In: Statistics and Probability
6.31. The exponential distribution can be used to solve Poisson-type problems in which the intervals are not time. The Air Travel Consumer Report published by the U.S. Department of Transportation reported that in a recent year, Virgin America led the nation in fewest occurrences of mishandled baggage, with a mean rate of 0.95 per 1,000 passengers. Assume that mishandled baggage occurrences are Poisson distributed. Using the exponential distribution to analyze this problem, determine the average number of passengers between occurrences. Suppose baggage has just been mishandled.
Book says answers are
.μ = 1052.6
a..5908
b..1899 but how?
We know that the number of occurrence of events between two Poisson process follows exponential distribution.
Rate = 0.95/1000 per passengers = 0.00095 per passengers
Thus, the number of passengers will have their baggage handled properly (X) before the next mishandling occurs (Poisson process) foloow exponential distribution with rate parameter = 0.00095 per passengers
= 1/0.00095 = 1052.6 passengers
Probability that at least 500 passengers will have their baggage handled properly before the next mishandling occurs
= P(X 500) = exp(-0.00095 * 500) P(X x) = exp(-x)
= 0.6219
Probability that the number will be fewer than 200 passengers = P(X < 200)
= 1 - exp(-0.00095 * 200) P(X < x) = 1 - exp(-x)
= 1 - 0.8270
= 0.1730