 admissions per day. Answer problem a for high-pollution days.

c) On a random day during the year, what is the probability there will be 4 admissions to the emergency ward, assuming there are 345 normal-pollution days and 20 high-pollution days?

The answer for letter a is five beds. I need help for answer b.

All problems are taken from Rosner’s Textbook, Fundamental of Biostatistics

Explanation / Answer

b) If the average number of admissions on a high-pollution day is denoted by , then the probability of exactly k admissions on any particular high-pollution day is given by:

P(k) = k * e- / k!

For a high pollution day, = 4. We are interested in finding the number of beds which will suffice the admission requirement on at least 95% of high-pollution days. In other words, we want to find k such that the cumulative probability for k is 95%.

From the Poisson cumulative distribution table,
   Cum. Probability ( = 4, k = 7) = 0.9489
and, Cum. Probability ( = 4, k = 8) = 0.9786

Therefore, required number of beds to suffice the purpose is 8 beds.

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