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Suppose small aircraft arrive at a certain airport according to a Poisson process with rate \(\alpha=8\) per hour, so that the number of arrivals during a time period of \(t\) hours is a Poisson rv with parameter \(\mu=8 t\). a. What is the probability that exactly 6 small aircraft arrive during a 1-hour 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 \(2.5\)-hour period? That at most 10 arrive during this period?

Short Answer

Expert verified
a. 0.1221, 0.3324, 0.0999; b. 12, 3.464; c. 0.5633, 0.0108.

Step by step solution

01

Set up the given Poisson Parameter for 1-hour period

Given that the rate of arrivals, \( \alpha = 8 \) per hour, then the parameter \( \mu \) for a 1-hour period is \( \mu = 8 \times 1 = 8 \). This will be used to calculate the various probabilities for part (a) of the question.
02

Calculate the probability for exactly 6 arrivals

The probability mass function (PMF) for a Poisson distribution is given by \( P(X = k) = \frac{e^{-\mu} \mu^k}{k!} \). Substitute \( \mu = 8 \) and \( k = 6 \):\[ P(X = 6) = \frac{e^{-8} \times 8^6}{6!} \approx \frac{0.00033546 \times 262144}{720} \approx 0.1221 \].
03

Calculate the probability for at least 6 arrivals

For at least 6 arrivals, we need to find \( P(X \geq 6) = 1 - P(X < 6) \). Compute \( P(X < 6) = P(X = 0) + P(X = 1) + ... + P(X = 5) \) using the PMF and subtract from 1:\[ P(X \geq 6) = 1 - \left( P(X = 0) + P(X = 1) + \ldots + P(X = 5) \right) \approx 0.3324 \].
04

Calculate the probability for at least 10 arrivals

Similarly, \( P(X \geq 10) = 1 - P(X < 10) \). Compute the cumulative probability up to \( X = 9 \):\[ P(X \geq 10) = 1 - \left( P(X = 0) + P(X = 1) + \ldots + P(X = 9) \right) \approx 0.0999 \].
05

Determine expected value and standard deviation for 90 minutes

For a 90-minute period (or 1.5 hours), \( \mu = 8 \times 1.5 = 12 \). The expected value for a Poisson distribution is \( E(X) = \mu = 12 \), and the standard deviation is \( \sigma = \sqrt{\mu} = \sqrt{12} \approx 3.464 \).
06

Calculate the probability for at least 20 arrivals in 2.5 hours

For a 2.5-hour period, \( \mu = 8 \times 2.5 = 20 \). \( P(X \geq 20) = 1 - P(X < 20) \).\[ P(X \geq 20) = 1 - \left( P(X = 0) + P(X = 1) + \ldots + P(X = 19) \right) \approx 0.5633 \].
07

Calculate the probability for at most 10 arrivals in 2.5 hours

\( P(X \leq 10) = P(X = 0) + P(X = 1) + \ldots + P(X = 10) \), using the PMF for \( X \) with \( \mu = 20 \):\[ P(X \leq 10) \approx 0.0108 \].

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Key Concepts

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

Probability Calculation
In Poisson distribution, probability calculation is paramount. A Poisson distribution is used to predict the number of events happening within a fixed interval of time or space. The key to these calculations is the probability mass function (PMF). It is defined as follows: \( P(X = k) = \frac{e^{-\mu} \mu^k}{k!} \), where \( \mu \) is the average number of occurrences, \( k \) is the number of occurrences, and \( e \) is Euler's number, approximately 2.71828.
For example, to find the probability of exactly 6 aircraft arriving within an hour when the rate is 8, we set \( \mu = 8 \) and \( k = 6 \) to find \( P(X = 6) \). Likewise, for probabilities like "at least" something, such as "at least 6 arrivals," it involves adding probabilities from 0 to 5 and subtracting from 1 to find \( P(X \geq 6) \).
This foundational formula helps calculate precise probabilities for varied outcomes, which makes the Poisson process a powerful statistical tool.
Expected Value
The expected value, often denoted as \( E(X) \), in a Poisson distribution, represents the average number of occurrences expected in the interval. This is simply equal to the parameter \( \mu \) of the distribution. It is a measure of central tendency that gives an intuitive understanding of what is "average" or "normal" for the distribution.
In our problem scenario, if we need the expected number of aircraft arrivals during a 90-minute period (which is 1.5 hours), the parameter \( \mu \) becomes \( 8 \times 1.5 = 12 \). Thus, \( E(X) = 12 \), meaning on average, 12 aircraft are expected to arrive.
Understanding expected value is crucial as it guides decision making and predictions in practical scenarios.
Standard Deviation
The standard deviation in a Poisson distribution measures the variation or spread around the average number of occurrences. Importantly, for a Poisson distribution, the standard deviation \( \sigma \) is the square root of the expected value (or parameter \( \mu \)). This is expressed as \( \sigma = \sqrt{\mu} \).
For instance, during a 90-minute period with \( \mu = 12 \), the standard deviation is \( \sqrt{12} \approx 3.464 \). This indicates that while the average is 12, the number of aircraft arrivals can commonly vary by about 3.464 around this mean.
This concept helps in assessing how variable the number of arrivals can be, providing insights into the consistency or volatility of events.
Cumulative Probability
Cumulative probability refers to the probability that a random variable takes on a value less than or equal to a certain number. In Poisson distribution, it is computed by summing up probabilities from the smallest possible value to the target value. For instance, \( P(X \leq 10) \) signifies the cumulative probability that "at most 10" aircraft arrive.
To find "at least" probabilities like \( P(X \geq k) \), subtract the cumulative probability up to \( (k-1) \) from 1; for example, \( P(X \geq 10) = 1 - P(X < 10) \).
In our exercise, calculating \( P(X \leq 10) \) and \( P(X \geq 20) \) across different periods (using appropriate \( \mu \)) involves applying these cumulative principles. It provides a broader view of possible outcomes, essential for predictions and risk assessments.

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Most popular questions from this chapter

Suppose that \(30 \%\) of all students who have to buy a text for a particular course want a new copy (the successes!), whereas the other \(70 \%\) want a used copy. Consider randomly selecting 25 purchasers. a. What are the mean value and standard deviation of the number who want a new copy of the book? b. What is the probability that the number who want new copies is more than two standard deviations away from the mean value? c. The bookstore has 15 new copies and 15 used copies in stock. If 25 people come in one by one to purchase this text, what is the probability that all 25 will get the type of book they want from current stock? [Hint: Let \(X=\) the number who want a new copy. For what values of \(X\) will all 25 get what they want?] d. Suppose that new copies cost \(\$ 100\) and used copies cost \(\$ 70\). Assume the bookstore currently has 50 new copies and 50 used copies. What is the expected value of total revenue from the sale of the next 25 copies purchased? Be sure to indicate what rule of expected value you are using. [Hint: Let \(h(X)=\) the revenue when \(X\) of the 25 purchasers want new copies. Express this as a linear function.]

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