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Answer :
The probability that across six months there would be zero months with no AMI IP hospitalizations is approximately 0.6328.
To answer these questions, we will assume that the number of AMI IP hospitalizations in Benton County follows a Poisson distribution with an average rate of 30 hospitalizations per year.
7) To calculate the probability of 2 AMI IP hospitalizations in Benton County over the course of one month, we need to adjust the rate to a monthly rate. Since there are 12 months in a year, the monthly rate would be 30/12 = 2.5 hospitalizations per month.
Using the Poisson probability formula:
P(X = 2) = (e^(-λ) * λ^2) / 2!
where λ is the average rate per month, we can calculate the probability:
P(X = 2) = (e^(-2.5) * 2.5^2) / 2!
≈ 0.1839 (rounded to 4 decimal places)
Therefore, the probability of 2 AMI IP hospitalizations in Benton County over the course of one month is approximately 0.1839.
8) To calculate the probability that there would be no AMI IP hospitalizations in Benton County for a given month, we again adjust the rate to a monthly rate of 2.5 hospitalizations per month.
Using the Poisson probability formula:
P(X = 0) = e^(-λ)
where λ is the average rate per month, we can calculate the probability:
P(X = 0) = e^(-2.5)
≈ 0.0821 (rounded to 4 decimal places)
Therefore, the probability that there would be no AMI IP hospitalizations in Benton County for a given month is approximately 0.0821.
9) To calculate the probability of exactly one AMI IP hospitalization in two months, we multiply the monthly rate by the number of months (2) to get a total rate of 2.5 * 2 = 5 hospitalizations.
Using the Poisson probability formula:
P(X = 1) = e^(-λ) * λ^1 / 1!
where λ is the average rate for two months (5 hospitalizations), we can calculate the probability:
P(X = 1) = e^(-5) * 5^1 / 1!
≈ 0.0337 (rounded to 4 decimal places)
Therefore, the probability of exactly one AMI IP hospitalization in two months is approximately 0.0337.
10) To calculate the probability that across six months there would be zero months with no AMI IP hospitalizations, we need to calculate the probability of at least one hospitalization in each month and subtract it from 1.
Using the Poisson probability formula:
P(X ≥ 1) = 1 - P(X = 0) = 1 - e^(-λ)
where λ is the average rate per month (2.5 hospitalizations), we can calculate the probability of at least one hospitalization in a month:
P(X ≥ 1) = 1 - e^(-2.5)
≈ 0.9179 (rounded to 4 decimal places)
Since we want all six months to have at least one hospitalization, we raise this probability to the power of 6:
P(all 6 months ≥ 1 hospitalization) = (0.9179)^6
≈ 0.6328 (rounded to 4 decimal places)
Therefore, the probability that across six months there would be zero months with no AMI IP hospitalizations is approximately 0.6328.
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