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Chapter 5: Problem 8

Basic Computation: Poisson Distribution Given a binomial experiment with \(n=150\) trials and probability of success on a single trial \(p=0.06\), find the value of \(\lambda\) and then use the Poisson distribution to estimate the probability of \(r \leq 2\) successes.

Short Answer

Expert verified
The value of \( \lambda \) is 9, and the probability of \( r \leq 2 \) successes is approximately 0.0062.

Step by step solution

01

Calculate the Poisson Parameter (位)

The parameter \( \lambda \) for a Poisson distribution is the product of the number of trials \(n\) and the probability of success \(p\) in a single trial. This is calculated as \( \lambda = n \times p \). Given \( n = 150 \) and \( p = 0.06 \), we have:\[ \lambda = 150 \times 0.06 = 9 \]
02

Identify the Poisson Probability Formula

The probability of exactly \( r \) successes in a Poisson distribution is given by:\[ P(X = r) = \frac{e^{-\lambda} \cdot \lambda^r}{r!} \]For this problem, we need to find \( P(r \leq 2) \), which means we need to calculate \( P(r=0) + P(r=1) + P(r=2) \).
03

Calculate P(r = 0)

Substitute \( r = 0 \) into the Poisson formula:\[ P(r = 0) = \frac{e^{-9} \cdot 9^0}{0!} = e^{-9} \cdot 1 = e^{-9} \]Using a calculator, \( e^{-9} \approx 0.000123 \).
04

Calculate P(r = 1)

Substitute \( r = 1 \) into the Poisson formula:\[ P(r = 1) = \frac{e^{-9} \cdot 9^1}{1!} = e^{-9} \cdot 9 \approx 0.001103 \]
05

Calculate P(r = 2)

Substitute \( r = 2 \) into the Poisson formula:\[ P(r = 2) = \frac{e^{-9} \cdot 9^2}{2!} = e^{-9} \cdot \frac{81}{2} \approx 0.004964 \]
06

Sum the Probabilities

Add the probabilities for \( r = 0 \), \( r = 1 \), and \( r = 2 \):\[ P(r \leq 2) = P(r=0) + P(r=1) + P(r=2) \approx 0.000123 + 0.001103 + 0.004964 \approx 0.006190 \]

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Key Concepts

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

Understanding a Binomial Experiment
A binomial experiment is a type of statistical experiment that has two possible outcomes in each trial, often labeled as "success" or "failure." Here are some key features:
  • The number of trials, denoted by \(n\), is fixed and known in advance.
  • Each trial is independent, meaning the outcome of one doesn鈥檛 affect another.
  • The probability of success, denoted by \(p\), is the same for each trial.
In our exercise, we have a binomial experiment with \(n=150\) trials. Each trial has a probability of success \(p=0.06\). We observe successes as events that fit this probability in the large number of trials.
Calculating Probability of Success
Probability of success refers to \(p\), the likelihood of achieving a desired outcome in a single trial. In the context of our binomial experiment, this success probability remains constant across all trials.
When considering a large number of trials with a small probability of success, using the binomial model directly may become cumbersome. This is because calculating probabilities for many outcomes is complex.
Here is where the Poisson distribution offers a practical approximation if the product of the number of trials \(n\) and the probability of success \(p\) (yielding \(\lambda\)) is relatively small.
Exploring the Poisson Probability Formula
The Poisson distribution is a powerful tool when working with rare events in large sample sizes. It simplifies calculations by approximating the binomial distribution under certain conditions. A crucial component of the Poisson distribution is the parameter \(\lambda\), which is the expected number of occurrences of the event.
The Poisson probability formula is expressed as:\[ P(X = r) = \frac{e^{-\lambda} \cdot \lambda^r}{r!} \]where:
  • \(e\) is Euler's number, approximately 2.71828.
  • \(\lambda\) is the average number of successes.
  • \(r!\) is the factorial of \(r\), representing how many ways \(r\) successes can occur.
In our exercise, with \(\lambda = 9\), we calculate the probability of \(r \leq 2\) by summing up \(P(r=0)\), \(P(r=1)\), and \(P(r=2)\). This gives us the total probability of having 2 or fewer successes in our scenario. Thus, Poisson distribution helps us convert complex binomial situations into more manageable calculations.

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Most popular questions from this chapter

Negative Binomial Distribution: Type A Blood Donors Blood type A occurs in about \(41 \%\) of the population (Reference: Laboratory and Diagnostic Tests by F. Fischbach). A clinic needs 3 pints of type A blood. A donor usually gives a pint of blood. Let \(n\) be a random variable representing the number of donors needed to provide 3 pints of type A blood. (a) Explain why a negative binomial distribution is appropriate for the random variable \(n\). Write out the formula for \(P(n)\) in the context of this application. Hint: See Problem 30 . (b) Compute \(P(n=3), P(n=4), P(n=5)\), and \(P(n=6)\). (c) What is the probability that the clinic will need from three to six donors to obtain the needed 3 pints of type A blood? (d) What is the probability that the clinic will need more than six donors to obtain 3 pints of type A blood? (e) What are the expected value \(\mu\) and standard deviation \(\sigma\) of the random variable \(n\) ? Interpret these values in the context of this application.

Law Enforcement: Burglaries The Honolulu Advertiser stated that in Honolulu there was an average of 661 burglaries per 100,000 households in a given year. In the Kohola Drive neighborhood there are 316 homes. Let \(r=\) number of these homes that will be burglarized in a year. (a) Explain why the Poisson approximation to the binomial would be a good choice for the random variable \(r\). What is \(n\) ? What is \(p ?\) What is \(\lambda\) to the nearest tenth? (b) What is the probability that there will be no burglaries this year in the Kohola Drive neighborhood? (c) What is the probability that there will be no more than one burglary in the Kohola Drive neighborhood? (d) What is the probability that there will be two or more burglaries in the Kohola Drive neighborhood?

Interpretation From long experience a landlord knows that the probability an apartment in a complex will not be rented is \(0.10 .\) There are 20 apartments in the complex, and the rental status of each apartment is independent of the status of the others. When a minimum of 16 apartment units are rented, the landlord can meet all monthly expenses. Which probability is more relevant to the landlord in terms of being able to meet expenses: the probability that there are exactly four unrented units or the probability that there are four or fewer unrented units? Explain.

Fishing: Lake Trout At Fontaine Lake Camp on Lake Athabasca in northern Canada, history shows that about \(30 \%\) of the guests catch lake trout over 20 pounds on a 4 -day fishing trip (Source: Athabasca Fishing Lodges, Saskatoon, Canada). Let \(n\) be a random variable that represents the first trip to Fontaine Lake Camp on which a guest catches a lake trout over 20 pounds. (a) Write out a formula for the probability distribution of the random variable \(n\). (b) Find the probability that a guest catches a lake trout weighing at least 20 pounds for the first time on trip number \(3 .\) (c) Find the probability that it takes more than three trips for a guest to catch a lake trout weighing at least 20 pounds. (d) What is the expected number of fishing trips that must be taken to catch the first lake trout over 20 pounds? Hint: Use \(\mu\) for the geometric distribution and round.

Critical Thinking In an experiment, there are \(n\) independent trials. For each trial, there are three outcomes, \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). For each trial, the probability of outcome \(\mathrm{A}\) is \(0.40 ;\) the probability of outcome \(\mathrm{B}\) is \(0.50 ;\) and the probability of outcome \(\mathrm{C}\) is \(0.10 .\) Suppose there are 10 trials. (a) Can we use the binomial experiment model to determine the probability of four outcomes of type A, five of type \(B\), and one of type C? Explain. (b) Can we use the binomial experiment model to determine the probability of four outcomes of type \(\mathrm{A}\) and six outcomes that are not of type \(\mathrm{A}\) ? Explain. What is the probability of success on each trial?
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