Question

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. (a) What is the probability that at least 490 passengers will have their baggage handled properly before the next mishandling occurs? (b) What is the probability that the number will be fewer than 190 passengers?

Answer 1

Answer:

a) $P(X \ge 490) = 0.628$

b) $P(X \leq 190) = 0.165$

Step-by-step explanation:

a) This is an exponential distribution question since it is a poisson type problem

Mean rate is 0.95 per 1000 passengers

Mean rate, $\lambda = 0.95/1000$

$\lambda = 0.00095$ per passenger

Probability that at least 490 passengers will have their baggage handled properly before the next mishandling occurs :

$P(X \ge x) = e^{- \lambda x}$

$P(X \ge 490) = e^{-490 * 0.00095}$

$P(X \ge 490) = e^{-0.4655}\\P(X \ge 490) = 0.628$

b)  Probability that the number will be fewer than 200 passengers

$P(X \leq 190) = 1 - P(X \geq 190)\\P(X \leq 190) = 1 - e^{-190 \lambda} \\P(X \leq 190) = 1 - e^{-190 *0.00095}$

$P(X \leq 190) = 1 - 0.835\\P(X \leq 190) = 0.165$