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

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

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) = (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) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-6)*6^(0))/((0)!) = 0.0025`

`P(X = 1) = (e^(-6)*6^(1))/((1)!) = 0.0149`

`P(X = 2) = (e^(-6)*6^(2))/((2)!) = 0.0446`

`P(X = 3) = (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.