E. Coli outbreak. Outbreaks of E. coli have occurred at a rate of 2.5 per 100,000. Assume this rate represents the mean for the number of outbreaks (as a Poisson random variable).
a. What is the probability that at most five cases of E. coli per 100,000 are reported in California this year?
b. What is the probability that more than five cases of E. coli are reported in California this year?
c. Approximately 95% of occurrences of E. coli involve at most how many cases?
To solve these questions, we will use the Poisson probability distribution formula, which is given by:
P(x;λ) = (e^(-λ) * λ^x) / x!
Where:
P(x;λ) is the probability of observing x events in a given interval,
λ is the average rate of events per interval,
e is a mathematical constant approximately equal to 2.71828, and
x is the number of events we are interested in.
Now let's solve each part of the question separately:
a. To find the probability that at most five cases of E. coli are reported in California this year, we need to calculate P(x ≤ 5) using the Poisson probability distribution formula. In this case, the average rate of E. coli outbreaks is 2.5 per 100,000, so λ = 2.5.
P(x ≤ 5) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5)
b. To find the probability that more than five cases of E. coli are reported in California this year, we need to calculate P(x > 5), which is the complement of P(x ≤ 5). So, P(x > 5) = 1 - P(x ≤ 5).
c. To find the approximate maximum number of cases involving at most a certain percentage of occurrences (in this case, approximately 95%), we can use the cumulative distribution function (CDF) of the Poisson distribution. Looking up the value in the table, we can determine the value of x that corresponds to a cumulative probability of 0.95. Let's denote this value as X.
P(x ≤ X) = 0.95
These calculations can be performed with the help of a calculator or statistical software, or by using tables specifically designed for the Poisson distribution.