Answer:
0.1723 = 17.23% probability that randomly selected region had exactly two hits.
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.
535 bombs hit the combined area of 570 regions
This means that the mean number of hits per region is given by:
[tex]\mu = \frac{535}{570} = 0.9386[/tex]
Find the probability that randomly selected region had exactly two hits
This is P(X = 2).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = x) = \frac{e^{-\0.9386}*(0.9386)^{2}}{(2)!} = 0.1723[/tex]
0.1723 = 17.23% probability that randomly selected region had exactly two hits.
