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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) Use Poisson function with \(\lambda=2\). b) Expected occurrences: 6. c) Use Poisson with \(\lambda=6\). d) Probability: 0.2707. e) Probability: 0.1606. f) Probability: 0.1563.

Step by step solution

01

Poisson Probability Function

The Poisson probability function is given by: \( P(X = x) = \frac{{e^{-\lambda} \lambda^x}}{{x!}} \), where \( \lambda \) is the average rate of occurrences and \( x \) is the number of occurrences.
02

Expected Number of Occurrences in Three Periods

The expected number of occurrences for a Poisson distribution in \( t \) time periods is given by \( \lambda_{total} = \lambda \times t \). Here, \( \lambda = 2 \) for one time period, so for three time periods, \( \lambda_{total} = 2 \times 3 = 6 \).
03

Poisson Probability Function for Three Time Periods

To find the probability of \( x \) occurrences in three periods, use \( \lambda_{total} = 6 \) in the Poisson probability function: \( P(X = x) = \frac{{e^{-6} 6^x}}{{x!}} \).
04

Probability of Two Occurrences in One Time Period

For \( x = 2 \) and \( \lambda = 2 \), apply the function: \( P(X = 2) = \frac{{e^{-2} 2^2}}{{2!}} = \frac{{e^{-2} imes 4}}{{2}} \). Calculate this to find the exact probability.
05

Probability Calculation for Two Occurrences

Calculating: \( P(X = 2) = \frac{{4e^{-2}}}{2} \approx 0.2707 \).
06

Probability of Six Occurrences in Three Time Periods

Use the function for three periods: \( P(X = 6) = \frac{{e^{-6} 6^6}}{{6!}} \). Calculate this to determine the probability.
07

Probability Calculation for Six Occurrences

Calculating: \( P(X = 6) = \frac{{e^{-6} imes 46656}}{{720}} \approx 0.1606 \).
08

Probability of Five Occurrences in Two Time Periods

For \( x = 5 \) in two periods, where \( \lambda_{total} = 4 \), use \( P(X = 5) = \frac{{e^{-4} 4^5}}{{5!}} \). Calculate this probability.
09

Probability Calculation for Five Occurrences

Calculating: \( P(X = 5) = \frac{{e^{-4} imes 1024}}{120} \approx 0.1563 \).

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Key Concepts

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

Probability Function
The Poisson probability function is a way to determine the likelihood of a certain number of events happening within a fixed interval of time or space. It's a popular tool when each event occurs independently, and we know the average number of events (called the rate) that occur per unit period.
The function is expressed mathematically as follows: - For a given number of occurrences \( x \), and a known average rate \( \lambda \), the probability \( P(X = x) \) is given by: \[ P(X = x) = \frac{{e^{-\lambda} \lambda^x}}{{x!}} \]
  • \( e \) is the base of the natural logarithm, approximately 2.71828.
  • \( x! \) ("x factorial") is the product of all positive integers up to \( x \).

Simply put, this formula calculates how likely it is to see exactly \( x \) events when we expect \( \lambda \) events on average.
Expected Value
In probability theory, the expected value is the average of all possible outcomes of a random variable, weighted by their probabilities. For a Poisson distribution, it's quite intuitive.
The expected value is simply the average number of occurrences \( \lambda \) that you expect to see within a specific period of time. So, if you know your rate is 2 occurrences per period, your expected value \( E[X] \) over one period is \( 2 \).
When considering multiple periods, you adjust the expected value accordingly. Thus, for 3 time periods: \[ \lambda_{total} = \lambda \times t = 2 \times 3 = 6 \text{ occurrences} \]
  • This reflects the total expected occurrences over the extended timeframe.
  • It's a straightforward calculation but highlights how accumulative expected values help in predicting outcomes over extended periods.
Time Period Analysis
Time period analysis in a Poisson distribution involves understanding how the rate of occurrences can change depending on how many time periods are being considered.
In our context, when the rate per time period \( \lambda \) is 2, analyzing three time periods becomes crucial when one needs to predict probabilities for varying occurrences:
  • For probability functions across multiple time periods, multiply the \( \lambda \) by the number of periods, thus converting it into \( \lambda_{total} \).
  • This total rate reflects the accumulated expected occurrences over the considered timeframe.
So, if you need to calculate the probability for, say, 6 occurrences over 3 periods, use \( \lambda_{total} = 6 \) in the Poisson formula: \[ P(X = x) = \frac{{e^{-6} 6^x}}{{x!}} \]
This expansion of the time frame is essential because it adjusts the probability calculations to consider the longer duration, which could otherwise skew the findings if treated as single-period events.

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Most popular questions from this chapter

The National Basketball Association (NBA) records a variety of statistics for each team. Two of these statistics are the percentage of field goals made by the team and the percentage of three-point shots made by the team. For a portion of the 2004 season, the shooting records of the 29 teams in the NBA showed that the probability of scoring two points by making a field goal was \(.44,\) and the probability of scoring three points by making a threepoint shot was .34 (NBA website, January 3,2004 ). a. What is the expected value of a two-point shot for these teams? b. What is the expected value of a three-point shot for these teams? c. If the probability of making a two-point shot is greater than the probability of making a three-point shot, why do coaches allow some players to shoot the three-point shot if they have the opportunity? Use expected value to explain your answer.

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A new automated production process averages 1.5 breakdowns per day. Because of the cost associated with a breakdown, management is concerned about the possibility of having three or more breakdowns during a day. Assume that breakdowns occur randomly, that the probability of a breakdown is the same for any two time intervals of equal length, and that breakdowns in one period are independent of breakdowns in other periods. What is the probability of having three or more breakdowns during a day?

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