Question

Please answer a-c

Suppose that the probability that a passenger will miss a flight is 0.09850. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flights because passengers must be "bumped" from the flight. Suppose that an airplane has a seating capacity of 56 passengers.

(a) If 5858 tickets are sold, what is the probability that 57 or 58 passengers show up for the flight resulting in an overbooked flight?

(b) Suppose that 62 tickets are sold. What is the probability that a passenger will have to be "bumped"?

(c) For a plane with seating capacity of 55 passengers, how many tickets may be sold to keep the probability of a passenger being "bumped" below

55 %?

Answer 1

Ans:

a)

Use binomial distribution with n=58 and p=1-0.0985=0.9015

(p=probability that a passenger shows up for the flight)

P(overbooked)=P(x>56)

=P(x=57)+P(x=58)

=58C₅₇*0.9015^57*0.0985+0.9015^58

=0.0179

b)Use binomial distribution with n=62 and p=1-0.0985=0.9015

P(overbooked or bumped)=P(x>56)

=1-P(x<=56)

=1-binomdist(56,62,0.9015,true)

=0.4195

c)

P(bumped)=P(X>55)=1-P(X<=55)

=1-binomdist(55,n,0.9015,true)

n	P(bumped)
56	0.0030
57	0.0196
58	0.0661

Number of tickets sold should be 57 to keep the probability of a passenger being "bumped" below 5%.