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 four persons per day. (a) What is the probability that the number of people entering the intensive care unit on a particular day is two? (Round your answer to three decimal places.) What is the probability that the number of people entering the intensive care unit on a particular day is less than or equal to two? (Round your answer to three decimal places.) (b) Is it likely that, on a given day, the number of people entering the intensive care unit, will exceed an x-value of 8? Calculate the mean and standard deviation of the probability distribution to explain your answer. O Yes it is likely because x is less than 2 standard deviations above the mean. O No it is not likely because x is more than 2 standard deviations above the mean.
![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 four persons per day.
(a) What is the probability that the number of people entering the intensive care unit on a particular day is two? (Round your answer to three decimal places.)
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What is the probability that the number of people entering the intensive care unit on a particular day is less than or equal to two? (Round your answer to three decimal places.)
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(b) Is it likely that, on a given day, the number of people entering the intensive care unit, will exceed an \( x \)-value of 8? Calculate the mean and standard deviation of the probability distribution to explain your answer.
- [ ] Yes, it is likely because \( x \) is less than 2 standard deviations above the mean.
- [ ] No, it is not likely because \( x \) is more than 2 standard deviations above the mean.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F675d7c68-68c2-40f4-bb09-6fc638745853%2F17980116-8079-4018-90c4-f15a449acdbe%2Fe6do6g_processed.jpeg&w=3840&q=75)

(a) The number of people entering the intensive care unit on a particular day follows a Poisson distribution with a mean (λ) equal to four persons per day. To calculate the probability:
Probability that x = 2:
P(x = 2) = (e^(-λ) * λ^x) / x!
P(x = 2) = (e^(-4) * 4^2) / 2!
P(x = 2) ≈ (0.01832 * 16) / 2 ≈ 0.1832 (rounded to three decimal places).
Probability that x ≤ 2:
P(x ≤ 2) = P(x = 0) + P(x = 1) + P(x = 2)
To calculate P(x = 0):
P(x = 0) = (e^(-λ) * λ^0) / 0!
P(x = 0) = (e^(-4) * 1) / 1 = 0.01832 (rounded to three decimal places).
To calculate P(x = 1):
P(x = 1) = (e^(-λ) * λ^1) / 1!
P(x = 1) = (e^(-4) * 4) / 1 = 0.1465 (rounded to three decimal places).
Now, add these probabilities:
P(x ≤ 2) = 0.01832 + 0.1465 + 0.1832 ≈ 0.348 (rounded to three decimal places)
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