1
$\begingroup$

The California Health Region states that the number of people per year in the Los Angeles area who contract flesh-eating disease is a random variable - the behavior of which can be modeled using the Poisson distribution with a mean of 5.1.

What is the probability that between two and four people, inclusive, will contract the flesh-eating disease in a four-month period?

I have done the following

Using the Poisson formula:

Lambda = 5.1

P(2 < X < 4) = e^-5.1 * 5.1^2 / 2! + e^-5.1 * 5.1^3 / 3! + e^-5.1 * 5.1^4 / 4!

However, with my answer of 0.385 I got it wrong, I am thinking that maybe the mean is wrong since they are asking 4 months and not a whole year (5.1) I thought of 5.1/12 since 12 months in a year. Then that's the mean for 1 month so I multiply by 4 to get 1.7 and just use 1.7 lambda in the formula to get the answer.

What is the correct solution and answer?

1 Answers 1

1

The annual rate is $5.1$ cases per year. For a four-month period, this corresponds to a rate of $\lambda = 5.1/3 = 1.7$ cases every four months. Then the desired probability is $$\Pr[2 \le X \le 4] = e^{-1.7} \left( \frac{(1.7)^2}{2!} + \frac{(1.7)^3}{3!} + \frac{(1.7)^4}{4!}\right).$$