Because not all airline passengers show up for their reserve

Because not all airline passengers show up for their reserved seat, an airline sells 127 tickets for a flight that holds only 121 passengers. The probability that a passenger does not show up is 0.16, and the passengers behave independently. Round your answers to two decimal places (e.g. 98.76).

(a) What is the probability that every passenger who shows up gets a seat?
(b) What is the probability that the flight departs with empty seats?
(c) What are the mean and (d) standard deviation of the number of passengers who show up?

Solution

Because not all airline passengers show up for their reserved seat, an airline sells 127 tickets for a flight that holds only 121 passengers. The probability that a passenger does not show up is 0.16, and the passengers behave independently. Round your answers to two decimal places (e.g. 98.76).

(a) What is the probability that every passenger who shows up gets a seat?

Here, we have to find the probability that every passenger who shows up gets a seat.

We are given,

Total passengers = 121

Total tickets sold = 127

Required probability = 121 / 127 = 0.95

(b) What is the probability that the flight departs with empty seats?

Here, we have to find the probability that the flight departs with empty seats.

The probability that a passenger does not show up is given as 0.16.

Here, we have n=121 and p = 0.16

Required probability = p^n = (0.16)^121 = 0.00 approximately.

(c) What are the mean and (d) standard deviation of the number of passengers who show up?

Here we have to find the mean and standard deviation.

Probability for passengers who not show up = 0.16

So, probability for passengers who show up = 1 – 0.16 = 0.84

The formula for mean is given as below:

Mean = n*p = 121*0.84 = 101.64

Standard deviation = sqrt(npq)

Standard deviation = sqrt(121*0.84*0.16) = 4.03