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Consider a Poisson random variable with \(\mu=3\). Use the Poisson formula to calculate the following probabilities: a. \(P(x=0)\) b. \(P(x=1)\) c. \(P(x>1)\)

Short Answer

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Question: Given a Poisson random variable with \(\mu=3\), calculate the probabilities of observing exactly 0 occurrences, exactly 1 occurrence, and more than 1 occurrence. Answer: The probabilities are as follows: a. The probability of observing exactly 0 occurrences is approximately \(0.0498\). b. The probability of observing exactly 1 occurrence is approximately \(0.1494\). c. The probability of observing more than 1 occurrence is approximately \(0.8008\).

Step by step solution

01

Calculate P(x=0)

First, apply the Poisson formula to find \(P(x=0)\). In this case, \(k=0\) and \(\mu=3\): \(P(x=0)=\frac{e^{-\mu}\mu^0}{0!}=\frac{e^{-3}3^0}{1}=\frac{e^{-3}}{1}= e^{-3}\approx 0.0498\)
02

Calculate P(x=1)

Now, apply the Poisson formula again to find \(P(x=1)\). This time, \(k=1\) and \(\mu=3\): \(P(x=1)=\frac{e^{-\mu}\mu^1}{1!}=\frac{e^{-3}3^1}{1}=\frac{3e^{-3}}{1}=3e^{-3}\approx 0.1494\)
03

Calculate P(x>1)

To find \(P(x>1)\), we can use the fact that the sum of all probabilities in a distribution is equal to 1, and we already have \(P(x=0)\) and \(P(x=1)\). Thus, we can calculate \(P(x>1)\) as follows: \(P(x>1)=1-P(x=0)-P(x=1)=1-0.0498-0.1494\approx 0.8008\) Now we have the calculated probabilities: a. \(P(x=0)\approx 0.0498\) b. \(P(x=1)\approx 0.1494\) c. \(P(x>1)\approx 0.8008\)

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Key Concepts

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

Probability Calculation
Calculating probabilities is a core aspect of understanding any probability distribution. When working with the Poisson distribution, probability calculations involve specific application of the Poisson probability formula. This formula helps to determine the likelihood of a given number of events happening in a fixed interval of time or space. In such scenarios, the formula is defined as: \[ P(x=k) = \frac{e^{-\mu}\mu^k}{k!} \] Here:
  • \(P(x=k)\) is the probability of observing \(k\) events in the given interval.
  • \(\mu\) represents the average number of events occurring in the interval.
  • \(e\) is the base of the natural logarithm, approximately equal to 2.71828.
  • \(k!\) is the factorial of \(k\), which is the product of all positive integers up to \(k\).
For instance, when calculating \(P(x=0)\) for a Poisson variable with \(\mu=3\), you plug in these values and do the computations. After using this formula, you can directly find specific probabilities such as \(P(x=0)\), \(P(x=1)\), or determine the probability of more than a certain number of events occurring.
Poisson Random Variable
A Poisson random variable helps us understand the distribution of counts of events in a specified interval, with specific relation to their average rate of occurrence, \(\mu\). These random variables are particularly useful in settings where events occur independently and at a constant average rate, such as:
  • Number of buses arriving at a bus stop in an hour.
  • Number of typos in a document.
  • Calls received by a call center per minute.
The Poisson random variable is essential in describing how often an event is likely to occur over a given period when that event's occurrences are independent of each other. It is especially useful in fields such as telecommunications, manufacturing, and retail, where understanding occurrences over time or space is critical. Using the Poisson distribution, you can compute not just the exact number of times an event occurs, but also analyze longer periods or larger spaces by adjusting the \(\mu\) accordingly.
Exponential Function
The exponential function is pivotal in the Poisson probability formula, represented by \(e^{-\mu}\). Exponential functions extrapolate the idea of growth and decay and are defined by the constant \(e\), which is a critical mathematical constant in calculus. In the Poisson formula, the exponential function serves as a weighting factor that accounts for the interval's length and event rate:
  • \(e\), approximated to 2.71828, is the base of natural logarithms, essential for continuous growth calculations.
  • In contexts where the time frame or space is large, \(e^{-\mu}\) modulates the decaying probabilities with more events being counted.
Understanding exponential functions is crucial for interpreting a Poisson distribution's behavior over continuous domains. They represent how likelihoods diminish exponentially with increasing event counts. Mastery of how to handle \(e^{-\mu}\) mathematically is necessary for accurately calculating and interpreting results from a Poisson distribution.

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Most popular questions from this chapter

Under what conditions can the Poisson random variable be used to approximate the probabilities associated with the binomial random variable? What application does the Poisson distribution have other than to estimate certain binomial probabilities?

A psychiatrist believes that \(80 \%\) of all people who visit doctors have problems of a psychosomatic nature. She decides to select 25 patients at random to test her theory.a. Assuming that the psychiatrist's theory is true, what is the expected value of \(x\), the number of the 25 patients who have psychosomatic problems? b. What is the variance of \(x\), assuming that the theory is true? c. Find \(P(x \leq 14)\). (Use tables and assume that the theory is true.) d. Based on the probability in part \(c\), if only 14 of the 25 sampled had psychosomatic problems, what conclusions would you make about the psychiatrist's theory? Explain.

Tay-Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carriers of the disease, the probability that their offspring will develop the disease is approximately .25. Suppose a husband and wife are both carriers of the disease and the wife is pregnant on three different occasions. If the occurrence of Tay-Sachs in any one offspring is independent of the occurrence in any other, what are the probabilities of these events? a. All three children will develop Tay-Sachs disease. b. Only one child will develop Tay-Sachs disease. c. The third child will develop Tay-Sachs disease, given that the first two did not.

The National Hockey League (NHL) has \(80 \%\) of its players born outside the United States, and of those born outside the United States, \(50 \%\) are born in Canada. \(^{2}\) Suppose that \(n=12\) NHL players were selected at random. Let \(x\) be the number of players in the sample who were born outside of the United States so that \(p=.8\). Find the following probabilities: a. At least five or more of the sampled players were born outside the United States. b. Exactly seven of the players were born outside the United States. c. Fewer than six were born outside the United States.

Americans are really getting away while on vacation. In fact, among small business owners, more than half \((51 \%)\) say they check in with the office at least once a day while on vacation; only \(27 \%\) say they cut the cord completely. \({ }^{7}\) If 20 small business owners are randomly selected, and we assume that exactly half check in with the office at least once a day, then \(n=20\) and \(p=.5 .\) Find the following probabilities. a. Exactly 16 say that they check in with the office at least once a day while on vacation. b. Between 15 and 18 (inclusive) say they check in with the office at least once a day while on vacation. c. Five or fewer say that they check in with the office at least once a day while on vacation. Would this be an unlikely occurrence?

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