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A binomial distribution models the number of successes in a fixed number of independent trials where the outcome of each trial can be either success or failure. In this problem, a success is defined as a passenger showing up for their reserved spot.
The important parameters in a binomial distribution are:

• Number of trials ()
• Probability of success in each trial ()

Here, the number of trials () is the total number of reservations made and the probability of success () is the probability that a passenger will show up, which is 0.9005.
We denote this distribution as: \(X ~ Binomial(n, 0.9005)\).

When dealing with a large number of trials in a binomial distribution, calculating probabilities can become cumbersome. Thankfully, we can use the normal approximation to simplify the process.
The normal approximation works well when the number of trials () is large, and both \(np\) and \(n(1−p)\) are greater than 5.

The binomial distribution \(Binomial(n, p)\) can be approximated by a normal distribution \(Normal(\mu, \sigma^2)\) where mean (\(\mu\)) and standard deviation (\(\sigma\)) are computed as follows:

• The mean \(\mu = np\)
• The standard deviation \(\sigma = \sqrt{np(1−p)}\)

For our problem, \(np = n * 0.9005\) and \(n(1-0.9005)\). This simplifies our calculations significantly.

Once we approximate the binomial distribution with a normal distribution, we often need to convert it to a standard normal distribution to use standard normal tables.
The standard normal distribution has a mean of 0 and a standard deviation of 1. To convert a normal random variable \( X \) to a standard normal random variable \( Z \), we use the formula:

\( Z = \frac{X - \mu}{\sigma}\).
The value \( Z \) allows us to use standard normal distribution tables to find probabilities.
In our problem, we need to calculate \( P(X \leq 213) \), where \( X ~ Normal(\mu, \sigma) \). We convert this to the standard normal variable and use the z-value corresponding to a 0.95 probability, which is approximately 1.645.

In probability and statistics, the mean and standard deviation are crucial for understanding distributions.

• The mean (\(\mu\)) provides the average outcome we can expect. For a binomial distribution, \( \mu = np \).
• The standard deviation (\( \sigma \)) quantifies the variability or dispersion of the outcomes. For a binomial distribution, \( \sigma = \sqrt{np(1−p)} \).

The mean tells us where the center of the data is, while the standard deviation gives us an idea of how spread out the data points are around the mean.
For the ticket reservation problem, the mean number of passengers showing up is computed as \( \mu = n * 0.9005 \). The standard deviation is calculated as \( \sigma = \sqrt{n * 0.9005 * 0.0995} \).

In a ticket reservation system, airlines often overbook flights to account for no-shows. No-show rates need to be carefully considered in the booking strategy.
Here, the probability that a passenger does not show up is 0.0995. Given the airplane capacity of 213 seats, airlines must determine the maximum number of reservations (\( n \)) to ensure that the probability of accommodating all passengers who show up is at least 0.95.
Using probabilities and distribution approximations, airlines can manage overbooking risks. By solving for \( n \) using the computed mean, standard deviation, and standard normal distribution, the airline ensures that the chance of excess passengers is minimal.