Question

According to geologists, the San Francisco Bay Area experiences five earthquakes with a magnitude of 6.5 or greater every 100 years. What is the probability that no earthquakes with a magnitude of 6.5 or greater strike the San Francisco Bay Area in the next 40 years?

Answer 1

Answer:

13.53% probability that no earthquakes with a magnitude of 6.5 or greater strike the San Francisco Bay Area in the next 40 years

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 time interval.

According to geologists, the San Francisco Bay Area experiences five earthquakes with a magnitude of 6.5 or greater every 100 years.

One earthquake each 100/5 = 20 years.

What is the probability that no earthquakes with a magnitude of 6.5 or greater strike the San Francisco Bay Area in the next 40 years?

40 years, so $\mu = 40/20 = 2$

This probability is P(X = 0).

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

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

13.53% probability that no earthquakes with a magnitude of 6.5 or greater strike the San Francisco Bay Area in the next 40 years