Answer:
21.77% probability that the antenna will be struck exactly once during this time period.
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 interval.
In this question:
[tex]\mu = 2.40[/tex]
Find the probability that the antenna will be struck exactly once during this time period.
This is P(X = 1).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 1) = \frac{e^{-2.40}*2.40^{1}}{(1)!} = 0.2177[/tex]
21.77% probability that the antenna will be struck exactly once during this time period.
