The problem you've provided is based on the scenario of overbooked flights, where the airline expects a certain proportion of people to not show up. This can be modeled using the binomial distribution, given that each passenger independently shows up for the flight with a certain probability. Let's go through it step by step.

Given Information:

• The airline has 165 seats available on the flight.
• 175 passengers have booked tickets.
• The probability of a person not showing up is 11%. Thus, the probability of a person showing up is: $p = 1 - 0.11 = 0.89$

Part (a) - Expected Number and Standard Deviation of People Showing Up:

We can treat this as a binomial problem with the following parameters:

• n = 175 (total passengers)
• p = 0.89 (probability of a passenger showing up)

The expected number of people showing up for the flight is given by: $E(X) = n \cdot p = 175 \cdot 0.89 = 155.75$

The standard deviation is given by the formula for binomial distribution: $\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{175 \cdot 0.89 \cdot 0.11} = \sqrt{17.1575} \approx 4.14$

So, the answers for part (a) are:

• Expected number of people: 155.75
• Standard deviation: 4.14

Part (b) - Probability that 6 Seats are Empty:

We want to calculate the probability that 6 seats are empty, meaning that exactly 159 passengers show up (since 165 - 6 = 159).

We need to calculate the binomial probability: $P(X = 159) = \binom{175}{159} \cdot (0.89)^{159} \cdot (0.11)^{16}$ This can be computed using statistical software or a calculator capable of handling binomial probabilities.

For the sake of time, I'll compute this using Python.

Part (c) - Chance of Overbooking:

The flight has 165 seats. If more than 165 people show up, the flight will be overbooked. So we need to calculate the probability that more than 165 people show up: $P(X > 165) = 1 - P(X \leq 165)$ This can also be computed using a cumulative binomial distribution function.

I'll compute both parts (b) and (c) and share the results shortly.### Results:

• Part (b): The probability that exactly 6 seats are empty (i.e., 159 passengers show up) is approximately 0.0752 or 7.52%.

• Part (c): The probability that the flight is overbooked (i.e., more than 165 passengers show up) is approximately 0.0054 or 0.54%.

If you'd like more details or have any questions, feel free to ask!

Here are 5 related questions to explore further:

1. What happens to the probability of overbooking if more passengers book tickets?
2. How would the results change if the no-show rate was 20% instead of 11%?
3. What is the probability that exactly 165 passengers show up for the flight?
4. How does the binomial distribution change if the flight involves multiple stopovers?
5. Can this binomial model be approximated by a normal distribution for large flights?

Tip: When working with binomial probabilities for large $n$, consider using the normal approximation for quicker calculations!