Answer to Question #127490 in Statistics and Probability for beenseoked

Question

The average numbers of traffic accidents on a certain section of highway is two per week. Assume that the number of accidents follows a Poisson distribution with µ = 2.
1. Find the probability of no accidents on this section of highway during a 1-week period.
2. Find the probability of at most three accidents on this section of highway during a 2-week period.

Answer 1

1.

Poisson distribution with mean µ = 2 is $P(k)=\dfrac{2^k e^{-2}}{k!}$ , where k is a number of accidents per week.

$P(0)=\dfrac{2^0 e^{-2}}{0!} \approx 0.135$

2.

If mean for 1 week is 2 accidents, then mean for 2 weeks is 4 accidents. Poisson distribution with mean µ = 4 is $P(k)=\dfrac{4^k e^{-4}}{k!}$, where k is a number of accidents per 2 weeks.

Probability of at most three accidents is

$P(0)+P(1)+P(2)+P(3)=\dfrac{4^0 e^{-4}}{0!}+\dfrac{4^1 e^{-4}}{1!}+\dfrac{4^2 e^{-4}}{2!} + \dfrac{4^3 e^{-4}}{3!} \approx 0.433$