posted by Jason L on .
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