Asked by Matt
A specially equipped trauma emergency room at a hospital has been in operation for 40 weeks and has been used a total of 240 times. Assuming the weekly pattern of demand for this facility is Poission, compute the following:
1) The probability that the room is not used in a given week,
2) The probability that the room is used seven or more times in a week, and
3) The mean demand for a two week period.
Poisson distribution with a mean of m is this: P(x) = e^(-m) m^x / x!
For #1: 240/40 = 6 (average per week)
P(0) = e^(-6) 6^0 / 0! = 0.0025 (rounded value)
For #2: P(>=7) = 1 - P(<7) = 1 - [P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6)]
You might be able to use a table instead of computing these by hand.
For #3: The mean demand for a two week period = 12.
1) The probability that the room is not used in a given week,
2) The probability that the room is used seven or more times in a week, and
3) The mean demand for a two week period.
Poisson distribution with a mean of m is this: P(x) = e^(-m) m^x / x!
For #1: 240/40 = 6 (average per week)
P(0) = e^(-6) 6^0 / 0! = 0.0025 (rounded value)
For #2: P(>=7) = 1 - P(<7) = 1 - [P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6)]
You might be able to use a table instead of computing these by hand.
For #3: The mean demand for a two week period = 12.
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