Question

Historical data shows that the average number of patient arrivals at the intensive care unit of General Hospital is 3 patients every 2 hours. Assume that the patient arrivals are distributed according to Poisson distribution. Determine the probability of 6 patients arriving in a five-hour period?

1) .137
2) .109
3) .246
4) .001

Answer 1

2) .109

The probability of 6 patients arriving in a five-hour period can be calculated using the Poisson distribution formula. The probability is approximately 0.109, which corresponds to option 2) in the choices.

Step-by-step explanation:

To determine the probability of 6 patients arriving in a five-hour period, we can use the Poisson distribution formula. The formula for the probability mass function (PMF) of the Poisson distribution is:

P(X=k) = (e^-lambda * lambda^k) / k!

Where lambda is the average number of arrivals in the given time period, and k is the number of arrivals we want to find the probability for.

In this case, the lambda value is 3 patients per 2 hours.

Let's calculate the probability:

1. First, we need to convert the lambda value to the five-hour period. Since 5 hours is 2.5 times the length of 2 hours, we multiply the lambda value by 2.5:
2. lambda = 3 * 2.5 = 7.5 patients
3. Now, substitute the lambda and k values into the formula:
4. P(X=6) = (e^-7.5 * 7.5^6) / 6!
5. P(X=6) = (e^-7.5 * 16807.5) / 720
6. Calculate the probability:
7. P(X=6) ≈ 0.109

Therefore, the probability of 6 patients arriving in a five-hour period is approximately 0.109, which corresponds to option 2) in the given choices.