Problem 126
Question
The time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour. (a) What is the probability that more than three aircraft arrive within an hour? (b) If 30 separate one-hour intervals are chosen, what is the probability that no interval contains more than three arrivals? (c) Determine the length of an interval of time (in hours) such that the probability that no arrivals occur during the interval is \(0.10 .\)
Step-by-Step Solution
Verified Answer
(a) \( P(X > 3) \approx 0.019
(b) P(Y = 30) \approx 0.54
(c) Interval length is approximately 2.3 hours.
1Step 1: Understand the Exponential Distribution
The exponential distribution is given by the probability density function (PDF) \( f(t; \lambda) = \lambda e^{-\lambda t} \). The mean, \( \mu \), is related to the rate parameter \( \lambda \) by \( \lambda = \frac{1}{\mu} \). Here, \( \mu = 1 \) hour, so \( \lambda = 1 \).
2Step 2: Convert to Poisson Distribution
The problem refers to the number of arrivals, which typically follows a Poisson distribution if arrivals are exponentially distributed. The Poisson parameter \( \lambda = 1 \text{ (mean arrivals per hour)} \) will be used. Now, let's solve: \( P(X > 3) \).
3Step 3: Calculate Probability for More Than Three Arrivals
Use the cumulative distribution function (CDF) for the Poisson distribution. First, calculate for 0, 1, 2, and 3 arrivals: \[ P(X \leq 3) = e^{-1} \sum_{k=0}^{3} \frac{1^k}{k!} = e^{-1} \left(1 + 1 + \frac{1}{2} + \frac{1}{6}\right) \]Then, find \( P(X > 3) = 1 - P(X \leq 3) \).
4Step 4: Probability for 30 Intervals
For a Poisson process, use the complement probability for having 3 or fewer arrivals in each interval. Let \( Y \) be the number of intervals with \( X \leq 3 \): \[ P(X > 3) = 1 - P(X \leq 3) \] The probability that no interval has more than 3 arrivals is \[ P(Y = 30) = (P(X \leq 3))^{30} \].
5Step 5: Find Interval Length for No Arrival with 0.1 Probability
To find time \( t \) such that no arrivals occur with probability 0.1, set up:\[ P(T = 0) = e^{-\lambda t} = 0.1 \]Solve for \( t \):\[ t = -\frac{\ln(0.1)}{\lambda} = \ln(10) \approx 2.3 \text{ hours} \] because \( \lambda = 1 \).
Key Concepts
Poisson DistributionProbability Density Function (PDF)Cumulative Distribution Function (CDF)Rate Parameter
Poisson Distribution
The Poisson Distribution is a fundamental concept when dealing with the modeling of count-based data over a fixed period. It's often used in scenarios where you need to predict the number of events occurring in fixed intervals of time or space, like the number of customer arrivals at a store within an hour. A key consideration for using the Poisson Distribution is that each event is independent of the last, and they must occur at a constant average rate.
- It is characterized by the parameter \( \lambda \), which is the average number of events in the interval.
- The Probability Mass Function (PMF) of the Poisson Distribution is given by: \[ P(X = k) = \frac{\lambda^k \cdot e^{-\lambda}}{k!} \]where \( k \) is a non-negative integer.
Probability Density Function (PDF)
The Probability Density Function (PDF) is a key concept in continuous probability distributions like the Exponential Distribution. The PDF for a given distribution describes the likelihood of different outcomes across a continuous random variable. In simple terms, it helps us understand how the probabilities are distributed over different values.
- For an Exponential Distribution, the PDF is given by: \[ f(t; \lambda) = \lambda e^{-\lambda t} \]where \( \lambda \) is the rate parameter, and \( t \) is the time between events.
- The total area under the PDF curve is equal to 1, which implies certainty that some event will happen.
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) is extremely useful when you want to calculate the probability that a random variable is less than or equal to a certain value. Think of it as a sum of probabilities from the minimum up to a certain threshold.
- For example, in a Poisson Distribution, the CDF helps us find the probability of having up to a certain number of arrivals within a period. You add up all the individual probabilities (from PMF) for having 0, 1, 2, ..., up to the desired number of events.
- Mathematically for the Poisson: \[ P(X \leq k) = \sum_{i=0}^{k} \frac{\lambda^i}{i!} e^{-\lambda} \]
- Using the CDF, you can determine probabilities for ranges, such as finding the likelihood of "3 or more" events by calculating \( 1 - P(X \leq 2) \).
Rate Parameter
The Rate Parameter, often represented as \( \lambda \), plays a crucial role in defining how often events occur in processes described by Exponential or Poisson distributions. Understanding this parameter is essential, as it provides insight into the expected frequency of events in a given timeframe.
- For the Exponential Distribution, the rate parameter determines the average rate at which events happen. The mean of the Exponential Distribution is \( \frac{1}{\lambda} \), which was given as 1 hour in our exercise, hence \( \lambda = 1 \).
- In a Poisson Distribution, \( \lambda \) signifies the average number of events within the specified period (e.g., per hour).
- Rate Parameters help convert between Exponential and Poisson distributions, as seen in scenarios where you derive \( \lambda \) from the mean of exponential occurrences to use in a Poisson model.
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