Answer:
18.04% probability that exactly 3 serious deviations and incursions will occur at LAX in a randomly selected year
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
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
Suppose the mean number of deviations and incursions per year at the Los Angeles International Airport (LAX) is 2.
This means that [tex]\mu = 2[/tex]
Find the probability that exactly 3 serious deviations and incursions will occur at LAX in a randomly selected year
This is P(X = 3).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804[/tex]
18.04% probability that exactly 3 serious deviations and incursions will occur at LAX in a randomly selected year
