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Chapter 5: Problem 26

Business Franchise: Shopping Center Chances: Risk and Odds in Everyday Life, by James Burke, reports that only \(2 \%\) of all local franchises are business failures. A Colorado Springs shopping complex has 137 franchises (restaurants, print shops, convenience stores, hair salons, etc.). (a) Let \(r\) be the number of these franchises that are business failures. Explain why a Poisson approximation to the binomial would be appropriate for the random variable \(r\). What is \(n\) ? What is \(p\) ? What is \(\lambda\) (rounded to the nearest tenth)? (b) What is the probability that none of the franchises will be a business failure? (c) What is the probability that two or more franchises will be business failures? (d) What is the probability that four or more franchises will be business failures?

Short Answer

Expert verified
(a) Poisson is appropriate because \(n=137\), \(p=0.02\), with \(\lambda=2.74\). (b) P(0 failures) = 0.0646. (c) P(\(\geq 2\) failures) = 0.7585. (d) P(\(\geq 4\) failures) = 0.2959.

Step by step solution

01

Understand the Context and Variables

We have a total of 137 franchises, and the probability of a franchise being a failure is given as 0.02 or 2%. The problem involves calculating probabilities concerning the number of business failures, where "failure" is a rare event (2% probability), and there are many trials (137 franchises).
02

Determine Appropriateness of Poisson Approximation

The Poisson approximation to the binomial distribution is appropriate when the number of trials \(n\) is large, and the probability of success \(p\) is small. Here, \(n = 137\) and \(p = 0.02\). The product \(np\) gives us the mean \(\lambda\) of the Poisson distribution. Calculate \(\lambda = np = 137 \times 0.02 = 2.74\). Given \(\lambda\), the Poisson distribution can be used.
03

Find Probability of Zero Failures

The probability of having exactly zero failures is calculated using the Poisson distribution formula: \( P(r) = \frac{e^{-\lambda} \cdot \lambda^r}{r!} \). For zero failures \( r = 0 \), substitute \( \lambda = 2.74 \): \( P(0) = \frac{e^{-2.74} \cdot 2.74^0}{0!} = e^{-2.74} \). Calculate this value.
04

Calculate Probability of Two or More Failures

First, find the probability of zero or one failure using the Poisson distribution and sum them. For one failure \( r = 1 \): \( P(1) = \frac{e^{-2.74} \cdot 2.74^1}{1!} \). Then, calculate the total probability for zero and one failure and subtract from 1 to find the probability of two or more failures: \( P(r \geq 2) = 1 - (P(0) + P(1)) \).
05

Calculate Probability of Four or More Failures

Now, compute probabilities for 0, 1, 2, or 3 failures using the Poisson formula for each and sum them up: \( P(2) = \frac{e^{-2.74} \cdot 2.74^2}{2!} \) and \( P(3) = \frac{e^{-2.74} \cdot 2.74^3}{3!} \). Subtract their sum from 1 to find \( P(r \geq 4) = 1 - (P(0) + P(1) + P(2) + P(3)) \).
06

Consolidate Results

Using a calculator or logarithmic tables: - Compute \( e^{-2.74} \approx 0.0646 \). - Step 3: \( P(0) = 0.0646 \). - Step 4: Compute \( P(1) = 0.0646 \times 2.74 \approx 0.1769 \), \( P(0+1) \approx 0.2415 \), so \( P(r \geq 2) = 1 - 0.2415 \approx 0.7585 \). - Step 5: \( P(2) = 0.0646 \times \frac{2.74^2}{2} \approx 0.2417 \) and \( P(3) = 0.0646 \times \frac{2.74^3}{6} \approx 0.2209 \). Total for 0,1,2,3 failures is \( 0.7041 \), so \( P(r \geq 4) \approx 0.2959 \).

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Key Concepts

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

Binomial Distribution
The binomial distribution is a specific type of probability distribution that is ideal for scenarios where we have several trials, each with two possible outcomes. In the context of the franchise problem, each franchise can either succeed or fail, making it a clear example of a binomial setting.

Key characteristics of the binomial distribution include:
  • Fixed number of trials: We know the total number of franchises, which equals 137.
  • Two possible outcomes: Each franchise can be a success or a failure.
  • Constant probability: The probability of a franchise being a failure is constant at 2% or 0.02 for every franchise.
  • Independence: The outcome of any one franchise does not affect the others.
In many cases with a high number of trials and a low probability of success, calculating probabilities using the binomial formula can be complex and cumbersome. That's where approximations like the Poisson distribution come in handy, simplifying the computation while still giving accurate results.
Probability Calculations
Probability calculations, in this context, help us determine the likelihood of a certain number of failures among the franchises. When using a Poisson approximation, It's vital to set up the problem correctly by determining the key parameters:

  • Mean ( lambda): The mean number of expected failures is given by lambda, calculated as the product of the total number of trials (137) and the probability of failure (0.02). Hence, lambda = 137 \times 0.02 = 2.74. This is the key parameter for the Poisson distribution.
For specific probabilities like zero, two or four failures, the Poisson probability mass function (PMF) is used:
  • Formula: The general Poisson formula is \( P(r) = \frac{e^{-\lambda} \cdot \lambda^r}{r!} \). For zero failures, r = 0, simplifying the calculation significantly.
By iterating calculations for different numbers of failures and summing results where needed, students can determine the likelihood of alternative outcomes, helping them understand real-world probabilities in business applications.
Statistical Approximation
Statistical approximation methods allow us to handle complex distributions efficiently. In situations involving a large number of trials, like the 137 franchises example, using approximations can simplify calculations.

The Poisson approximation to the binomial distribution is particularly useful here when:
  • The number of trials is large (n): As with 137 franchises.
  • The probability of success (or in this case failure, p) is small: Such as the 2% failure rate.
The essential step is computing the mean number of expected events (failures, in this case), denoted as lambda = np, which is used in Poisson's formula. This allows us to approximate probabilities for variables like zero, two, or four+ failures by simplifying the normally complex binomial calculations into more manageable Poisson calculations.

Understanding these approximation methods is crucial in statistical education and helps in making quicker and accurate decisions in fields like business and economics, where timely data analysis is often necessary.

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Most popular questions from this chapter

Law: Bar Exam Bob is a recent law school graduate who intends to take the state bar exam. According to the National Conference on Bar Examiners, about \(57 \%\) of all people who take the state bar exam pass (Source: The Book of Odds by Shook and Shook, Signet). Let \(n=1,2,3, \ldots\) represent the number of times a person takes the bar exam until the first pass. (a) Write out a formula for the probability distribution of the random variable \(n\). (b) What is the probability that Bob first passes the bar exam on the second try \((n=2) ?\) (c) What is the probability that Bob needs three attempts to pass the bar exam? (d) What is the probability that Bob needs more than three attempts to pass the bar exam? (e) What is the expected number of attempts at the state bar exam Bob must make for his (first) pass? Hint: Use \(\mu\) for the geometric distribution and round.

Criminal Justice: Parole USA Today reports that about \(25 \%\) of all prison parolees become repeat offenders. Alice is a social worker whose job is to counsel people on parole. Let us say success means a person does not become a repeat offender. Alice has been given a group of four parolees. (a) Find the probability \(P(r)\) of \(r\) successes ranging from 0 to 4 . (b) Make a histogram for the probability distribution of part (a). (c) What is the expected number of parolees in Alice's group who will not be repeat offenders? What is the standard deviation? (d) Quota Problem How large a group should Alice counsel to be about \(98 \%\) sure that three or more parolees will not become repeat offenders?

Statistical Literacy In a binomial experiment, is it possible for the probability of success to change from one trial to the next? Explain.

Health Care: Office Visits What is the age distribution of patients who make office visits to a doctor or nurse? The following table is based on information taken from the Medical Practice Characteristics section of the Statistical Abstract of the United States (116th Edition). \begin{tabular}{l|ccccc} \hline Age group, years & Under 15 & \(15-24\) & \(25-44\) & \(45-64\) & 65 and older \\ \hline Percent of office visitors & \(20 \%\) & \(10 \%\) & \(25 \%\) & \(20 \%\) & \(25 \%\) \\ \hline \end{tabular} Suppose you are a district manager of a health management organization (HMO) that is monitoring the office of a local doctor or nurse in general family practice. This morning the office you are monitoring has eight office visits on the schedule. What is the probability that (a) at least half the patients are under 15 years old? First, explain how this can be modeled as a binomial distribution with 8 trials, where success is visitor age is under 15 years old and the probability of success is \(20 \%\). (b) from 2 to 5 patients are 65 years old or older (include 2 and 5 )? (c) from 2 to 5 patients are 45 years old or older (include 2 and 5 )? Hint: Success is 45 or older. Use the table to compute the probability of success on a single trial. (d) all the patients are under 25 years of age? (e) all the patients are 15 years old or older?

Statistical Literacy What does the random variable for a binomial experiment of \(n\) trials measure?
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