91Ó°ÊÓ

A common application of exponential distributions (see Problem 18 ) is waiting time between Poisson events (e.g., successes in the Poisson distribution; see Section 5.4 ). In our study of the Poisson distribution (Example 11, Section 5.4), we saw that the mean success rate per hour of catching a fish at Pyramid Lake is \(\lambda=0.667\) fish/hour. From this we see that the mean waiting time between fish can be thought of as \(\beta=1 / \lambda=1 / 0.667 \approx 1.5\) hours/fish. Remember, the fish at Pyramid Lake tend to be large. Suppose you have just caught a fish. Let \(x\) be a random variable representing the waiting time (in hours) to catch the next fish. Use the exponential distribution to determine the probability that the waiting time is (a) less than half an hour (i.e., \(0 < x < 0.5)\) (b) more than 3 hours (i.e., \(3 < x < \infty\) ). Hint: \(e^{-\infty}=0\) (c) between 1 and 3 hours (i.e., \(1

Step by step solution

01

Understanding the Exponential Distribution

The exponential distribution is defined by a rate parameter \( \lambda \). In this problem, \( \lambda = 0.667 \) fish/hour. The waiting time \( x \) to catch the next fish is exponentially distributed, and we are interested in finding probabilities for different time intervals. The probability density function is \( f(x) = \lambda e^{-\lambda x} \).

02

Exponential Distribution for (a) Less Than Half an Hour

The probability that \( x < 0.5 \) hours is given by the cumulative distribution function (CDF) \( F(x) = 1 - e^{-\lambda x} \). For \( x = 0.5 \), the probability is: \[ P(x < 0.5) = 1 - e^{-0.667 \times 0.5} \approx 1 - e^{-0.3335} \approx 1 - 0.7165 \approx 0.2835. \]

03

Probability for (b) More Than 3 Hours

We find the probability \( P(x > 3) \) using \( 1 - F(3) \):\[ P(x > 3) = 1 - (1 - e^{-0.667 \times 3}) = e^{-2.001} \approx 0.1353. \]

04

Probability for (c) Between 1 and 3 Hours

The probability \( P(1 < x < 3) \) is found by \( F(3) - F(1) \):\[ P(1 < x < 3) = (1 - e^{-0.667 \times 3}) - (1 - e^{-0.667 \times 1}) = e^{-0.667} - e^{-2.001} \approx 0.5134 - 0.1353 \approx 0.3781. \]

05

Mean and Standard Deviation

For an exponential distribution with rate \( \lambda \), the mean \( \mu \) is \( \frac{1}{\lambda} \) and the standard deviation \( \sigma \) is also \( \frac{1}{\lambda} \). Here, \( \beta = \frac{1}{0.667} \approx 1.5 \). Hence, both the mean and standard deviation are approximately 1.5 hours.

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Key Concepts

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

Poisson Distribution

The Poisson distribution is often used to model events that occur randomly over a specified interval of time or space. It's characterized by the average number of events, \( \, \lambda \, \), happening in a fixed period. For instance, this could be the number of fish caught in an hour at a specific location. Its significance lies in predicting the probability of a given number of events happening in a set interval.To summarize the core properties of the Poisson distribution:

• It is defined by the parameter \lambda\, which is both the rate and the mean number of events.
• It assumes that events occur independently.
• The standard deviation is \sqrt{\lambda}\.

This distribution is used to model rare events, and as demonstrated here, to understand the happenings over time or space efficiently.

Probability Density Function

In probability theory, a Probability Density Function (PDF) helps describe the likelihood of a random variable to take on a specific value. For continuous random variables, it represents the relative likelihood for this variable to be near a given measure.The exponential distribution's PDF is given by:\[ f(x) = \lambda e^{-\lambda x} \]where \( \, \lambda \, \) is the rate parameter. The exponential distribution tells us about the "waiting time" till the next event, such as catching a fish.A few highlights about PDFs:

• The area under the PDF curve for a range of values gives the probability that the random variable falls within this range.
• The total area under a PDF curve is always equal to 1.
• This allows us to calculate specific probabilities, such as waiting times being less than a certain number of hours.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is key when analyzing the probabilities in a continuous distribution like the exponential distribution. Unlike the PDF, which is about the density, the CDF provides the probability that a random variable \( \, X \, \) is less than or equal to a particular value \( \, x \, \).For the exponential distribution, the CDF is given by:\[ F(x) = 1 - e^{-\lambda x} \]This function helps us find probabilities over a range of values. For example, with catching fish in the given time range, we can determine the probability:

• Less than or equal to a certain time using \( F(x) \, \).
• More than a certain time by calculating \( 1 - F(x) \, \).
• Between two times via the difference \( F(b) - F(a) \, \).

Using the CDF's properties, students can intuitively determine probabilities related to waiting times in real-world scenarios.

Mean and Standard Deviation

The mean and standard deviation are fundamental statistics that describe the average and variability of a distribution. For exponential distributions, both metrics provide essential insights into the typical waiting times and the variability of those times.For an exponential distribution, the mean \( \, \mu \, \) and the standard deviation \( \, \sigma \, \) share the same formula:\[ \text{Mean and Standard Deviation} = \frac{1}{\lambda} \]In our context:

• The mean waiting time between catching two fish is about \( 1.5 \, \text{hours} \), calculated as \( \frac{1}{0.667} \).
• The standard deviation, representing the spread of potential waiting times, is also \( 1.5 \, \text{hours} \).

This outcome means that on average, the expectation is to wait 1.5 hours between events. This also suggests variability around this average. Comprehending these values aids in understanding how predictable or varied the waiting times are likely to be in real settings.