A large airline keeps track of the number of no-shows for one of its most import

Question

A large airline keeps track of the number of no-shows for one of its most important com­muter flights. Over time the airline has found that the number of ticketed passengers who do not show up is a random variable with the following probability distribution: x 0 1 2 3 4 5 6 7 8 p(x) 0.05 0.08 0.13 0.23 0.18 0.13 0.08 0.06 0.06 1. What is the probability that at least 3 ticketed passengers do not show up for the flight? 2. What is the probability that not more than 6 passengers do not show up for the flight? 3. The aircraft used for the flight has 35 seats. If the airline routinely overbooks the flight by 4 passengers, what is the probability that on any given day every ticketed passenger who shows up will get a seat? 4. Find the mean and standard deviation for the number of no shows.

Explanation / Answer

1. A. the probability that at least 3 ticketed passengers do not show up for the flight

P(at least 3) = 1 - P(at most 2) = 1 - (0.05 + 0 + 0.08)

2. No more than 6=0.05+0.08+0.13+0.23+0.18+0.13+0.08

4. Mean=x P(x)=189/50

Sd=sqrt(sumx^2p(x)-mean^2)=2.0424

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