# For any given flight, an airline tries to sell as many tickets as possible. Suppose that on average, 10% 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 should sell on a flight with 300 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.

## To determine the maximum number of tickets an airline should sell on a flight with 300 seats, let's work through the problem step by step.

1. First, we need to find the proportion of ticket holders who are likely to show up. Given that, on average, 10% of ticket holders fail to show up, the proportion of ticket holders who do not show up is 0.10.

2. To determine the proportion of ticket holders who show up, subtract the proportion who do not show up from 1: 1 - 0.10 = 0.90.

3. Using the Central Limit Theorem, we can assume that the distribution of the number of ticket holders who show up follows a normal distribution with mean np and variance np(1-p), where n is the number of tickets sold and p is the proportion of ticket holders who show up.

4. We want to be approximately 99% confident that all ticket holders who show up will be able to board the plane. This means we want to find the value of n such that the probability of more than 300 ticket holders showing up is less than or equal to 0.01.

5. To apply the de Moivre-Laplace 1/2-correction, we adjust the boundaries of our normal distribution by adding or subtracting 0.5: (300 + 0.5) and (300 - 0.5).

6. The probability of more than 300 ticket holders showing up can be found using a normal distribution table or calculator. We calculate this probability using the adjusted boundaries (300 - 0.5) and (300 + 0.5), the mean np (which is 300 * 0.90), and the variance np(1-p).

7. Let's set up the equation using the cumulative distribution function (CDF) for a normal distribution: P(X > 300) = 1 - P(X ≤ 300), where X is the number of ticket holders who show up.

8. The equation becomes: 1 - P(X ≤ 300) = 0.01.

9. Solving this equation numerically for n, the number of tickets sold, will give us the maximum number of tickets the airline should sell.

To solve the equation numerically, you can use methods such as the Newton-Raphson method or trial and error by calculating the CDF for different values of n until you find the closest value that satisfies the equation.

Alternatively, you can use statistical software or online calculators that provide the option to solve this equation for you.

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