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Consider a Poisson distribution with a mean of two occurrences per time period. a. Write the appropriate Poisson probability function. b. What is the expected number of occurrences in three time periods? c. Write the appropriate Poisson probability function to determine the probability of \(x\) occurrences in three time periods. d. Compute the probability of two occurrences in one time period. e. Compute the probability of six occurrences in three time periods. f. Compute the probability of five occurrences in two time periods.

Short Answer

Expert verified
a. Poisson function: \( P(X = x) = \frac{e^{-2} \cdot 2^x}{x!} \). b. Expected value: 6 occurrences. c. For 3 periods: \( P(X = x) = \frac{e^{-6} \cdot 6^x}{x!} \). d. Probability of 2 occurrences: 0.2707. e. Probability of 6 occurrences: 0.1606. f. Probability of 5 occurrences: 0.1563.

Step by step solution

01

Define the Poisson Probability Function

For a Poisson distribution with mean \( \lambda \), the probability of observing \( x \) occurrences is given by the Poisson probability function: \[ P(X = x) = \frac{e^{-\lambda} \cdot \lambda^x}{x!} \]Here, \( \lambda = 2 \) is the mean number of occurrences per time period.
02

Expected Number in Three Time Periods

The expected number of occurrences in multiple time periods is the product of the mean per period and the number of periods. For three time periods: \[ E = \lambda \times 3 = 2 \times 3 = 6 \]Thus, the expected number of occurrences is 6.
03

Poisson Probability Function for Three Time Periods

For three time periods, the mean becomes \( \lambda = 6 \). The probability of \( x \) occurrences is given by: \[ P(X = x) = \frac{e^{-6} \cdot 6^x}{x!} \]
04

Probability of Two Occurrences in One Time Period

Using the Poisson probability function with \( \lambda = 2 \) for one period:\[ P(X = 2) = \frac{e^{-2} \cdot 2^2}{2!} = \frac{e^{-2} \cdot 4}{2} = 0.2707 \]
05

Probability of Six Occurrences in Three Time Periods

For three periods, mean \( \lambda = 6 \). Using the Poisson function:\[ P(X = 6) = \frac{e^{-6} \cdot 6^6}{6!} = \frac{e^{-6} \cdot 46656}{720} = 0.1606 \]
06

Probability of Five Occurrences in Two Time Periods

For two periods, mean \( \lambda = 4 \). Using the Poisson formula:\[ P(X = 5) = \frac{e^{-4} \cdot 4^5}{5!} = \frac{e^{-4} \cdot 1024}{120} = 0.1563 \]

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Key Concepts

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

Probability Function
In a Poisson distribution, the probability function is an essential element. It defines the likelihood of a specific number of occurrences within a defined time period. The function is expressed as follows:
  • The number of occurrences, represented as \( x \), is any whole number (0, 1, 2, etc.).
  • The mean number of occurrences in a single time period is denoted by \( \lambda \).
The formula for the Poisson probability function is:\[ P(X = x) = \frac{e^{-\lambda} \cdot \lambda^x}{x!} \]Here:
  • \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
  • Factors include exponentials and factorials, which can seem complex at first glance.
  • Exponentials like \( e^{-\lambda} \) handle the decay function, showing the decline in probability as occurrences increase.
  • Factorials, \( x! \), depict permutations of various numbers.
This function allows us to compute the probability for different values of \( x \) given a known \( \lambda \), which translates to the mean occurrences in the given time period.
Expected Value
The expected value is a critical concept in the context of the Poisson distribution, representing the average occurrence in a specific time span. It simplifies finding an average number of events. If you know the mean number of occurrences \( \lambda \) in one time period, the expected number for multiple periods is straightforward. The formula is:
\[ E = \lambda \times n \]
  • Where \( n \) is the number of time periods we're analyzing.
For example, if the mean number of occurrences in one period is 2, and you wish to consider three periods, the expected value would be: \[ E = 2 \times 3 = 6 \]This result indicates we can anticipate six occurrences over three periods on average. The expected value provides a basis for determining likelihoods and probabilities in the context of multiple periods.
Mean Occurrences
The mean occurrences is a measure signifying the average number of events within a specific time period. Symbolized by \( \lambda \), it serves as the backbone of the Poisson distribution. To grasp this better:
  • Mean occurrences \( \lambda \) are typically assigned before analyzing the data, based on previous experience or statistical analysis.
  • This parameter remains consistent across calculations, defining the probability function and guiding expectations for varying values of \( x \).
The mean is essential because it directly influences the results of the Probability Function. When applied over multiple periods, the mean is multiplied by the number of periods to find an adjusted mean, thus extending its utility and influence:\[ \text{Adjusted mean for } n \text{ periods} = \lambda \times n \]Understanding the mean occurrences helps form expectations of the distribution and anchors predictions for the observed events in specified time frames.
Time Periods
In the realm of Poisson distribution, time periods are the segments over which occurrences are measured. Their role is pivotal in adjusting calculations relative to the span of observation. Here's a bit more insight into their importance:
  • A time period can be any definable segment of time: minutes, hours, days, or otherwise.
  • The flexibility of time periods means the same principles apply, irrespective of whether you're looking at short or long durations.
  • Time periods influence the Expected Value directly as they multiply the mean \( \lambda \).
For example, if the mean is 2 for one period and we consider three periods, the revised mean becomes \( 2 \times 3 = 6 \), offering a renewed calculation base.
Recognizing how time periods interact with mean occurrences and probability functions enhances the understanding and application of the Poisson distribution.

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Most popular questions from this chapter

A technician services mailing machines at companies in the Phoenix area. Depending on the type of malfunction, the service call can take \(1,2,3,\) or 4 hours. The different types of malfunctions occur at about the same frequency. a. Develop a probability distribution for the duration of a service call. b. Draw a graph of the probability distribution. c. Show that your probability distribution satisfies the conditions required for a discrete probability function. d. What is the probability a service call will take three hours? e. A service call has just come in, but the type of malfunction is unknown. It is 3: 00 P.M. and service technicians usually get off at 5: 00 P.M. What is the probability the service technician will have to work overtime to fix the machine today?

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