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The number of stork sightings on a route in South Carolina follows a Poisson process with a mean of 2.3 per year. (a) What is the mean time between sightings? (b) What is the probability that there are no sightings within three months \((0.25\) years \() ?\) (c) What is the probability that the time until the first sighting exceeds six months? (d) What is the probability of no sighting within three years?

Short Answer

Expert verified
(a) 0.435 years; (b) 0.562; (c) 0.316; (d) 0.001.

Step by step solution

01

Calculate the Mean Time Between Sightings

The mean time between sightings in a Poisson process is calculated as the reciprocal of the rate (mean sightings per year). Given the rate \( \lambda = 2.3 \) sightings per year, the mean time between sightings is \( \frac{1}{2.3} \approx 0.435 \) years.
02

Probability of No Sightings Within Three Months

For a Poisson process, the probability of \( k \) events (sightings) in time \( t \) can be calculated using \( P(X = k) = \frac{(\lambda t)^k \cdot e^{-\lambda t}}{k!} \). For zero sightings within 0.25 years, \( \lambda t = 2.3 \times 0.25 = 0.575 \). Thus, \( P(X = 0) = e^{-0.575} \approx 0.562 \).
03

Probability of First Sighting Exceeding Six Months

The time until the first event in a Poisson process is exponential. The probability that the time \( T \) exceeds 0.5 years for \( \lambda = 2.3 \) is \( P(T > 0.5) = e^{-\lambda t} = e^{-2.3 \times 0.5} = e^{-1.15} \approx 0.316 \).
04

Probability of No Sightings Within Three Years

For \( \lambda t = 2.3 \times 3 = 6.9 \), the probability of no sightings in three years is \( P(X = 0) = e^{-6.9} \approx 0.001 \).

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Key Concepts

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

Mean Time Between Events
Understanding the concept of "Mean Time between Events" is crucial when dealing with Poisson processes. It essentially tells us the average time that will elapse between two consecutive events, such as sightings in our example. In a Poisson process, this is calculated as the inverse of the rate of occurrence.So, if you know the mean number of events (like sightings) per unit time, you can easily find the mean time between those events. For instance, if storks are spotted an average of 2.3 times per year, the mean time between these sightings is given by the reciprocal of this rate. Mathematically, this is:\[ \text{Mean Time Between Events} = \frac{1}{\lambda} \]where \( \lambda \) is the event rate. In our scenario, \( \lambda = 2.3 \) sightings per year, leading to a mean time of approximately 0.435 years between stork sightings.
Exponential Distribution
The Exponential Distribution comes into play when we're interested in understanding the time between events in a Poisson process. It's the probability distribution that describes the time between consecutive events.When we talk about the time until the first event or sighting, the exponential distribution is used. This is because the time to the next event is memoryless - past events do not affect future events. The probability that the time until the next sighting is greater than some time \( t \) is given by:\[ P(T> t) = e^{-\lambda t} \]For example, if you wanted to find the probability that no sightings occur within the first 6 months, you would use this formula. With \( \lambda = 2.3 \) and \( t = 0.5 \) years (since 6 months is half a year), you calculate it as:\[ P(T > 0.5) = e^{-2.3 \times 0.5} \approx 0.316 \]This implies there's about a 31.6% chance that you won't see a stork in the first six months of observation.
Probability Calculations
In a Poisson process, we often need to calculate the probability of observing a certain number of events in a specified time frame. These kinds of calculations use the Poisson distribution formula:\[ P(X = k) = \frac{(\lambda t)^k \cdot e^{-\lambda t}}{k!} \]where:
  • \( X \) is the number of events during time \( t \)
  • \( \lambda \) is the rate of events
  • \( k \) is the actual number of events you're interested in
In the case where we seek the probability of zero events (such as zero sightings over three months, or 0.25 years), we set \( k = 0 \). Thus, the formula simplifies to:\[ P(X = 0) = e^{-\lambda t} \]For a quarter of a year with \( \lambda \cdot t = 2.3 \times 0.25 \), we find:\[ P(X = 0) = e^{-0.575} \approx 0.562 \]This result means there's a roughly 56.2% probability of not seeing any storks in that period.
Event Rate Estimation
Event Rate Estimation in a Poisson process helps us make informed predictions about occurrences over time. This is particularly helpful when planning or expecting specific outcomes in a given time period.The event rate \( \lambda \) provides how frequently an event, such as sightings here, occurs per unit time. If the average number of sightings per year is known (2.3 in our case), we can estimate various probabilities for different durations and occurrences.For example, looking at longer durations, like 3 years, we use:\[ \lambda t = 2.3 \times 3 = 6.9 \]The probability calculation for no events in this time frame, again using:\[ P(X = 0) = e^{-6.9} \approx 0.001 \]reassures us that there's nearly no chance of no sightings within that longer time span. Event rate estimation thus becomes a powerful tool in anticipating future occurrences and making informed decisions based on these probabilities.

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Most popular questions from this chapter

Errors caused by contamination on optical disks occur at the rate of one error every \(10^{5}\) bits. Assume that the errors follow a Poisson distribution. (a) What is the mean number of bits until five errors occur? (b) What is the standard deviation of the number of bits until five errors occur? (c) The error-correcting code might be ineffective if there are three or more errors within \(10^{5}\) bits. What is the probability of this event?

Asbestos fibers in a dust sample are identified by an electron microscope after sample preparation. Suppose that the number of fibers is a Poisson random variable and the mean number of fibers per square centimeter of surface dust is \(100 .\) A sample of 800 square centimeters of dust is analyzed. Assume that a particular grid cell under the microscope represents \(1 / 160,000\) of the sample. (a) What is the probability that at least one fiber is visible in the grid cell? (b) What is the mean of the number of grid cells that need to be viewed to observe 10 that contain fibers? (c) What is the standard deviation of the number of grid cells that need to be viewed to observe 10 that contain fibers?

Raw materials are studied for contamination. Suppose that the number of particles of contamination per pound of material is a Poisson random variable with a mean of 0.01 particle per pound. (a) What is the expected number of pounds of material required to obtain 15 particles of contamination? (b) What is the standard deviation of the pounds of materials required to obtain 15 particles of contamination?

The time between arrivals of customers at an automatic teller machine is an exponential random variable with a mean of five minutes. (a) What is the probability that more than three customers arrive in 10 minutes? (b) What is the probability that the time until the fifth customer arrives is less than 15 minutes?

Suppose that \(X\) has a Weibull distribution with \(\beta=2\) and \(\delta=8.6 .\) Determine the following: (a) \(P(X<10)\) (b) \(P(X>9)\) (c) \(P(8x)=0.9\)

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