Question

In analyzing hits by certain bombs in a war, an area was partitioned into 570 regions, each with an area of 0.75 km^ 2 . A total of 535 bombs hit the combined area of 570 regions Assume that we want to find the probability that randomly selected region had exactly two hits the Poisson distribution formula

Answer 1

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ 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:

$\mu = \frac{535}{570} = 0.9386$

Find the probability that randomly selected region had exactly two hits

This is P(X = 2).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = x) = \frac{e^{-\0.9386}*(0.9386)^{2}}{(2)!} = 0.1723$

0.1723 = 17.23% probability that randomly selected region had exactly two hits.