
A certain airline has 167 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 13% of the time, the airline always overbooks this flight. That is, there are more passengers that have tickets on the flight than there are seats. Suppose the airline has 177 passengers booked for 167 seats. Assume one person showing up for the flight does not affect others who may, or may not, show up for this flight.What is the chance that a passenger with a flight reservation will not make it to LAX due to overbooking? Use four decimals in your answer.

To solve this problem, we can model the number of passengers who show up for the flight as a **binomial distribution**. The key is to calculate the probability that more than 167 passengers show up, leading to the possibility that a passenger won't get a seat due to overbooking.

### Step-by-step solution:

1. **Define the problem variables:**
   - \( n = 177 \): the total number of passengers with reservations.
   - \( p = 0.87 \): the probability that a passenger shows up (since 13% do not show up, 87% do).
   - \( 1 - p = 0.13 \): the probability that a passenger does not show up.

2. **Modeling as a binomial distribution:**
   The number of passengers that show up for the flight follows a binomial distribution:
   \[
   X \sim \text{Binomial}(n = 177, p = 0.87)
   \]
   We are interested in the probability that more than 167 passengers show up. That is, we need to compute \( P(X > 167) \).

3. **Approximating with the normal distribution:**
   Since \( n \) is relatively large, we can approximate the binomial distribution with a normal distribution using the central limit theorem. The mean \( \mu \) and standard deviation \( \sigma \) for the binomial distribution are given by:
   \[
   \mu = n \times p = 177 \times 0.87 = 154.29
   \]
   \[
   \sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{177 \times 0.87 \times 0.13} = 4.788
   \]

   The binomial distribution can now be approximated as:
   \[
   X \sim N(\mu = 154.29, \sigma = 4.788)
   \]

4. **Finding the probability using the normal approximation:**
   We are interested in \( P(X > 167) \). First, we convert this to a standard normal variable \( Z \):
   \[
   Z = \frac{167 - \mu}{\sigma} = \frac{167 - 154.29}{4.788} = \frac{12.71}{4.788} \approx 2.655
   \]
   Now, we look up the standard normal table to find the probability for \( Z = 2.655 \). The cumulative probability \( P(Z \leq 2.655) \) is approximately 0.9960.

   Thus, the probability that 167 or fewer passengers show up is approximately 0.9960. Therefore, the probability that more than 167 passengers show up is:
   \[
   P(X > 167) = 1 - 0.9960 = 0.0040
   \]

### Final Answer:
The probability that a passenger with a reservation will not make it to LAX due to overbooking is approximately **0.0040** (or 0.40%).

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### Relative Questions:
1. What is a binomial distribution, and when is it used?
2. How does the central limit theorem apply to binomial distributions?
3. Why can a normal distribution be used to approximate a binomial distribution in this case?
4. How do you calculate the mean and standard deviation for a binomial distribution?
5. What is the significance of using the standard normal table in this calculation?

### Tip:
When approximating a binomial distribution with a normal distribution, remember to apply a continuity correction if you're dealing with discrete data, though in this problem we skipped it for simplicity.

A certain airline has 167 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 13% of the time, the airline always overbooks this flight. That is, there are more passengers that have tickets on the flight than there are seats. Suppose the airline has 177 passengers booked for 167 seats. Assume one person showing up for the flight does not affect others who may, or may not, show up for this flight. What is the chance that a passenger with a flight reservation will not make it to LAX due to overbooking? Use four decimals in your answer.

Learn how to calculate the probability that a passenger won't make it to their destination due to overbooking on a flight. This problem involves key statistical concepts such as binomial distribution, normal approximation, and the central limit theorem, using a real-world scenario of 177 passengers booked for 167 seats. The step-by-step solution includes the use of a binomial distribution model, normal distribution approximation, and calculation of probabilities. Ideal for university-level statistics and probability learners.

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