Posted by **Jason L** on Friday, March 2, 2007 at 8:23am.

A specially equipped trauma emergency room at a hospital has been in operation for 40 weeks and has been used a total of 108 times. Assuming the weekly pattern of demand for this facility is poisson compute the following -

probability room is not used in a given week

probability that the room is used 7 or more times in a week

the mean demand for a 2 week period

Assume the true probablity of uses per week is L = 108/40 = 2.7, as established by the rather large sample.

The probability that there will be k occurences is

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

Where p is the Poisson distribution function.

(See http://en.wikipedia.org/wiki/Poisson_distribution )

For k = 0, P = e^-2.7 = 0.067

That is the probability there will be zero uses in one week.

Fot k = 7, P = e^-2.7*2.7^7/7! = 0.0139

for k = 8, P = 0.0047

For k = 9, P = 0.0014

P for k>9 is negligible. The total probability for k equal to or greater than 7 will be 0.020

The average number of uses in 2 weeks is just 2.7/week x 2 weeks = 5.4