Respuesta :
Answer:
a) 0.3293 = 32.93% probability that the number of major hurricane to strike U.S mainland in any any give year is exactly one.
b) 0.878 = 87.8% probability that the number of major hurricane to strike U.S mainland in any any give year is at most one.
c) 0.122 = 12.2% probability that the number of major hurricane to strike U.S mainland in any any give year is more than one.
Step-by-step explanation:
We have the mean during an interval, which means that the Poisson distribution is used to solve this question.
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
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
During the 20th century the mean number of the major hurricanes to strike U.S mainland per year was 0.6.
This means that [tex]\mu = 0.6[/tex]
Find the probability that the number of major hurricane to strike U.S mainland in any any give year is:
a) Exactly one
This is P(X = 1). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 1) = \frac{e^{-0.6}*0.6^{1}}{(1)!} = 0.3293[/tex]
0.3293 = 32.93% probability that the number of major hurricane to strike U.S mainland in any any give year is exactly one.
b) At most one
This is:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
Then
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.6}*0.6^{0}}{(0)!} = 0.5488[/tex]
[tex]P(X = 1) = \frac{e^{-0.6}*0.6^{1}}{(1)!} = 0.3293[/tex]
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.5488 + 0.3292 = 0.878[/tex]
0.878 = 87.8% probability that the number of major hurricane to strike U.S mainland in any any give year is at most one.
c) More than one
This is:
[tex]P(X > 1) = 1 - P(X \leq 1) = 1 - 0.878 = 0.122[/tex]
0.122 = 12.2% probability that the number of major hurricane to strike U.S mainland in any any give year is more than one.
