Part 1: Calculating the Probability

To determine the probability that at least 165 passengers will show up, we use the normal approximation to the binomial distribution.

Step 1: Define the binomial distribution parameters

• The number of tickets sold (trials, $n$) = 176
• Probability that a passenger shows up ($p$) = 0.91 (since 9% don't show up, 91% do)
• The number of passengers that show up is binomial: $X \sim \text{Binomial}(n = 176, p = 0.91)$.

Step 2: Find the mean and standard deviation of the binomial distribution

• Mean $\mu = n \cdot p = 176 \cdot 0.91 = 160.16$
• Standard deviation $\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{176 \cdot 0.91 \cdot 0.09} \approx 3.97$

Step 3: Apply the normal approximation

We can use the normal approximation to estimate the probability of at least 165 passengers showing up:

• Standardize the value $X = 165$ using a continuity correction. The corrected value is 164.5.
• $Z = \frac{164.5 - 160.16}{3.97} \approx 1.09$

Step 4: Use the normal distribution

The probability of getting a $Z$-score of 1.09 or higher can be found using a standard normal table or calculator:

• $P(X \geq 165) = P(Z \geq 1.09) \approx 0.137$

Thus, the probability of at least 165 passengers showing up is approximately 0.127, or 12.7%.

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Part 2: Should the airline change the number of tickets they sell?

The answer to this depends on how the airline evaluates the risk of bumping passengers vs. the cost of empty seats.

• Option A is the most appropriate:
  "The probability is fairly low, so it is likely that they should not change the number of tickets they sell. However, the decision also depends on the relative costs of not selling seats and bumping passengers."

  This option acknowledges that the probability (12.7%) of having more passengers than seats is relatively low but also recognizes that other factors, like the cost of bumping passengers and the cost of empty seats, should be considered before making a decision.

Conclusion:

• Part 1: The probability of at least 165 passengers showing up is 0.127.
• Part 2: The airline should consider both the probability and the relative costs, making Option A the best response.

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Would you like a breakdown of how we use normal approximation in more detail or help with another part of the problem?

Follow-up questions:

1. What is the effect of selling more or fewer tickets on this probability?
2. How would changing the no-show rate (e.g., 8% or 10%) affect the result?
3. What are the implications of using a binomial approximation vs. the exact binomial probability in this case?
4. How might the costs of bumping passengers compare to leaving seats unsold?
5. How would you model this problem if the airline increased its fleet to larger planes?

Tip:

Whenever using normal approximation for binomial probabilities, don't forget to apply the continuity correction!