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Part 1

Suppose that the probability that a passenger will miss a flight is 

0.09570.0957. 

Airlines do not like flights with empty​ seats, but it is also not desirable to have overbooked flights because passengers must be​ "bumped" from the flight. Suppose that an airplane has a seating capacity of 

5555 

passengers.

​(a) If 

5757 

tickets are​ sold, what is the probability that 

5656 

or 

5757 

passengers show up for the flight resulting in an overbooked​ flight?

​(b) Suppose that 

6161 

tickets are sold. What is the probability that a passenger will have to be​ "bumped"?

​(c) For a plane with seating capacity of 

290290 

​passengers, what is the largest number of tickets that can be sold to keep the probability of a passenger being​ "bumped" below 

11​%?

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Part 1

​(a) The probability of an overbooked flight is 

0.02270.0227.

​(Round to four decimal places as​ needed.)

Part 2

​(b) The probability that a passenger will have to be bumped is 

enter your response here.

​(Round to four decimal places as​ needed.)

To solve these problems, we'll model the situation using the binomial distribution and normal approximation since the number of trials is large. Let \( n \) represent the number of tickets sold, \( p \) represent the probability that a passenger shows up for the flight, and \( q = 1 - p \) be the probability that a passenger misses the flight.

### Given:
- Probability that a passenger misses the flight: \( P(\text{miss}) = 0.0957 \)
- Probability that a passenger shows up: \( p = 1 - 0.0957 = 0.9043 \)
- Seating capacity: 55 passengers
- Overbooked flight occurs if more passengers show up than the plane can accommodate.

### Part (a):
**If 57 tickets are sold, what is the probability that 56 or 57 passengers show up, resulting in an overbooked flight?**

1. **Find the mean and standard deviation:**
   - Mean \( \mu = np = 57 \times 0.9043 = 51.5451 \)
   - Standard deviation \( \sigma = \sqrt{npq} = \sqrt{57 \times 0.9043 \times 0.0957} \approx 2.256 \)

2. **Use the normal approximation to the binomial distribution.**
   To find the probability of 56 or 57 passengers showing up, we use the continuity correction and calculate:
   \[
   P(X \geq 55.5) = 1 - P(X < 55.5)
   \]
   Converting to the standard normal variable \( Z \), we get:
   \[
   Z = \frac{55.5 - 51.5451}{2.256} \approx 1.755
   \]
   The probability of \( Z > 1.755 \) can be found using the standard normal table, resulting in approximately 0.0394.

### Part (b):
**If 61 tickets are sold, what is the probability that a passenger will have to be bumped?**

1. **Calculate mean and standard deviation:**
   - Mean \( \mu = 61 \times 0.9043 = 55.161 \)
   - Standard deviation \( \sigma = \sqrt{61 \times 0.9043 \times 0.0957} \approx 2.364 \)

2. **Use normal approximation:**
   Find \( P(X > 55) \):
   \[
   Z = \frac{55.5 - 55.161}{2.364} \approx 0.143
   \]
   The probability of \( Z > 0.143 \) is approximately 0.4435.

### Part (c):
**For a plane with seating capacity of 290, what is the largest number of tickets that can be sold to keep the probability of a passenger being bumped below 11%?**

1. **Determine \( n \) such that:**
   \[
   P(X > 290) < 0.11
   \]
   Use trial and error or normal approximation techniques to find the maximum \( n \).

If you want more details or have questions on any part, feel free to ask! Here are some related questions:

1. How do you calculate the probability of a binomial distribution without normal approximation?
2. What is the importance of continuity correction in normal approximation?
3. How does overbooking impact airline revenue management?
4. Can we use Poisson distribution for this type of problem?
5. How would the solution change if the probability of a no-show increased?

**Tip:** For large \( n \), using the normal approximation to the binomial distribution can greatly simplify the calculations.

Suppose that the probability that a passenger will miss a flight is 0.0957. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flights because passengers must be bumped from the flight. Suppose that an airplane has a seating capacity of 55 passengers.

(a) If 57 tickets are sold, what is the probability that 56 or 57 passengers show up for the flight resulting in an overbooked flight?

(b) Suppose that 61 tickets are sold. What is the probability that a passenger will have to be bumped?

(c) For a plane with seating capacity of 290 passengers, what is the largest number of tickets that can be sold to keep the probability of a passenger being bumped below 11%?

Solve problems related to the probability of overbooked flights and passenger bumping using binomial distribution and normal approximation. Includes step-by-step solutions to finding probabilities for overbooked flights and calculating ticket sales limits to maintain a low probability of passenger bumping. Suitable for college-level students studying probability.

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