3. Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. Compute the probability of at least one arrival in a 15-second period.
3. Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. Compute the probability of at least one arrival in a 15-second period.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Problem Statement:**
Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. Compute the probability of at least one arrival in a 15-second period.
**Solution Outline:**
To tackle this problem, we can use the Poisson distribution, which is generally used for modeling the number of times an event happens in a fixed interval of time or space.
1. **Convert the Time Unit:**
- Mean arrival rate per minute: \( \lambda = 10 \) passengers.
- Convert the rate to a 15-second period.
Since there are 60 seconds in a minute, the rate per 15 seconds is:
\[
\lambda_{15s} = \frac{10 \text{ passengers}}{60 \text{ seconds}} \times 15 \text{ seconds} = 2.5 \text{ passengers}
\]
2. **Poisson Probability Formula:**
The Poisson probability of observing \( k \) events in a given interval is:
\[
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
\]
where \( \lambda \) is the average number of events (in this case, 2.5), and \( k \) is the actual number of events.
3. **Probability of at Least One Arrival:**
The probability of at least one arrival is 1 minus the probability of no arrivals.
- Probability of no arrivals (k=0):
\[
P(X = 0) = \frac{e^{-2.5} \cdot 2.5^0}{0!} = e^{-2.5}
\]
- Therefore, the probability of at least one arrival:
\[
P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-2.5}
\]
4. **Compute the Exponential:**
Using the value of \( e^{-2.5} \approx 0.0821 \):
\[
P(X \geq 1) = 1 - 0.0821 = 0.9179
\]
**Conclusion:**
The probability of at least one passenger arriving in a 15-second period at the passenger](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F53c8a81f-874c-4d59-9446-0d95f9a2a58d%2Fc5e38c10-7678-4a0e-b5e0-888381ff3e64%2F5gssha3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. Compute the probability of at least one arrival in a 15-second period.
**Solution Outline:**
To tackle this problem, we can use the Poisson distribution, which is generally used for modeling the number of times an event happens in a fixed interval of time or space.
1. **Convert the Time Unit:**
- Mean arrival rate per minute: \( \lambda = 10 \) passengers.
- Convert the rate to a 15-second period.
Since there are 60 seconds in a minute, the rate per 15 seconds is:
\[
\lambda_{15s} = \frac{10 \text{ passengers}}{60 \text{ seconds}} \times 15 \text{ seconds} = 2.5 \text{ passengers}
\]
2. **Poisson Probability Formula:**
The Poisson probability of observing \( k \) events in a given interval is:
\[
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
\]
where \( \lambda \) is the average number of events (in this case, 2.5), and \( k \) is the actual number of events.
3. **Probability of at Least One Arrival:**
The probability of at least one arrival is 1 minus the probability of no arrivals.
- Probability of no arrivals (k=0):
\[
P(X = 0) = \frac{e^{-2.5} \cdot 2.5^0}{0!} = e^{-2.5}
\]
- Therefore, the probability of at least one arrival:
\[
P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-2.5}
\]
4. **Compute the Exponential:**
Using the value of \( e^{-2.5} \approx 0.0821 \):
\[
P(X \geq 1) = 1 - 0.0821 = 0.9179
\]
**Conclusion:**
The probability of at least one passenger arriving in a 15-second period at the passenger
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