[tck44]

Hi,

Just wondering if someone can help me understand this example that was given on a practice midterm for a Statistics class I'm taking:

"The average number of accidents on a certain section of highway is two per week. Find the probability of at most one accident on this section of the highway during a two week period"

The answer given is 0.092, but no solution is presented, so I don't know how the prof got to this answer.

Thanks in advance.

[tck44]

Figured it out.. it's the Poisson probability formula.

There are an average of two accidents per week. So, in a two week period, there will be an average of four accidents. Using 4 as our Poisson parameter, we can start calculating:

P(X <= 1) = P(X = 0) + P(X = 1)

By Poisson,

P(X = 0) = e^(-4) x (4^0 / 0!) = 0.0183
P(X = 1) = e^(-4) x (4^1 / 1!) = 0.0732

Thus, P(X <= 1) = 0.0183 + 0.0732 = 0.0915, which rounds to 0.092.