Question

In a given region, the number of tornadoes in a one-week period is modeled by a Poisson distribution with mean 2. The numbers of tornadoes in different weeks are mutually independent. Calculate the probability that fewer than four tornadoes occur in a three-week period.

Answer 1

Answer:

0.1512 = 15.12% probability that fewer than four tornadoes occur in a three-week period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

In a given region, the number of tornadoes in a one-week period is modeled by a Poisson distribution with mean 2

Three weeks, so $\mu = 2*3 = 6$

Calculate the probability that fewer than four tornadoes occur in a three-week period.

This is:

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.0025$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.0149$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.0446$

$P(X = 3) = \frac{e^{-6}*6^{3}}{(3)!} = 0.0892$

Then

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0025 + 0.0149 + 0.0446 + 0.0892 = 0.1512$

0.1512 = 15.12% probability that fewer than four tornadoes occur in a three-week period.