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Suppose small aircraft arrive at an airport according to a Poisson process with rate \(\alpha=8 / \mathrm{h}\), so that the number of arrivals during a time period of \(t\) hours is a Poisson rv with parameter \(\hat{\lambda}=8 t\). a. What is the probability that exactly 6 small aircraft arrive during a 1 -h period? At least 6 ? At least 10 ? b. What are the expected value and standard deviation of the number of small aircraft that arrive during a 90 -min period? c. What is the probability that at least 20 small aircraft arrive during a \(2 \frac{1}{2} h\) period? That at most 10 arrive during this period?

Short Answer

Expert verified
a) P(exactly 6) ≈ 0.122, P(at least 6) ≈ 0.687, P(at least 10) ≈ 0.332; b) E(X) = 12, SD(X) ≈ 3.464; c) P(at least 20) ≈ 0.441, P(at most 10) ≈ 0.010.

Step by step solution

01

Define the Poisson Distribution Parameter

For part (a), the parameter for the Poisson distribution during a 1-hour period is given by \( \hat{\lambda} = 8 \times 1 = 8 \). Therefore, \( \lambda = 8 \).
02

Probability of Exactly 6 Arrivals

Using the Poisson probability formula \( P(X = k) = \frac{{\lambda^k e^{-\lambda}}}{k!} \), we calculate \( P(X = 6) = \frac{8^6 e^{-8}}{6!} \).
03

Calculate the Probability for 6 Arrivals

First calculate: \( 8^6 = 262144 \) and \( 6! = 720 \). Then, \( P(X = 6) = \frac{262144 \cdot e^{-8}}{720} \approx 0.122 \).
04

Probability of At Least 6 Arrivals

To find \( P(X \geq 6) \), use the complement rule: \[ P(X \geq 6) = 1 - P(X < 6) = 1 - \sum_{k=0}^{5} P(X = k) \].
05

Calculate the Probabilities for Less Than 6 Arrivals

Calculate \( P(X = k) \) for \( k = 0, 1, 2, 3, 4, 5 \) and sum them: \[P(X = 0) + P(X = 1) + ... + P(X = 5) \approx 0.313 \].
06

Find Probability for At Least 6 Arrivals

\( P(X \geq 6) = 1 - 0.313 = 0.687 \)
07

Probability of At Least 10 Arrivals

Use the same complement method: \( P(X \geq 10) = 1 - P(X < 10) = 1 - \sum_{k=0}^{9} P(X = k) \approx 0.332 \).
08

Expected Value and Standard Deviation for 90-minute Period

Convert 90 minutes to hours: \( t = 1.5 \ \text{hours} \). Then set \( \lambda = 8 \times 1.5 = 12 \). The expected value is \( 12 \) and the standard deviation is \( \sqrt{12} \approx 3.464 \).
09

Probability of At Least 20 Arrivals in 2.5 Hours

Set \( \lambda = 8 \times 2.5 = 20 \). Calculate \( P(X \geq 20) = 1 - P(X < 20) \). Use a Poisson table or software to find \( P(X < 20) \approx 0.559 \). Thus, \( P(X \geq 20) \approx 0.441 \).
10

Probability of At Most 10 Arrivals in 2.5 Hours

Again, with \( \lambda = 20 \), find \( P(X \leq 10) = \sum_{k=0}^{10} P(X = k) \approx 0.010 \) using software or tables.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
In a Poisson distribution, the expected value is a key concept that measures the central tendency, or average number of events, that you can expect in a given time period. It is denoted by \( \lambda \), which is the product of the rate of occurrence \( \alpha \) and the length of the time period \( t \).
For example, if small aircraft arrive at a rate of 8 per hour, the expected value of arrivals in a 90-minute period (which is 1.5 hours) would be \( \lambda = 8 \times 1.5 = 12 \).

This means, on average, you can expect 12 aircraft to land in that period. The expected value is particularly useful because it provides a straightforward interpretation of what's typically anticipated, helping airport staff plan for resources like runway availability and ground crew allocation.
Standard Deviation
Standard deviation is another important measure in statistics that tells us how much the number of arrivals can vary from the expected value. In a Poisson distribution, the standard deviation is simply the square root of the expected value, \( \sqrt{\lambda} \).
For the 90-minute period example where the expected value is 12, the standard deviation would be \( \sqrt{12} \approx 3.464 \). This means that the actual number of aircraft arrivals is likely to fluctuate around 12 by about 3 to 4 aircraft.

Understanding standard deviation is crucial for risk assessment. It gives a sense of reliability and predictability to the expected outcomes. Smaller standard deviations imply more predictability and less variance from the expected value.
Probability Calculations
Calculating probabilities in a Poisson distribution involves using the Poisson formula: \( P(X = k) = \frac{{\lambda^k e^{-\lambda}}}{k!} \), where \( k \) is the number of events (such as aircraft arrivals).
For instance, to find the probability of exactly 6 aircraft arriving in a 1-hour period (where \( \lambda = 8 \)), we plug these values into the formula: \( P(X = 6) = \frac{8^6 e^{-8}}{6!} \approx 0.122 \).

Moreover, knowing how to calculate probability for scenarios such as "at least" or "at most" a certain number of arrivals often involves cumulative probabilities, where values are summed across a range.
Complement Rule
The complement rule is a fundamental principle in probability that can simplify calculations. It states that the probability of an event happening is one minus the probability of it not happening: \( P(A) = 1 - P(A^c) \). In the context of Poisson distributions, it's very helpful for calculating probabilities like "at least".
For example, to find the probability that at least 6 aircraft arrive \( P(X \geq 6) \), you can calculate it as \( 1 - P(X < 6) \). By finding the sum of probabilities for 0 to 5 arrivals, you can use the complement rule to quickly obtain \( P(X \geq 6) \).

This method streamlines many probability calculations, making it easier and more efficient, especially when dealing with distributions that extend over a range of possible outcomes.

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