Airline overbooking
For any given flight, an airline tries to sell as many tickets as possible. Suppose that on average, 20% of ticket holders fail to show up, all independent of one another. Knowing this, an airline will sell more tickets than there are seats available (i.e., overbook the flight) and hope that there is a sufficient number of ticket holders who do not show up, to compensate for its overbooking. Using the Central Limit Theorem, determine n, the maximum number of tickets an airline can sell on a flight with 400 seats so that it can be approximately 99% confident that all ticket holders who do show up will be able to board the plane. Use the de Moivre-Laplace 1/2-correction in your calculations. Hint: You may have to solve numerically a quadratic equation.
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To determine the maximum number of tickets an airline can sell, we need to calculate the confidence interval using the Central Limit Theorem and the de Moivre-Laplace 1/2-correction.
First, let's calculate the mean and standard deviation of the number of ticket holders who do show up on a given flight. We know that, on average, 20% of ticket holders fail to show up, so the probability of a ticket holder showing up is 1 - 0.20 = 0.80.
Let X be the number of ticket holders who show up. X follows a binomial distribution with parameters n (number of tickets sold) and p (probability of showing up). The mean of X is given by λ = np, and the standard deviation of X is given by σ = sqrt(np(1-p)).
For our problem, n is the number of tickets sold, which is what we want to determine. We know p = 0.80, and the number of seats available is 400.
λ = np = 400 * 0.80 = 320
σ = sqrt(np(1-p)) = sqrt(400 * 0.80 * (1 - 0.80)) = sqrt(64) = 8
Next, we can use the Central Limit Theorem to approximate the distribution of X as a normal distribution. Since n is large (the number of tickets sold), we can use the normal distribution to calculate the confidence interval.
The confidence interval can be calculated using the formula:
CI = λ ± z * (σ/sqrt(n))
Where CI is the confidence interval, λ is the mean, z is the z-score corresponding to the desired confidence level, σ is the standard deviation, and n is the number of tickets sold.
For our problem, we want to be 99% confident that all ticket holders who show up can board the plane. This means we need to find the z-score corresponding to a 99% confidence level.
Using a standard normal distribution table or a calculator, we can find the z-score corresponding to a 99% confidence level is approximately 2.576.
Plugging in the values into the formula, we have:
2.576 * (8/sqrt(n)) = 400 - 320
Solving this equation numerically, we find that n is approximately 414.
Therefore, the maximum number of tickets the airline can sell on a flight with 400 seats to be approximately 99% confident that all ticket holders who do show up will be able to board the plane is 414 (rounded up).
To determine the maximum number of tickets an airline can sell on a flight with 400 seats, we need to use the Central Limit Theorem. The Central Limit Theorem states that the distribution of the sample mean approaches a normal distribution as the sample size increases.
Let's denote the maximum number of tickets the airline can sell as n. The proportion of ticket holders who do not show up is 20%, which means that 80% of ticket holders will show up on average.
We want to be approximately 99% confident that all ticket holders who do show up will be able to board the plane. This means that we need to determine the critical value for a 99% confidence interval.
Since the distribution is approximately normal, we can use the inverse standard normal distribution to find the critical value. The critical value for a 99% confidence interval is approximately 2.576.
Now, using the de Moivre-Laplace 1/2-correction, we have:
1/2 - sqrt(n * 0.8 * 0.2 / 400) = -2.576
Simplifying the equation, we have:
sqrt(n * 0.8 * 0.2 / 400) = 1/2 + 2.576
Squaring both sides of the equation, we have:
n * 0.8 * 0.2 / 400 = (1/2 + 2.576)^2
Solving for n, we have:
n = (0.128 * 400) / (0.8 * (1/2 + 2.576)^2)
n = 51.2 / (0.8 * (1/2 + 2.576)^2)
n = 51.2 / (0.8 * 3.576^2)
n ≈ 51.2 / 9.185
n ≈ 5.57
Therefore, the maximum number of tickets the airline can sell on a flight with 400 seats is 5. Note that selling fractions of tickets is not possible, so the airline can sell a maximum of 5 tickets in this case.