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Let's break down the problem step by step to find the correct answers:

We have the following details:
- A flight from Calgary (YYC) to Los Angeles (LAX) with **168 seats available**.
- The airline overbooks by **178 passengers**.
- There's an **11% chance** that a person with a ticket does **not show up**.

Let the number of passengers who actually show up be a **binomial random variable**, since each passenger has a probability of showing up or not, independent of others.

The **probability of showing up** = \( p = 1 - 0.11 = 0.89 \).

Let’s solve each part carefully:

---

### (a) Expectation and Standard Deviation

We are asked to find:
- The **expected number** of passengers showing up.
- The **standard deviation** of the number of passengers showing up.

Since this is a binomial distribution \( X \sim \text{Binomial}(n=178, p=0.89) \):

- The **expected number** of people showing up is given by:
  \[
  E[X] = n \cdot p = 178 \cdot 0.89
  \]
  \[
  E[X] = 158.42
  \]
  So, **the expected number of passengers showing up is 158.42**.

- The **standard deviation** for a binomial distribution is given by:
  \[
  \sigma[X] = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{178 \cdot 0.89 \cdot 0.11}
  \]
  \[
  \sigma[X] = \sqrt{17.4578} \approx 4.18
  \]
  So, **the standard deviation is 4.18**.

### (b) Probability that 6 seats are empty

For 6 seats to be empty, **162 passengers** need to show up (since there are 168 seats, 168 - 6 = 162).

We need to calculate \( P(X = 162) \), where \( X \) is binomial with parameters \( n = 178 \) and \( p = 0.89 \).

The binomial probability mass function (PMF) is given by:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
In this case:
\[
P(X = 162) = \binom{178}{162} (0.89)^{162} (0.11)^{16}
\]

To calculate this probability:
\[
P(X = 162) = \binom{178}{162} (0.89)^{162} (0.11)^{16}
\]

Using a calculator for binomial probability:
\[
P(6 \text{ seats empty}) \approx 0.0559
\]

So, **the probability is approximately 0.0559** (to four decimal places).

### (c) Probability that a passenger won’t make the flight due to overbooking

This is the probability that **more than 168 passengers show up**, which means **169 or more passengers** will try to board, but only 168 seats are available.

We want \( P(X \geq 169) \), which is:
\[
P(X \geq 169) = 1 - P(X \leq 168)
\]

We sum the binomial probabilities for \( X = 0 \) to \( X = 168 \) and subtract the result from 1.

Using cumulative binomial probabilities:
\[
P(X \leq 168) \approx 0.9964
\]
Thus,
\[
P(X \geq 169) = 1 - 0.9964 = 0.0036
\]

So, **the probability that a passenger will not make it is approximately 0.0036**.

---

### Final Answers:
- **(a)** Expected number of passengers showing up: **158.42**, Standard deviation: **4.18**.
- **(b)** Probability of 6 seats being empty: **0.0559**.
- **(c)** Probability that a passenger won’t make it due to overbooking: **0.0036**.

Let me know if you need further clarification or want to explore any details!

---

Here are 5 related questions to expand your understanding:

1. What is the probability that exactly 160 passengers will show up?
2. If the no-show probability changes to 15%, how does this affect the expected number of passengers?
3. How does the airline’s overbooking policy affect customer satisfaction?
4. What is the probability that more than 170 passengers show up?
5. What is the impact of reducing the number of booked passengers to 175 on the probability of overbooking?

**Tip:** In binomial distributions, the expected value is always \( np \), and the standard deviation depends on both \( p \) and \( 1-p \), showing how variability changes with the success probability.

A certain airline has 168 seats available for a flight from YYC (Calgary International Airport) to LAX (Los Angeles International Airport). Because people with reservations do not show up for their flight 11% of the time, the airline always overbooks this flight. Suppose the airline has 178 passengers booked for 168 seats. Assume one person showing up for the flight does not affect others who may or may not show up. (a) How many people (with tickets) does the airline expect to show up for the flight? (b) What is the probability that there will be 6 seats empty? (c) What is the chance that a passenger will not make it to LAX due to overbooking?

Learn how to calculate expected passengers, empty seats, and overbooking risks for a flight from Calgary to Los Angeles using binomial distribution and probability. This problem demonstrates key statistical concepts like expected value, standard deviation, and cumulative binomial probabilities. Suitable for high school and college-level students.

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