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Chapter 5: Problem 83

Let \(x\) be a Poisson random variable. Using the Poisson probabilities table, write the probability distribution of \(x\) for each of the following. Find the mean, variance, and standard deviation for each of these probability distributions. Draw a graph for each of these probability distributions. a. \(\lambda=1.3\) b. \(\lambda=2.1\)

Short Answer

Expert verified
For \(\lambda=1.3\), the computed mean and variance are both 1.3, with a standard deviation of \(\sqrt{1.3}\). For \(\lambda=2.1\), the computed mean and variance are both 2.1, with a standard deviation of \(\sqrt{2.1}\). The probability distribution graph for each lambda would vary based on the computed outcomes.

Step by step solution

01

Poisson Probability for \(\lambda=1.3\)

The probability mass function of a Poisson random variable is given by \(P(X=k)= \frac{\lambda^k e^{-\lambda}}{k!}\). Here, for \(\lambda=1.3\) and \(k=0,1,2,...\) we calculate probabilities and create a distribution.
02

Mean, Variance and Standard Deviation for \(\lambda=1.3\)

For a Poisson distribution, the mean and variance are equal to the parameter, \(\lambda\). So, for \(\lambda=1.3\), the mean and variance both are 1.3. Standard deviation is the square root of variance, so we calculate \(\sqrt{1.3}\) for this.
03

Draw Probability distribution graph for \(\lambda=1.3\)

On a graph paper or software, plot the probabilities against the possible outcomes (0,1,2,...) with \(k\) on the x-axis and \(P(X=k)\) on the y-axis for \(\lambda=1.3\). The graph will show the distributions of probabilities.
04

Poisson Probability for \(\lambda=2.1\)

Repeat step 1 for \(\lambda=2.1\). Using same equation, calculate probabilities for all \(k=0,1,2,...\) and create a distribution.
05

Mean, Variance and Standard Deviation for \(\lambda=2.1\)

For this \(\lambda=2.1\), the mean and variance both are 2.1. Calculate standard deviation as \(\sqrt{2.1}\).
06

Draw Probability distribution graph for \(\lambda=2.1\)

Plot the probabilities against possible outcomes (0,1,2,...) for \(\lambda=2.1\), showing the distributions of the probabilities.

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Key Concepts

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

Probability Mass Function
In probability theory, the Probability Mass Function (PMF) is vital when working with discrete random variables such as the Poisson distribution. For a Poisson random variable with rate parameter \( \lambda \), the PMF is expressed mathematically as \( P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} \). This function allows us to calculate the probability of obtaining exactly \( k \) occurrences in a fixed interval of time or space.Let's break this down:
  • The term \( \lambda^k \) represents the rate raised to the power of \( k \) occurrences.
  • \( e^{-\lambda} \) is the natural exponential function which accounts for the probability that no events occur for the rate \( \lambda \).
  • \( k! \) (k factorial) is the product of all positive integers up to \( k \), which normalizes the probability.
Understanding and computing the PMF is the first step in analyzing a Poisson distribution, allowing us to build the full probability distribution.
Mean and Variance
The mean and variance are fundamental characteristics of any probability distribution. For the Poisson distribution, a special property is that both the mean and variance are equal to the parameter \( \lambda \). This simplifies calculations greatly, providing insights into the behavior of the distribution without complex computations.Let's clarify these points:
  • Mean (\( \mu \)): In the context of the Poisson distribution, the mean is typical of the expected number of events in the specified interval or space.
  • Variance (\( \sigma^2 \)): Reflects the expected degree of dispersion from the mean, indicating how much the outcomes vary.
Given \( \lambda = 1.3 \), both the mean and variance are 1.3. Similarly, for \( \lambda = 2.1 \), both are 2.1. The uniformity of mean and variance in Poisson makes it unique and straightforward for analysis.
Standard Deviation
While variance gives us an idea of how data is spread around the mean, the standard deviation provides a more intuitive measure by expressing this spread in the same units as the data itself. To calculate the standard deviation for a Poisson distribution, take the square root of the variance or \( \lambda \).Some key points:
  • For \( \lambda = 1.3 \), calculate the standard deviation as \( \sqrt{1.3} \approx 1.14 \).
  • For \( \lambda = 2.1 \), it becomes \( \sqrt{2.1} \approx 1.45 \).
The standard deviation is crucial as it helps gauge the variability in the frequency of events, offering a clearer sense of typical fluctuation around the mean in practical terms.
Probability Distribution Graph
Visualizing the probability distribution of a Poisson variable is insightful. The probability distribution graph provides a visual representation of probabilities for each potential outcome \( k \) on the x-axis, with the corresponding probabilities \( P(X=k) \) on the y-axis.Key aspects of drawing these graphs:
  • For \( \lambda = 1.3 \), the graph showcases a peak close to zero with a rapid decline in probabilities for larger \( k \).
  • For \( \lambda = 2.1 \), the curve shifts slightly to the right, indicating a higher average number of events.
Such graphs not only make the numerical results more comprehensible but also reveal characteristics such as skewness or how spread out the possible outcomes are. They are especially useful for seeing how changes in \( \lambda \) affect the shape of the distribution.

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Most popular questions from this chapter

Let \(x\) be a discrete random variable that possesses a binomial distribution. Using the binomial formula, find the following probabilities. a. \(P(x=0)\) for \(n=5\) and \(p=.05\) b. \(P(x=4)\) for \(n=7\) and \(p=.90\) c. \(P(x=7)\) for \(n=10\) and \(p=.60\) Verify your answers by using Table I of Appendix \(\mathrm{C}\).

An office supply company conducted a survey before marketing a new paper shredder designed for home use. In the survey, \(80 \%\) of the people who used the shredder were satisfied with it. Because of this high acceptance rate, the company decided to market the new shredder. Assume that \(80 \%\) of all people who will use it will be satisfied. On a certain day, seven customers bought this shredder. a. Let \(x\) denote the number of customers in this sample of seven who will be satisfied with this shredder. Using the binomial probabilities table (Table I, Appendix C), obtain the probability distribution of \(x\) and draw a graph of the probability distribution. Find the mean and standard deviation of \(x\). b. Using the probability distribution of part a, find the probability that exactly four of the seven customers will be satisfied.

According to a Harris Interactive poll, \(52 \%\) of American college graduates have Facebook accounts (http://www.harrisinteractive.com/harris_poll/pubs/Harris_Poll \(200904_{-16}\).pdf). Suppose that this result. is true for the current population of American college graduates. a. Let \(x\) be a binomial random variable that denotes the number of American college graduates in a random sample of 15 who have Facebook accounts. What are the possible values that \(x\) can assume? b. Find the probability that exactly 9 American college graduates in a sample of 15 have Facebook accounts.

According to a survey, \(30 \%\) of adults are against using animals for research. Assume that this result holds true for the current population of all adults. Let \(x\) be the number of adults who are against using animals for research in a random sample of two adults. Obtain the probability distribution of \(x\). Draw a tree diagram for this problem.

According to Case Study \(4-2\) in Chapter 4 , the probability that a baseball player will have no hits in 10 trips to the plate is \(.0563\), given that this player has a batting average of \(.250\). Using the binomial formula, show that this probability is indeed \(.0563 .\)
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