Question

In: Statistics and Probability

An airline knows that in the long run only 90% of passengers who book a seat...

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.

Expert Solution

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.

orchestra answered 2 years ago

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