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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. Exactly 6: 0.122; At least 6: 0.686; At least 10: 0.332. b. Expected: 12, Std Dev: 3.464. c. At least 20: 0.422; At most 10: 0.013.

Step by step solution

01

Understand the Poisson Distribution

The number of arrivals in a given time frame follows a Poisson distribution with the parameter \(\mu = \alpha \times t\). For the given problem, the rate \(\alpha = 8\) arrivals per hour.
02

Calculate Part (a): Probability for 1-Hour Period

For a 1-hour period, \(\mu = 8 \times 1 = 8\). The probability of exactly 6 arrivals: \(P(X=6) = \frac{e^{-8} \cdot 8^6}{6!}\). The probabilities for at least 6 and at least 10 arrivals can be found by summing probabilities of all outcomes greater than or equal to those numbers.
03

Compute Exact Probability for 6 Arrivals

Use the formula for Poisson probability, \(P(X=k) = \frac{e^{-\mu} \cdot \mu^k}{k!}\), and calculate: \(P(X=6) = \frac{e^{-8} \cdot 8^6}{720} \approx 0.122\).
04

At Least 6 Arrivals: Use Complement Rule

Use the complement to compute \(P(X \geq 6) = 1 - P(X < 6)\). This involves calculating the sum of probabilities for \(X = 0, 1, 2, 3, 4, 5\) and subtracting from 1.
05

At Least 10 Arrivals: Sum Probabilities

Compute \(P(X \geq 10) = 1 - P(X < 10)\), summing \(P(X=0)\) to \(P(X=9)\) and subtracting the result from 1 to find the cumulative probability.
06

Expected Value and Standard Deviation for 90-min Period

A 90-min period corresponds to \(1.5\) hours. Calculate \(\mu = 8 \times 1.5 = 12\). The expected value is the mean, \(E(X) = \mu = 12\), and the standard deviation, \(\sigma = \sqrt{12} \approx 3.464\).
07

At Least 20 Arrivals in 2.5 Hours

For 2.5 hours, \(\mu = 8 \times 2.5 = 20\). Compute \(P(X \geq 20) = 1 - P(X < 20)\), summing the probabilities from \(X=0\) to \(X=19\) and using the Poisson formula.
08

At Most 10 Arrivals in 2.5 Hours

Similarly, use the Poisson distribution to find \(P(X \leq 10)\) directly, which involves calculating each \(P(X=k)\) for \(k=0, 1, 2, ..., 10\) and summing these probabilities.

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Key Concepts

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

Probability Theory
Probability theory serves as the mathematical framework for analyzing random events and outcomes. It helps us understand the likelihood of various events occurring.
In the context of the Poisson distribution, it describes the probability of a given number of events occurring within a fixed interval of time or space.
In our exercise, understanding the probabilities related to aircraft arrivals are key.
  • The basic probability ranges from 0 (impossible event) to 1 (certain event).
  • When calculating probabilities for specific outcomes, such as exactly 6 aircraft arrivals, we use established formulas like the one for Poisson distribution.
  • Probabilities can also describe ranges, such as finding the probability of at least or at most a certain number of events, often using complementary probabilities.
Grasping these concepts not only helps with such exercises but also in gauging eventualities in real-world scenarios, such as predicting busy airport hours.
Random Variables
Random variables (RV) are numerical outcomes of random phenomena. They can take various values, each with an associated probability.
In our example, the number of aircraft arrivals in a certain period is a random variable.
  • Random variables could be discrete (like the number of aircraft) or continuous (like the time between arrivals).
  • A Poisson random variable is particularly useful for counting events that happen independently over constant intervals of time or space.
  • For a Poisson RV, such as arrivals per hour, the parameter \( \mu\) represents the average number of events in one interval.
With clear understanding, you can manage randomness in various contexts, including predicting events in engineering, finance, and data science.
Expected Value
The expected value is a fundamental concept in probability theory, representing the average or mean value of a random variable.
In simpler terms, it's the long-run average result of an experiment if you repeated it many times.
  • In a Poisson distribution, the expected value is equal to the parameter \( \mu \).
  • For example, in a 90-minute period with a mean arrival rate of 8 aircrafts per hour, the expected value is \(12\), reflecting that on average, 12 aircrafts will arrive.
  • The expected value is crucial in planning and decision-making, as it provides a baseline for what to "expect".
Understanding expected values helps in anticipating future outcomes and setting realistic expectations.
Standard Deviation
The standard deviation measures the amount of variation or dispersion in a set of values. In probability, it tells you how much individual outcomes differ from the mean.
  • For a Poisson random variable, the standard deviation is the square root of its mean, \( \sigma = \sqrt{\mu}\).
  • In our exercise, for a 90-minute period with an expected value of 12 aircraft, the standard deviation is approximately \(3.464\).
  • This gives insight into the variability of events around the mean.
Knowing the standard deviation helps in understanding normal variations and in making probabilistic forecasts more robust by accounting for variability.

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Most popular questions from this chapter

Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair \((1,2)\) represents the selection of boards 1 and 2 for inspection. a. List the ten different possible outcomes. b. Suppose that boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define \(X\) to be the number of defective boards observed among those inspected. Find the probability distribution of \(X\). c. Let \(F(x)\) denote the odf of \(X\). First determine \(F(0)=\) \(P(X \leq 0), F(1)\), and \(F(2)\); then obtain \(F(x)\) for all other \(x\).

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