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
To calculate the probability that the room is not used in a given week, use the Poisson distribution formula, where L is the average number of uses per week:
P(k) = e^-L * L^k / k!
In this case, L is 2.7 (calculated as the total uses divided by the number of weeks), and we want to find the probability when k = 0:
P(0) = e^-2.7 = 0.067
So, the probability that the room is not used in a given week is 0.067 or 6.7%.
To calculate the probability that the room is used 7 or more times in a week, calculate the probabilities for k = 7, 8, 9, and any values greater than 9, and sum them up:
P(k = 7) = e^-2.7 * 2.7^7 / 7! = 0.0139
P(k = 8) = e^-2.7 * 2.7^8 / 8! = 0.0047
P(k = 9) = e^-2.7 * 2.7^9 / 9! = 0.0014
P(k > 9) = negligible probability
The total probability for k equal to or greater than 7 will be 0.020 or 2.0%.
To find the mean demand for a 2-week period, simply multiply the average number of uses per week (L) by 2:
Average demand for a 2-week period = 2.7/week * 2 weeks = 5.4