Question

An airline knows that in the long run only 90% of passengers who book a seat show up for their flight. On a particular flight with 193 seats there are 225 reservations. Let X denote the number of passengers that show up. (a) (4 pts) Identify the probability distribution of X by name and the parameters needed. (b) (6 pts) Assuming passengers make independent decisions, what is the exact probability that the flight will be over-booked? Do not simplify your answer! (c) (8 pts) Approximate the probability in the previous part using an appropriate distribution. Justify your approximation, and find the numerical value of that.

Answer 1

a) The number of people who show up out of the 225 reservations could be modelled here as a binomial distribution given as:

Therefore it is a binomial distribution with parameters: n = 225 and probability of showing up, p = 0.9

b) The exact probability that the flight will be overbooked is computed here as:
P(X > 193) as there are only 193 seats on plane

= 1 - P(X <= 193)

This is computed in EXCEL as:
=1-binom.dist(193,225,0.9,TRUE)

0.9729 is the output here.

Therefore 0.9729 is the required probability here.

c) The binomial distribution is approximated to a normal distribution here as:

The required probability here is:
P(X > 193)

Applying the continuity correction, we have here:
P(X > 193.5)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal tables, we have here:

Therefore 0.9772 is the required probability here.