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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Would you like more details on any part of this solution?

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.