Respuesta :
Answer:
There is a 0.58% probability of Florida being struck by four or more hurricanes in the same year.
Step-by-step explanation:
Since we have only the mean during the interval, we can solve this problem using the Poisson probability distribution.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
Poisson probability distribution
[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.
In this problem, we have that:
Since 1851, exactly 119 hurricanes have hit Florida (this includes the years 1851 and 2019). Counting 1851 and 2019, there are 169 years in this interval. This means that [tex]\mu = \frac{119}{169} = 0.704[/tex]
If the probability of hurricane strikes has remained the same since 1851, what is the probability of Florida being struck by four or more hurricanes in the same year?
This is [tex]P(X \geq 4)[/tex].
Either Florida is struck by less than four hurricanes in a given year, or it is struck by 4 or more. The sum of these probabilities is decimal 1.
[tex]P(X < 4) + P(X \geq 4) = 1[/tex]
[tex]P(X \geq 4) = 1 - P(X < 4)[/tex]
In which
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.704}*(0.704)^{0}}{(0)!} = 0.4946[/tex]
[tex]P(X = 1) = \frac{e^{-0.704}*(0.704)^{1}}{(1)!} = 0.3482[/tex]
[tex]P(X = 2) = \frac{e^{-0.704}*(0.704)^{2}}{(2)!} = 0.1226[/tex]
[tex]P(X = 3) = \frac{e^{-0.704}*(0.704)^{3}}{(3)!} = 0.0288[/tex]
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.4946 + 0.3482 + 0.1226 + 0.0288 = 0.9942[/tex]
Finally
[tex]P(X \geq 4) = 1 - P(X < 4) = 1 - 0.9942 = 0.0058[/tex]
There is a 0.58% probability of Florida being struck by four or more hurricanes in the same year.
