Answer :
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:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
[tex]e = 2.71828[/tex] is the Euler number
[tex]\mu[/tex] 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 [tex]\mu = 40/20 = 2[/tex]
This probability is P(X = 0).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353[/tex]
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