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Consider a Poisson distribution with \(\mu=3\) a. Write the appropriate Poisson probability function. b. \(\quad\) Compute \(f(2)\) c. \(\quad\) Compute \(f(1)\) d. Compute \(P(x \geq 2)\).

Short Answer

Expert verified
a. \( f(x; \mu) = \frac{e^{-\mu} \mu^x}{x!} \) b. \( f(2) \approx 0.224 \) c. \( f(1) \approx 0.149 \) d. \( P(x \geq 2) \approx 0.8012 \)

Step by step solution

01

Identify the Poisson Probability Function

The Poisson probability function is used to determine the probability of a given number of events happening in a fixed interval of time or space. For a Poisson distribution with parameter \( \mu \), the probability function is given by:\[ f(x; \mu) = \frac{e^{-\mu} \mu^x}{x!} \]Here, \( e \) is the base of the natural logarithm, \( \mu \) is the average number of occurrences, and \( x \) is the number of occurrences.
02

Substitute Values into the Poisson Function for f(2)

To compute \( f(2) \), substitute \( x = 2 \) and \( \mu = 3 \) into the Poisson probability function:\[ f(2; 3) = \frac{e^{-3} \cdot 3^2}{2!} \]Calculating further:- \( e^{-3} \approx 0.0498 \)- \( 3^2 = 9 \)- \( 2! = 2 \)Thus, \( f(2; 3) = \frac{0.0498 \cdot 9}{2} \approx 0.224 \).
03

Substitute Values into the Poisson Function for f(1)

To compute \( f(1) \), substitute \( x = 1 \) and \( \mu = 3 \) into the Poisson probability function:\[ f(1; 3) = \frac{e^{-3} \cdot 3^1}{1!} \]Calculating further:- \( e^{-3} \approx 0.0498 \)- \( 3^1 = 3 \)- \( 1! = 1 \)Thus, \( f(1; 3) = 0.0498 \cdot 3 \approx 0.149 \).
04

Compute Probability of x Greater Than or Equal to 2

The probability \( P(x \geq 2) \) can be found using the complement rule. First, calculate for \( P(x < 2) \) which includes \( P(0) + P(1) \). We know \( f(1) \) already:- \( f(0; 3) = \frac{e^{-3} \cdot 3^0}{0!} = e^{-3} \approx 0.0498 \)Adding up probabilities for \( P(x < 2) \):- \( P(x < 2) = f(0) + f(1) = 0.0498 + 0.149 = 0.1988 \)Finally, using the complement:\[ P(x \geq 2) = 1 - P(x < 2) = 1 - 0.1988 \approx 0.8012 \]

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Key Concepts

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

Probability Function
A probability function is a mathematical tool that provides the likelihood of a certain outcome. In the realm of probability and statistics, it defines how probabilities are distributed over various outcomes. One of the most important aspects of a probability function is that the sum of probabilities of all possible outcomes must equal 1, as this represents the certainty that one of the possible outcomes will happen. For example, if we consider the roll of a die, the probability function assigns a probability of 1/6 to each of the six possible outcomes, since there is an equal chance of rolling any number between 1 and 6. When dealing with continuous variables, we refer to similar tools called probability density functions. These are integral in fields like physics and engineering, where a precise definition of outcome distributions is crucial. For discrete variables, such as the Poisson distribution we've been discussing, the probability function helps calculate the probability for specific discrete outcomes.
Complement Rule
The complement rule is a fundamental principle in probability theory, particularly useful for calculating probabilities when direct computation is challenging. The rule states that the probability of an event not occurring is equal to 1 minus the probability of that event occurring. This can be expressed as:\[ P(A') = 1 - P(A) \]where \( P(A') \) is the probability of the event not occurring and \( P(A) \) is the probability of the event occurring.This rule becomes particularly handy when dealing with cumulative probabilities. For example, in our given problem, we calculated \( P(x \geq 2) \) by first determining \( P(x < 2) \). We then used the complement rule \[ P(x \geq 2) = 1 - P(x < 2) \].By using the complement of the series of events where \( x < 2 \), we effectively found the probability for all values equal to or greater than 2. This makes it much simpler and more efficient, especially when dealing with large number distributions.
Poisson Probability Function
The Poisson probability function is a powerful tool for modeling the number of times an event occurs within a given time period. It is particularly useful when dealing with rare events or when sampling is done in a fixed interval, such as counting the number of phone calls received at a call center per hour, or the number of typos on a page.The Poisson probability function is mathematically defined as:\[ f(x; \mu) = \frac{e^{-\mu} \mu^x}{x!} \]where:- \( e \) is the base of the natural logarithm,- \( \mu \) is the average number of occurrences,- \( x \) is the exact number of occurrences for which we're calculating the probability,- \( x! \) is the factorial of \( x \).In our exercise, with \( \mu = 3 \), we applied this function to compute probabilities for certain values of \( x \). It's important to note that the Poisson distribution is discrete, meaning it can only take on integer values. And because \( e^{- ext{some value}} \) decreases rapidly as that value increases, probabilities calculated using the Poisson probability function often decline with larger values of \( x \). This is reflective of the typical behavior of such phenomena in real life, where higher frequencies of rare events become increasingly unlikely.

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Most popular questions from this chapter

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