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What are the conditions that must be satisfied to apply the Poisson probability distribution?

Short Answer

Expert verified
The Poisson Distribution can be employed if four conditions are met: the events are not fixed and rather have a fixed interval such as time or space, events occur independently, the rate of event occurrence is constant, and the probability of an event occurring is proportional to the length of the interval.

Step by step solution

01

Define Poisson Distribution

The Poisson Distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known constant mean rate and independently of the time since the last event.
02

Identify Conditions for Poisson Distribution

To apply the Poisson Distribution, below conditions must be satisfied: \n1. The number of successes (events) that result from the experiment is not fixed. Instead, we have a fixed interval such as time or space.\n2. The events are independent of each other. The occurrence of one event does not affect the probability of the occurrence of an additional event. This is also known as the ‘rare event condition’.\n3. The rate of occurrence is constant. That is, the events are occurring at a constant rate.\n4. The probability of an event occurring is proportional to the length of the interval.
03

Apply Poisson Distribution

Once these conditions have been identified and satisfied, you can invoke the Poisson Distribution formula to calculate probabilities involving the number of events in a given interval. The formula is:\n\[ P(k, \lambda) = \frac{e^{-\lambda} \lambda^k}{k!} \]\nwhere: \n\(\lambda\) is the rate of occurrence (mean number of events in an interval), \ne is the base of the natural logarithm (approx. 2.71828), \nk is the actual number of successes (events)

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Key Concepts

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

Discrete Probability Distribution
The Poisson distribution is a crucial example of a discrete probability distribution. In statistics, a probability distribution describes how likely different outcomes are in an experiment. Discrete probability distributions, like the Poisson distribution, concern themselves with outcomes that are countable and distinct.

This is in contrast to continuous distributions, which deal with outcomes in a continuum.
  • Each test or event in a Poisson distribution results in a fixed outcome.
  • The number of events or "successes" is not predetermined and can vary.
  • The probability of each different number of occurrences is calculated based on a known average rate.
A common application of discrete probability distributions is counting events, such as the number of emails received in an hour or the number of decay events per unit time from a radioactive source. In such cases, Poisson distribution helps analyze scenarios with a large number of small probability events occurring independently over a specific domain.
Independent Events
In any probability distribution, an understanding of independent events is essential. The independence of events implies that the occurrence of one event does not alter the probability of another event occurring. When applying the Poisson distribution, independent events mean each event happens in isolation, with no influence from past events.

This can be better understood through real-world examples:
  • If you watch meteors at night, one meteor's appearance does not increase or decrease the likelihood of the next one appearing. Each are separate events.
  • Server requests in a minute — one customer's activity does not influence the likelihood of another customer's request.
A vital aspect of Poisson's application is ensuring that events are truly independent. If they influence each other, then the Poisson distribution might not be the best model.
Constant Mean Rate
A significant condition for utilizing the Poisson distribution is the constant mean rate. This implies that the average rate at which events occur remains constant over time or space. It's also known as the parameter λ (lambda). You determine the mean number of occurrences in any chosen interval of time or space with this rate.

Some pertinent aspects include:
  • This rate does not fluctuate, so the overall occurrence rate remains the same across different time intervals or regions.
  • It allows practitioners to predict events with more accuracy over fixed intervals.
For instance, if a website consistently has an average of 3 server downtimes per month, the prediction that there might be 3 downtimes in any subsequent month relies on the assumption that this rate is stable.

Ensuring the constant mean rate condition helps to use Poisson effectively in various statistical modeling applications.
Rare Event Condition
A distinct characteristic of the Poisson distribution is its capacity to model rare events. The rare event condition addresses situations with low probabilities of occurrence per trial or small time frames. As a rule of thumb, individual events should have low probabilities yet occur frequently enough over a long period to provide a predictable average rate.

This property makes Poisson distribution suitable for datasets where observations are unlikely, such as:
  • The probability of a particular day having no server outages given an average outage rate of once per week.
  • The frequency of a rare species being spotted within a specified geographical area.
By accommodating the rare event condition, the Poisson distribution offers a powerful tool for analyzing unlikely phenomena distributed in time or space.

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Most popular questions from this chapter

Consider the following three games. Which one would you be most likely to play? Which one would you be least likely to play? Explain your answer mathematically. Game I: You toss a fair coin once. If a head appears you receive \(\$ 3\), but if a tail appears you have to pay \(\$ 1\). Game II: You buy a single ticket for a raffle that has a total of 500 tickets. Two tickets are chosen without replacement from the \(500 .\) The holder of the first ticket selected receives \(\$ 300\), and the holder of the second ticket selected receives \(\$ 150 .\) Game III: You toss a fair coin once. If a head appears you receive \(\$ 1,000,002\), but if a tail appears you have to pay \(\$ 1,000,000\).

A contractor has submitted bids on three state jobs: an office building, a theater, and a parking garage. State rules do not allow a contractor to be offered more than one of these jobs. If this contractor is awarded any of these jobs, the profits earned from these contracts are $$\$ 10$$ million from the office building, $$\$ 5$$ million from the theater, and $$\$ 2$$ million from the parking garage. His profit is zero if he gets no contract. The contractor estimates that the probabilities of getting the office building contract, the theater contract, the parking garage contract, or nothing are \(.15, .30, .45\), and 10, respectively. Let \(x\) be the random variable that represents the contractor's profits in millions of dollars. Write the probability distribution of \(x\). Find the mean and standard deviation of \(x\). Give a brief interpretation of the values of the mean and standard deviation.

Explain the meaning of the probability distribution of a discrete random variable. Give one example of such a probability distribution. What are the three ways to present the probability distribution of a discrete random variable?

Customers arrive at the checkout counter of a supermarket at an average rate of 10 per hour, and these arrivals follow a Poisson distribution. Using each of the following two methods, find the probability that exactly 4 customers will arrive at this checkout counter during a 2 -hour period. a. Use the arrivals in each of the two nonoverlapping 1 -hour periods and then add these. (Note that the numbers of arrivals in two nonoverlapping periods are independent of each other.) b. Use the arrivals in a single 2 -hour period.

A high school history teacher gives a 50 -question multiple-choice examination in which each question has four choices. The scoring includes a penalty for guessing. Each correct answer is worth I point, and each wrong answer costs \(1 / 2\) point. For example, if a student answers 35 questions correctly, 8 questions incorrectly, and does not answer 7 questions, the total score for this student will be \(35-(1 / 2)(8)=31\) a. What is the expected score of a student who answers 38 questions correctly and guesses on the other 12 questions? Assume that the student randomly chooses one of the four answers for each of the 12 guessed questions. b. Does a student increase his expected score by guessing on a question if he has no idea what the correct answer is? Explain. c. Does a student increase her expected score by guessing on a question for which she can eliminate one of the wrong answers? Explain.

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