Question

Increased research and discussion have focused on the number of illnesses involving the organism Escherichia coli, which causes a breakdown of red blood cells and intestinal hemorrhages in its victims. Sporadic outbreaks of E.coli have occurred in Colorado at a rate of 2.5 per 100,000 for a period of 2 years. Let us suppose this rate has not changed. What is the probability that at most five cases of E. coli per 100,000 are reported in Colorado in a given year?

Answer 1

The probability that at most five cases of E. coli per 100,000 are reported in Colorado in a given year can be calculated using the Poisson distribution. The rate of E. coli outbreaks in Colorado is 1.25 per 100,000 people per year.

Step-by-step explanation:

To calculate the probability that at most five cases of E. coli per 100,000 are reported in Colorado in a given year, we can use the Poisson distribution. The Poisson distribution is commonly used to model the number of events occurring in a fixed interval of time or space when these events occur with a known average rate and independently of the time since the last event.

In this case, the rate of E. coli outbreaks in Colorado is 2.5 per 100,000 for a period of 2 years. We can convert this rate to a yearly rate by dividing it by 2. So, the yearly rate is 2.5 / 2 = 1.25 outbreaks per 100,000 people.

Using the Poisson distribution formula, we can calculate the probability of at most five outbreaks in a given year:

P(X <= 5) = e^(-λ) * (λ^0 / 0!) + e^(-λ) * (λ^1 / 1!) + e^(-λ) * (λ^2 / 2!) + ... + e^(-λ) * (λ^5 / 5!)

where λ is the average rate of outbreaks per 100,000 people in a year.

Substituting the value of λ as 1.25, we can calculate the probability using a calculator or by using a software like Microsoft Excel.