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Suppose \(X\) has a Poisson distribution with a mean of \(4 .\) Determine the following probabilities: (a) \(P(X=0)\) (b) \(P(X \leq 2)\) (c) \(P(X=4)\) (d) \(P(X=8)\)

Short Answer

Expert verified
(a) 0.0183, (b) 0.2381, (c) 0.1954, (d) 0.0255

Step by step solution

01

Understanding Poisson Distribution

The Poisson distribution is used to model the number of times an event occurs in a fixed interval of time or space. It is defined by a single parameter \( \lambda \), which is both the mean and variance of the distribution. In this exercise, \( \lambda = 4 \).
02

Formula for Probability in Poisson Distribution

The probability of observing \( k \) events in a Poisson distribution is given by the formula \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \), where \( e \) is approximately equal to 2.71828, \( \lambda \) is the mean, and \( k! \) is the factorial of \( k \).
03

Calculating \( P(X=0) \)

Substitute \( k=0 \) and \( \lambda=4 \) into the formula: \[ P(X=0) = \frac{e^{-4} \times 4^0}{0!} = e^{-4} \approx 0.0183. \]
04

Calculating \( P(X \leq 2) \): Finding \( P(X=1) \) and \( P(X=2) \)

First, find \( P(X=1) \): \[ P(X=1) = \frac{e^{-4} \times 4^1}{1!} = \frac{4e^{-4}}{1}. \] Approximate value is \( 0.0733 \). Then, find \( P(X=2) \): \[ P(X=2) = \frac{e^{-4} \times 4^2}{2!} = \frac{16e^{-4}}{2}. \] Approximate value is \( 0.1465 \).
05

Summing Probabilities for \( P(X \leq 2) \)

Calculate \( P(X \leq 2) \) by summing the individual probabilities: \[ P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) \approx 0.0183 + 0.0733 + 0.1465 = 0.2381. \]
06

Calculating \( P(X=4) \)

Substitute \( k=4 \) into the formula: \[ P(X=4) = \frac{e^{-4} \times 4^4}{4!} = \frac{256e^{-4}}{24}. \] Approximate value is \( 0.1954 \).
07

Calculating \( P(X=8) \)

Substitute \( k=8 \) into the formula: \[ P(X=8) = \frac{e^{-4} \times 4^8}{8!} = \frac{65536e^{-4}}{40320}. \] Approximate value is \( 0.0255 \).

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Key Concepts

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

Probability Calculation
The process of probability calculation is essential in understanding the likelihood of specific outcomes when using the Poisson distribution. The Poisson distribution is conceptually straightforward as it deals with counting the number of events occurring within a fixed interval. The formula to calculate the probability of observing exactly \( k \) events is:\[P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}\]In this formula:
  • \( \lambda \) is the average number of events (mean), and in your exercise, it is 4.
  • \( e \) is a constant approximately equal to 2.71828.
  • \( k \) is the number of occurrences you are looking for.
  • \( k! \) is the factorial of \( k \), which is the product of all positive integers less than or equal to \( k \).
To find specific probabilities, like \( P(X=0) \), \( P(X=4) \), or \( P(X=8) \), you substitute these respective values of \( k \) into the formula and compute the result.
Mean and Variance
In a Poisson distribution, the mean and variance are of utmost importance since they help define the distribution's properties. For a Poisson random variable, both the mean and variance are equal to \( \lambda \). This means:
  • Mean \( = \lambda = 4 \)
  • Variance \( = \lambda = 4 \)

This equality of mean and variance is unique to the Poisson distribution. Understanding this helps to predict how data from real-world phenomena might fluctuate around the mean. With a low variance, you might expect events to cluster around the mean value more tightly.
Factorial Computation
Computing factorials is a key step in determining probabilities within the Poisson framework. A factorial, denoted as \( n! \), is the product of all positive integers less than or equal to \( n \). For example:
  • \( 0! = 1 \)
  • \( 1! = 1 \times 1 = 1 \)
  • \( 2! = 2 \times 1 = 2 \)
  • \( 3! = 3 \times 2 \times 1 = 6 \)
  • And so on...
Factorials grow very quickly with larger numbers. In the probability calculations, factorials appear in the denominator, impacting the probability values significantly. Understanding how to compute factorials is crucial since they are used to normalize the probability that the model gives to any given number of events \( k \).
Event Probability
Event probability in the context of a Poisson distribution is all about determining how likely it is that a certain number of events will occur. In practical terms, it's about finding \( P(X=k) \).

The Poisson distribution assumes that:
  • Events occur independently within a fixed space or time interval.
  • The average rate (\[ \lambda \]) of occurrence is constant.
  • Two or more events cannot occur at the exact same instant.
By calculating event probabilities like \( P(X \leq 2) \), \( P(X=4) \), or \( P(X=8) \), one can assess how often certain outcomes are likely. For instance, finding \( P(X \leq 2) \) involves summing individual probabilities \( P(X=0) \), \( P(X=1) \), and \( P(X=2) \), leading to a comprehensive understanding of probabilities up to two events. This highlights the cumulative aspect of calculating event probabilities in contexts where multiple outcomes are possible.

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Most popular questions from this chapter

Samples of 20 parts from a metal punching process are selected every hour. Typically, \(1 \%\) of the parts require rework. Let \(X\) denote the number of parts in the sample of 20 that require rework. A process problem is suspected if \(X\) exceeds its mean by more than three standard deviations. (a) If the percentage of parts that require rework remains at \(1 \%,\) what is the probability that \(X\) exceeds its mean by more than three standard deviations? (b) If the rework percentage increases to \(4 \%,\) what is the probability that \(X\) exceeds \(1 ?\) (c) If the rework percentage increases to \(4 \%,\) what is the probability that \(X\) exceeds 1 in at least one of the next five hours of samples?

Let the random variable \(X\) have a discrete uniform distribution on the integers \(1 \leq x \leq 3\). Determine the mean and variance of \(X\).

A shipment of chemicals arrives in 15 totes. Three of the totes are selected at random, without replacement, for an inspection of purity. If two of the totes do not conform to purity requirements, what is the probability that at least one of the nonconforming totes is selected in the sample?

An installation technician for a specialized communication system is dispatched to a city only when three or more orders have been placed. Suppose orders follow a Poisson distribution with a mean of 0.25 per week for a city with a population of 100,000 , and suppose your city contains a population of 800,000 . (a) What is the probability that a technician is required after a one-week period? (b) If you are the first one in the city to place an order, what is the probability that you have to wait more than two weeks from the time you place your order until a technician is dispatched?

A fault-tolerant system that processes transactions for a financial services firm uses three separate computers. If the operating computer fails, one of the two spares can be immediately switched online. After the second computer fails, the last computer can be immediately switched online. Assume that the probability of a failure during any transaction is \(10^{-8}\) and that the transactions can be considered to be independent events. (a) What is the mean number of transactions before all computers have failed? (b) What is the variance of the number of transactions before all computers have failed?

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