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, 14% of their prospective guests will not claim their reservation. The hotel accepted 215 reservations and there are only 200 rooms in the hotel. You are to find out the approximate probability that all guests who arrive to claim a room will receive one.

a) Let x be the number of guests who arrive to claim a room out of the 215 people who reserved a room in the hotel. What kind of distribution does x follow?

b) Find the mean and standard deviation of x.

c) Approximate the probability that all guests who arrive to claim a room will receive one.

Answer 1

The distribution followed by the number of guests who arrive to claim a room out of the 215 people who reserved a room in the hotel is the binomial distribution. The mean and standard deviation of x are approximately 185.9 and 8.94, respectively. The approximate probability that all guests who arrive to claim a room will receive one is approximately 0.9438.

Step-by-step explanation:

a) The distribution followed by the number of guests who arrive to claim a room out of the 215 people who reserved a room in the hotel is the binomial distribution.
b) To find the mean and standard deviation of the random variable x, we use the formulas:
Mean (μ) = n * p = 215 * 0.86 ≈ 185.9
Standard Deviation (σ) = sqrt(n * p * (1 - p)) = sqrt(215 * 0.86 * 0.14) ≈ 8.94
c) To approximate the probability that all guests who arrive to claim a room will receive one, we can use the normal approximation to the binomial distribution. We can calculate the z-score using the formula: z = (x - μ) / σ = (200 - 185.9) / 8.94 ≈ 1.579. We can then use a standard normal distribution table or calculator to find the probability associated with the z-score of 1.579, which is approximately 0.9438.