Question

Infected patients with a certain virus arrive in a hospital according to a Poisson process with an average rate of 22.7 per day. The hospital management is planning the level of resources that they need in the coming weeks. As a part of this planning effort, they are interested to know what is the probability of receiving 3 patients in 2.1 hours? (Assume each day is 24 hours) state your answer in decimal format not percentage

Answer 1

The probability of receiving 3 patients in 2.1 hours at the hospital where patients arrive according to a Poisson process with an average rate of 22.7 per day is approximately 0.1792.

Step-by-step explanation:

The student asked what is the probability of receiving 3 patients in 2.1 hours in a hospital where patients arrive according to a Poisson process with an average rate of 22.7 per day. To solve this, we need to calculate the rate of patient arrivals per 2.1 hours. Since there are 24 hours in a day, the rate per 2.1 hours is (22.7 patients/day) * (2.1 hours / 24 hours/day) = 1.9875 patients per 2.1 hours. We denote this as λ (lambda).

The probability of observing exactly k events in a Poisson distribution is given by the formula P(X=k) = (λ^k * e^-λ) / k!, where 'e' is the base of the natural logarithm, λ is the average rate, and k is the number of events (patients in this case).

For k=3, we have:

P(X=3) = ((1.9875)^3 * e^-1.9875) / 3!
= (7.8541 * 0.1366) / 6
= 0.1792

Therefore, the probability of receiving 3 patients in 2.1 hours is approximately 0.1792.