To solve this problem, we need to calculate the probability that more than 22 passengers show up when Air-USA books 24 passengers. The probability that a booked passenger actually shows up is 86% (or 0.86).

This scenario can be modeled using the binomial distribution where:

• $n = 24$ (the total number of bookings),
• $p = 0.86$ (the probability a passenger shows up),
• We are interested in the probability that more than 22 passengers show up, which means we want to calculate $P(X > 22)$ where $X$ is the number of passengers that show up.

Step-by-step process:

1. Mean and Standard Deviation:

   • The expected number of passengers showing up (mean) $\mu = np = 24 \times 0.86 = 20.64$.
   • The standard deviation $\sigma = \sqrt{np(1-p)} = \sqrt{24 \times 0.86 \times 0.14} \approx 1.71$.

2. Use Normal Approximation to Binomial: Since $n$ is reasonably large, we can approximate the binomial distribution using the normal distribution with the calculated mean and standard deviation.

   Convert $X = 22.5$ (using continuity correction) to the standard normal variable $Z$: $Z = \frac{X - \mu}{\sigma} = \frac{22.5 - 20.64}{1.71} \approx 1.09$

3. Find the probability using the Z-score: The Z-score of 1.09 corresponds to a cumulative probability of approximately 0.8621. This means the probability that fewer than 22.5 passengers show up is 0.8621. So, the probability that more than 22 passengers show up is: $P(X > 22) = 1 - 0.8621 = 0.1379$

So, the probability that not enough seats will be available (i.e., more than 22 passengers show up) is approximately 0.1379 or 13.79%.

Answering the questions:

• Is this probability low enough so that overbooking is not a concern if you define unusual as 5% or less?

  • No, it is not low enough since 13.79% is greater than 5%.

• What about defining unusual as 10% or less?

  • No, it is still not low enough since 13.79% is greater than 10%.

Would you like more details or have any questions? Here are some related questions for further understanding:

1. How do you determine when it's appropriate to use the normal approximation to the binomial distribution?
2. What is the impact of increasing the probability of showing up on the overbooking issue?
3. How can airlines mitigate the risks of overbooking given these probabilities?
4. What is the probability that exactly 22 passengers will show up?
5. How does the choice of the threshold for "unusual" affect decision-making in practical scenarios?

Tip: Using the continuity correction improves the accuracy when using the normal approximation to the binomial distribution.