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The problem you have uploaded involves a Poisson probability model, which is typically used to model the number of events (in this case, people entering a security check-in lineup) occurring within a fixed interval of time.

Let's break down the parts:

### Part (a) Explanation:
The parameter 19.8 in the Poisson model is \(\lambda\), the expected or average number of people entering the security check-in lineup in a 15-minute interval. So, the correct explanation is:
- **C.** 19.8 represents the average number of people who enter the security check-in lineup every 15 minutes.

### Part (b) - Compute \(P(X = 18)\):
The probability mass function (PMF) of a Poisson distribution is given by:
\[
P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}
\]
For \(\lambda = 19.8\) and \(x = 18\), we can compute:
\[
P(X = 18) = \frac{e^{-19.8} \times 19.8^{18}}{18!}
\]
Let me calculate this for you.

### Part (c) - Probability for 5 minutes:
For this, we need to scale the mean \(\lambda\) to the 5-minute interval. Since 5 minutes is one-third of 15 minutes, we divide 19.8 by 3:
\[
\lambda_{5 \text{min}} = \frac{19.8}{3} = 6.6
\]
Now, we need to calculate the probability that at least 5 people enter, \(P(X \geq 5)\). This is given by:
\[
P(X \geq 5) = 1 - P(X \leq 4)
\]
Where \(P(X \leq 4)\) can be found by summing the probabilities for \(X = 0, 1, 2, 3, 4\).

### Part (d) - Probability that between 13 and 19 people entered:
This is the probability that \(13 \leq X \leq 19\), which is:
\[
P(13 \leq X \leq 19) = P(X = 13) + P(X = 14) + \dots + P(X = 19)
\]
Each of these probabilities can be computed using the Poisson PMF formula with \(\lambda = 19.8\).

Let me calculate all these parts for you.Here are the results for the computations:

### Part (b):
The probability that exactly 18 people enter the security check-in lineup in a 15-minute interval is:
\[
P(X = 18) = 0.0860
\]

### Part (c):
The probability that at least 5 people will enter the security check-in lineup in a 5-minute interval is:
\[
P(X \geq 5) = 0.7873
\]

### Part (d):
The probability that between 13 and 19 people, inclusive, entered the security lineup in a 15-minute interval is:
\[
P(13 \leq X \leq 19) = 0.4454
\]

Let me know if you'd like further details or explanations on any part!

Here are some related questions to deepen your understanding:
1. How would the results change if the average number of people entering the lineup per 15 minutes was 25 instead of 19.8?
2. What is the probability that no people enter the lineup in a 5-minute interval?
3. How can you adjust the Poisson distribution model for intervals longer than 15 minutes?
4. What is the variance of the Poisson distribution, and how does it relate to the average?
5. Can you find the cumulative probability for more than 20 people entering in 15 minutes?

**Tip:** In a Poisson distribution, the mean and variance are the same, which helps in understanding the spread of the distribution around the mean.

The number of people entering a security check-in lineup in a 15-minute interval at a medium-sized airport can be modeled by the following probability model: P(X=x) = (e^(-19.8) * (19.8^x)) / x!. X = 0, 1, 2, … What does 19.8 represent in the probability model? Compute the probability that 18 people enter the lineup in a 15-minute interval and that at least 5 people will enter in a 5-minute interval. Also compute the probability that between 13 and 19 people entered the security lineup in a 15-minute interval.

Learn how to model the number of people entering a security check-in lineup at a medium-sized airport using a Poisson probability model with an average rate of 19.8. This problem covers the Poisson distribution formula and computes probabilities for 15-minute and 5-minute intervals.

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