Question

Airlines and hotels often grant reservations in excess of capacity to minimize losses due to no-shows. Suppose the records of a hotel show that, on the average, 10% of its prospective guests will not claim their reservations. If the hotel accepts 254 reservations and there are only 230 rooms in the hotel, what is the probability that all guests who arrive to claim a room will receive one? (Use the normal approximation. Round your answer to four decimal places.)

Answer 1

P(Do not show up) = 0.10

P(Show up ) = 1 - 0.10 = 0.90 (Probability of success)

We have to calculate all guests who arrive to claim a room, that is guests should be at most 230.

P( X <= 230) = ?

Given, n = 254 , p = 0.90

Mean = np = 254 * 0.90 = 228.6

Standard deviation = sqrt(np(1-p) )

= sqrt(254 * 0.90 * 0.10)

= 4.7812

Using normal approximation to binomial distribution with continuity correction,

P( X <= x) = P( Z < x+0.5 - Mean / Standard deviation )

So,

P( X <= 230) = P( Z < 230.5 - 228.6 / 4.7812)

= P( Z < 0.3974)

= 0.6545