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Chapter 5: Problem 36

Consider a Poisson random variable with \(\mu=3\). Calculate the following probabilities using the table below. $$ \begin{array}{|l|l|l|} \hline \text { Probability } & \text { Formula } & \text { Calculated Value } \\\ \hline P(x=0) & \frac{\mu^{k} e^{-\mu}}{k !}= & \\ \hline P(x=1) & \frac{\mu^{k} e^{-\mu}}{k !}= & \\ \hline P \text { (more than } 1 \text { success }) & 1-[P(x=\longrightarrow)+P(x=\longrightarrow & \\ \hline \end{array} $$

Short Answer

Expert verified
Question: In an experiment, a random variable follows a Poisson distribution with a mean of 3. Calculate the probabilities of the following events: 1) no success occurs, 2) exactly one success occurs, 3) more than one success occurs. Provide the answers in a table format. Answer: $$ \begin{array}{|l|l|l|} \hline \text { Probability } & \text { Formula } & \text { Calculated Value } \\\ \hline P(x=0) & \frac{3^{0} e^{-3}}{0 !}= & 0.0498 \\\ \hline P(x=1) & \frac{3^{1} e^{-3}}{1 !}= & 0.1494 \\\ \hline P \text { (more than } 1 \text { success }) & 1-[P(x=0)+P(x=1)] = & 0.8008 \\\ \hline \end{array} $$

Step by step solution

01

Calculate P(x=0)

To find the probability when no success occurs, let \(k=0\). Using the Poisson formula, we have: $$P(x=0) = \frac{\mu^0 e^{-\mu}}{0!} = \frac{3^0 e^{-3}}{1} = e^{-3} \approx 0.0498.$$
02

Calculate P(x=1)

To find the probability when one success occurs, let \(k=1\). Using the Poisson formula, we get: $$P(x=1) = \frac{\mu^1 e^{-\mu}}{1!} = \frac{3^1 e^{-3}}{1} = 3e^{-3} \approx 0.1494.$$
03

Calculate P(more than 1 success)

To find the probability when more than one success occurs, we can find the probability of the complementary event and subtract it from 1. The complementary event is when 0 or 1 success occurs. We already calculated P(x=0) and P(x=1), so we can simply add them: $$P(\text{more than } 1\text { success}) = 1 - [P(x=0) + P(x=1)] = 1 - [0.0498 + 0.1494] \approx 0.8008.$$ Now, we can fill the table with our calculated values: $$ \begin{array}{|l|l|l|} \hline \text { Probability } & \text { Formula } & \text { Calculated Value } \\\ \hline P(x=0) & \frac{3^{0} e^{-3}}{0 !}= & 0.0498 \\\ \hline P(x=1) & \frac{3^{1} e^{-3}}{1 !}= & 0.1494 \\\ \hline P \text { (more than } 1 \text { success }) & 1-[P(x=0)+P(x=1)] = & 0.8008 \\\ \hline \end{array} $$

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Key Concepts

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

Poisson Random Variable
When dealing with probability, we often encounter different types of random variables, and the Poisson random variable is a critical concept in the realm of statistics.

A Poisson random variable is used to model the number of times an event occurs in a fixed interval of time or space. This type of distribution is appropriate for stochastic processes where events occur independently and at a constant average rate, denoted as \( \mu \). An excellent example of where a Poisson distribution can be applied is in counting the number of emails a person receives in a day or the number of customer arrivals at a store in an hour.

To calculate probabilities using a Poisson distribution, we utilize the formula \( P(x=k) = \frac{\mu^k e^{-\mu}}{k!} \) where \( e \) is the base of the natural logarithm (approximately equal to 2.71828), \( k \) is the number of successes we want to find the probability for, and \( k! \) is the factorial of \( k \).
Probability Calculation
Calculating probabilities is a fundamental skill in statistics, often requiring specific formulas and an understanding of variables at play. In the context of a Poisson random variable, probability calculation typically involves plugging values into the Poisson formula and carrying out the arithmetic.

For instance, using the Poisson distribution formula provided earlier, we can find specific probabilities by substituting the values of \( k \) and \( \mu \), and performing the calculation. It's essential not only to compute these values correctly but also to understand what they represent. The probability \( P(x=k) \) answers the question: 'What is the likelihood that exactly \( k \) events occur?'

As seen in the textbook example, different probabilities were calculated for \( k = 0 \) and \( k = 1 \), signifying the probabilities of zero occurrences and exactly one occurrence, respectively.
Complementary Events Probability
The concept of complementary events is incredibly useful for simplifying probability calculations. In essence, the probability of the complement of event A is equal to one minus the probability of event A itself. This principle is often symbolized as \( P(A') = 1 - P(A) \).

Applying this rule to Poisson distribution, if we want to find the probability of getting more than a certain number of occurrences, we can calculate the probabilities of the complementary events 鈥� having fewer than or equal to that number of occurrences 鈥� and subtract the sum from one.

In the given exercise, to calculate the probability of having more than one success (more than one occurrence), the probabilities of zero and one success are added together and subtracted from one, as demonstrated in the solution. This practice effectively streamlines the probability calculation process and eliminates the need to directly calculate the probability of complex events.

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Most popular questions from this chapter

Many employers provide workers with sick/personal days as well as vacation days. Among workers who have taken a sick day when they were not sick, \(49 \%\) say that they needed a break! \(^{13}\) Suppose that a random sample of \(n=12\) workers who took a sick day is selected. Rounding \(49 \%\) to \(p=.5\), find the probabilities of the following events. a. What is the probability that more than six workers say that they took a sick day because they needed a break? b. What is the probability that fewer than five of the workers needed a break? c. What is the probability that exactly 10 of the workers took a sick day because they needed a break?

Find the mean and standard deviation for a binomial distribution with these values: a. \(n=1000, p=.3\) b. \(n=400, p=.01\) c. \(n=500, p=.5\) d. \(n=1600, p=.8\)

High gas prices may keep some American vacationers closer to home. However, when given a choice of getaway spots, \(66 \%\) of U.S. leisure travelers indicated that they would like to visit national parks. \({ }^{15}\) A random sample of \(n=100\) leisure travelers is selected. a. What is the average of \(x\), the number of travelers in the sample who indicate they would like to visit national parks? What is the standard deviation of \(x ?\) b. Would it be unlikely to find only 50 or fewer of those sampled who indicated they would like to visit national parks? Use the applet to find the probability of this event. c. How many standard deviations from the mean is the value \(x=50 ?\) Does this confirm your answer in part b?

A preliminary investigation reported that approximately \(30 \%\) of locally grown poultry were infected with an intestinal parasite that, though not harmful to those consuming the poultry, decrease the usual weight growth rates in the birds. A diet supplement believed to be effective against this parasite was added to the bird's food. Twenty-five birds were examined after having the supplement for at least two weeks, and three birds were still found to be infested with the parasite. a. If the diet supplement is ineffective, what is the probability of observing three or fewer birds infected with the intestinal parasite? b. If in fact the diet supplement was effective and reduced the infection rate to \(10 \%,\) what is the probability observing three or fewer infected birds? a. If the diet supplement is ineffective, what is the probability of observing three or fewer birds infected with the intestinal parasite?

According to the Humane Society of the United States, there are approximately 65 million owned dogs in the United States, and approximately \(40 \%\) of all U.S. households own at least one dog. \({ }^{4}\) Suppose that the \(40 \%\) figure is correct and that 15 households are randomly selected for a pet ownership survey. a. What is the probability that exactly eight of the households have at least one dog? b. What is the probability that at most four of the households have at least one dog? c. What is the probability that more than 10 households have at least one dog?
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