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Chapter 5: Problem 13

The Poisson distribution with parameter \(\lambda=.3\) has been assigned for the outcome of an experiment. Let \(X\) be the outcome function. Find \(P(X=0)\), \(P(X=1),\) and \(P(X>1)\)

Short Answer

Expert verified
\(P(X=0)\approx0.7408\), \(P(X=1)\approx0.22224\), \(P(X>1)\approx0.03696\)

Step by step solution

01

Understanding the Poisson Distribution

The Poisson distribution models the number of events occurring within a given time interval, which is denoted by the parameter \(\lambda\). In this exercise, \(X\) follows a Poisson distribution with parameter \(\lambda = 0.3\). The probability mass function (PMF) for a Poisson distribution is given by \(P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}\), where \(k\) is the number of events.
02

Calculating P(X=0)

To find \(P(X = 0)\), use the PMF of the Poisson distribution: \[ P(X=0) = \frac{e^{-0.3} \cdot 0.3^0}{0!} = e^{-0.3} \times 1 \] Since \(0! = 1\), it simplifies to \[ P(X=0) = e^{-0.3} \] By calculating \(e^{-0.3} \approx 0.7408\), we get \(P(X=0) \approx 0.7408\).
03

Calculating P(X=1)

To find \(P(X=1)\), use the PMF of the Poisson distribution: \[ P(X=1) = \frac{e^{-0.3} \cdot 0.3^1}{1!} = e^{-0.3} \cdot 0.3 \] By calculating \(e^{-0.3} \cdot 0.3 \approx 0.22224\), we have \(P(X=1) \approx 0.22224\).
04

Calculating P(X>1)

To find \(P(X > 1)\), we first find the probabilities of \(X=0\) and \(X=1\), then use them to find \(P(X>1)\). Since the total probability is 1, \[ P(X>1) = 1 - [P(X=0) + P(X=1)] \] Based on previous calculations, \[ P(X>1) = 1 - (0.7408 + 0.22224) \approx 1 - 0.96304 \] Thus, \(P(X>1) \approx 0.03696\).

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Key Concepts

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

Probability Mass Function
The Probability Mass Function (PMF) is a fundamental concept in the study of discrete random variables like those in a Poisson distribution. It's a function that gives the probability that a discrete random variable equals a specific value. In simple terms, it helps us understand the likelihood of a particular number of events occurring. For a Poisson distribution, the PMF is expressed as follows:
\[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
Here, \(k\) is the number of events we are interested in. The parameter \(e\) represents the base of the natural logarithm, roughly equal to 2.71828. The role of \(k!\) is a mathematical operation called a factorial. It involves multiplying all integers from 1 to \(k\). Thus, the PMF helps us calculate exact probabilities for different values that a random variable can take.
Poisson Process
The Poisson Process is a statistical process that models a sequence of events occurring randomly over a fixed period of time or space. It is incredibly useful in scenarios where we wish to predict the number of occurrences, such as the arrival of buses at a station or the number of emails received in an hour.
For the Poisson process to be applicable, these events should meet specific criteria:
  • Events are independent 鈥� the occurrence of one does not affect another.
  • Two events cannot occur at exactly the same time.
  • The rate of occurrence is constant.
These conditions ensure that a Poisson Process is a good fit for predicting events over intervals. It鈥檚 considered a continuous-time process and each event is counted as it happens, adhering to the distributions defined using the Poisson PMF.
Parameter Lambda
In the Poisson distribution, the parameter \(\lambda\) is essential, as it defines the average rate of events occurring within a specific time frame. It's a positive real number, indicating the expected number of times an event is likely to happen. This parameter is crucial in shaping the distribution curve, directing how probability is spread across the number of possible events.
For example, if \(\lambda = 0.3\), it means, on average, 0.3 events are expected within the defined interval. Higher values of \(\lambda\) indicate more frequent events, leading to a wider spread of probabilities across different potential outcomes. In practice, specifying \(\lambda\) helps tailor the Poisson distribution to match the characteristics of a real-world process or situation, making it an integral part of modeling such phenomena.

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Most popular questions from this chapter

Jones puts in two new lightbulbs: a 60 watt bulb and a 100 watt bulb. It is claimed that the lifetime of the 60 watt bulb has an exponential density with average lifetime 200 hours \((\lambda=1 / 200)\). The 100 watt bulb also has an exponential density but with average lifetime of only 100 hours \((\lambda=1 / 100)\). Jones wonders what is the probability that the 100 watt bulb will outlast the 60 watt bulb. If \(X\) and \(Y\) are two independent random variables with exponential densities \(f(x)=\lambda e^{-\lambda x}\) and \(g(x)=\mu e^{-\mu x}\), respectively, then the probability that \(X\) is less than \(Y\) is given by $$P(X

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