Intensive Care. The number X of people entering the intensive care unit at a particular hospital on any one day has a Poisson probability distribution with mean equal to five persons per day.
a. What is the probability that the number of people enterin the intensive care unit on a particular day is two? Less than or equal to two?
b. Is it likely that X will exceed 10? Explain.
To find the probabilities in a Poisson distribution, we can use the formula:
P(x; λ) = (e^(-λ) * λ^x) / x!
where P(x; λ) is the probability of x events occurring, λ is the mean number of events per unit of time, e is the base of the natural logarithm (approximately 2.71828), and x! denotes the factorial of x.
a) Probability that the number of people entering the intensive care unit on a particular day is two:
To find this probability, we can substitute x = 2 and λ = 5 into the Poisson probability formula:
P(2; 5) = (e^(-5) * 5^2) / 2!
P(2; 5) = (e^(-5) * 25) / 2
Using a calculator, we can calculate the value of e^(-5) ≈ 0.00674.
P(2; 5) ≈ (0.00674 * 25) / 2
P(2; 5) ≈ 0.0337
Therefore, the probability that exactly two people enter the intensive care unit on a particular day is approximately 0.0337.
Probability that the number of people entering the intensive care unit on a particular day is less than or equal to two:
To find this probability, we need to calculate the sum of the probabilities of having 0, 1, and 2 people entering the ICU:
P(X ≤ 2) = P(0; 5) + P(1; 5) + P(2; 5)
Using the same formula for each value and adding the results together, we get:
P(X ≤ 2) = (e^(-5) * 5^0) / 0! + (e^(-5) * 5^1) / 1! + (e^(-5) * 5^2) / 2!
Calculating each term separately, we find:
P(X ≤ 2) = (0.00674 * 1) / 1 + (0.00674 * 5) / 1 + (0.00674 * 25) / 2
P(X ≤ 2) ≈ 0.0337 + 0.0337 + 0.0843
P(X ≤ 2) ≈ 0.1517
Therefore, the probability that the number of people entering the intensive care unit on a particular day is less than or equal to two is approximately 0.1517.
b) Probability that X will exceed 10:
To find this probability, we need to calculate the sum of the probabilities of having 11, 12, 13, and so on, up to infinity. Since calculating the infinite sum is not practical, we can use the complement rule:
P(X > 10) = 1 - P(X ≤ 10)
We've already found P(X ≤ 10) to be approximately 0.1517 from the previous calculation.
P(X > 10) ≈ 1 - 0.1517
P(X > 10) ≈ 0.8483
Therefore, it is likely that the number of people entering the intensive care unit on a particular day will exceed 10, with a probability of approximately 0.8483.