The number of planes arriving per day at a small private airport is a random variable having a poison distribution with $\lambda= 28.8$. What's the probability that the time between two such arrivals is at least one hour?
I'm thinking that this is an exponential distribution with $\theta = \frac{1}{28.8}$. So then it would follow that I would have to find $P(X\ge \frac{1}{24}) = 1- P(X \lt \frac{1}{24})= 1- \int_0 ^{\frac{1}{24}} 28.8e^{-28.8x} dx \approx .301$. I think this is equivalent to saying the complement of the probability that the time between arrivals is less than an hour.
Any help would be appreciated.
Best Answer
Your approach is correct!
Equivalently, you can notice that the required probability is equal to the probability that there will be no arrivals in an hour. Since, the arrivals per hour $X_h$ (and not per day) follow the Poisson distribution with $$λ_h=\frac{λ}{24}=\frac{28.8}{24}=1.2$$ you find that the requested probability is equal to $$P(X_h=0)=e^{-λ_h}\frac{λ_h^0}{0!}=e^{-1.2}=0.301194211$$