This problem is based on a Poisson distribution model for the number of people entering a security check-in lineup at an airport. Let me guide you through the solution step by step.

Part (a):

The Poisson distribution is defined as: $P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}, \quad x = 0, 1, 2, \dots$ Where $\lambda$ is the rate parameter, which is the average number of occurrences in a given time period. Here, the value 15 represents the rate parameter $\lambda$, which is the average number of people entering the security check-in lineup every 15 minutes. Thus, the most appropriate answer is:

• C. 15 represents the average number of people who enter the security check-in lineup every 15 minutes.

Part (b):

You need to compute the probability that exactly 18 people enter the security check-in lineup in a 15-minute interval. Using the Poisson formula: $P(X = 18) = \frac{e^{-15} 15^{18}}{18!}$

Let's compute this:

$P(X = 18) = \frac{e^{-15} \cdot 15^{18}}{18!}$ We will calculate this numerically.

Part (c):

For this part, you are asked to compute the probability that at least 5 people enter the security check-in lineup in a 5-minute interval.

Since the Poisson distribution for the entire 15-minute period has an average rate of 15, we scale it down for a 5-minute interval. The new rate $\lambda$ for 5 minutes is:

$\lambda = \frac{15}{3} = 5$

So, we need to calculate: $P(X \geq 5) = 1 - P(X \leq 4)$ Where $P(X \leq 4)$ is the cumulative probability for $X = 0$ to $X = 4$. This involves summing the probabilities from the Poisson distribution formula for $X = 0$ to $X = 4$ with $\lambda = 5$.

Part (d):

You're asked to compute the probability that the number of people entering the security lineup is 17, given that the number lies between 15 and 18 (inclusive). This is a conditional probability problem.

We need to compute: $P(X = 17 \mid 15 \leq X \leq 18) = \frac{P(X = 17)}{P(15 \leq X \leq 18)}$

Where: $P(15 \leq X \leq 18) = P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)$

We will compute these individual probabilities using the Poisson distribution with $\lambda = 15$.

I'll now calculate the values for Parts (b), (c), and (d). Let me work through the math.Here are the results of the calculations:

Part (b):

The probability that 18 people enter the security check-in lineup in a 15-minute interval is: $P(X = 18) = 0.0706$

Part (c):

The probability that at least 5 people enter the security check-in lineup in a 5-minute interval is: $P(X \geq 5) = 0.5595$

Part (d):

The conditional probability that exactly 17 people entered the security lineup, given that the number lies between 15 and 18 (inclusive), is: $P(X = 17 \mid 15 \leq X \leq 18) = 0.2395$

Would you like any further details or clarifications? Here are five related questions that may help expand your understanding:

1. How is the Poisson distribution applied to different time intervals?
2. What happens to the Poisson distribution as the average rate (λ) increases?
3. How is conditional probability calculated in more complex distributions?
4. What is the significance of calculating cumulative probabilities in Poisson distribution problems?
5. How can you apply the Poisson distribution to real-life scenarios like queues in customer service?

Tip: In the Poisson distribution, always check if the rate parameter $\lambda$ needs to be adjusted when considering different time intervals.