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Chapter 5: Problem 23

Criminal Justice: Drunk Drivers Harper's Index reported that the number of (Orange County, California) convicted drunk drivers whose sentence included a tour of the morgue was 569, of which only 1 became a repeat offender. (a) Suppose that of 1000 newly convicted drunk drivers, all were required to take a tour of the morgue. Let us assume that the probability of a repeat offender is still \(p=1 / 569 .\) Explain why the Poisson approximation to the binomial would be a good choice for \(r=\) number of repeat offenders out of 1000 convicted drunk drivers who toured the morgue. What is \(\lambda\) to the nearest tenth? (b) What is the probability that \(r=0 ?\) (c) What is the probability that \(r>1 ?\) (d) What is the probability that \(r>2 ?\) (e) What is the probability that \(r>3 ?\)

Short Answer

Expert verified
(a) 位=1.8. (b) P(r=0)鈮�0.165. (c) P(r>1)鈮�0.538. (d) P(r>2)鈮�0.271. (e) P(r>3)鈮�0.111.

Step by step solution

01

Understand Poisson Approximation

The Poisson approximation is suitable when the number of trials (n) is large, and the probability of success (p) is small. In this problem, n = 1000, and p = 1/569, which is approximately 0.00176. Both conditions for using the Poisson approximation are met.
02

Calculate Lambda (位)

The parameter 位 of the Poisson distribution is calculated as 位 = n * p. Given n = 1000 and p = 1/569, \[ 位 = 1000 \times \frac{1}{569} \approx 1.8 \] (rounded to the nearest tenth).
03

Probability for r=0 Repeat Offenders

Using the Poisson distribution: \( P(r=0) = \frac{\lambda^0 \cdot e^{-\lambda}}{0!} = e^{-1.8} \). Calculating directly in a calculator, \[ P(r=0) \approx 0.165 \]
04

Probability for r>1 Repeat Offenders

First, calculate \( P(r\leq1) = P(r=0) + P(r=1) \). For \( P(r=1) \), \[ P(r=1) = \frac{1.8^1 \cdot e^{-1.8}}{1!} = 1.8 \cdot e^{-1.8} \approx 0.297 \] Thus, \[ P(r\leq1) \approx 0.165 + 0.297 = 0.462. \] So, \( P(r>1) = 1 - P(r\leq1) \approx 0.538. \)
05

Probability for r>2 Repeat Offenders

First, calculate \( P(r\leq2) = P(r=0) + P(r=1) + P(r=2) \). For \( P(r=2) \), \[ P(r=2) = \frac{1.8^2 \cdot e^{-1.8}}{2!} = \frac{3.24 \cdot e^{-1.8}}{2} \approx 0.267. \] Thus, \[ P(r\leq2) \approx 0.165 + 0.297 + 0.267 = 0.729. \] So, \( P(r>2) = 1 - P(r\leq2) \approx 0.271. \)
06

Probability for r>3 Repeat Offenders

First, calculate \( P(r\leq3) = P(r=0) + P(r=1) + P(r=2) + P(r=3) \). For \( P(r=3) \), \[ P(r=3) = \frac{1.8^3 \cdot e^{-1.8}}{3!} = \frac{5.832 \cdot e^{-1.8}}{6} \approx 0.160. \] Thus, \[ P(r\leq3) \approx 0.165 + 0.297 + 0.267 + 0.160 = 0.889. \] So, \( P(r>3) = 1 - P(r\leq3) \approx 0.111. \)

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Key Concepts

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

Understanding Probability
Probability is a way of quantifying how likely an event is to happen. Imagine if you toss a coin; there is a probability of 0.5 (or 50%) that it will land on heads. In formal terms, the probability of an event is a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

In this exercise, the focus is on the probability of a driver becoming a repeat offender. Given that the past data showed 1 out of 569 convicted drivers became repeat offenders, this results in a probability of approximately 0.00176. It鈥檚 a small probability, which makes it suitable for special types of probability models like the Poisson distribution. By evaluating the calculated probabilities, students can understand how likely various outcomes are when a large group of individuals go through the exercise of touring the morgue.
Exploring the Binomial Distribution
The binomial distribution is used to model the number of successes in a fixed number of independent trials of a binary experiment. For example, each trial could be a single coin flip, which has two outcomes: heads or tails. Similarly, in our drunk driving scenario, each convicted individual's outcome is either becoming a repeat offender or not.

A key part of the binomial distribution is that each trial is independent with the same probability of success. In our case, this model has parameters \( n = 1000 \) (the number of convicted drivers) and \( p = \frac{1}{569} \) (the probability of a repeat offender). The binomial setup can inform how we expect a certain number of drivers to become repeat offenders. However, because calculating exact probabilities with a binomial formula for large \( n \) can be complex, approximations like the Poisson distribution are sometimes employed.
The Role of Approximation with Poisson Distribution
The Poisson distribution is useful when we're handling a scenario with many trials but with small probabilities per event. This often happens in cases of rare events, such as someone becoming a repeat offender in our context. It serves as an approximate model to the binomial distribution when certain conditions are satisfied: large number of trials and small probability of occurrence.

In such situations, this approximation simplifies the calculation of probabilities considerably by using a parameter called \( \lambda \). Here, \( \lambda \) is the mean number of successes expected in a given interval, calculated as \( \lambda = n \times p \). For this problem, \( \lambda \approx 1.8 \).

In practice, students calculate the probability of different outcomes such as zero repeat offenders or more than one repeat offender using this distribution. These computations provide insights without complexities of the binomial model, especially for larger datasets or when precise data manipulation is crucial for analysis.

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Most popular questions from this chapter

Marketing: Photograpby Does the kid factor make a difference? If you are talking photography, the answer may be yes! The following table is based on information from American Demographics (Vol. 19, No. 7 ). \begin{tabular}{l|cc} \hline Ages of children in household, years & Under 2 & None under 21 \\ \hline Percent of U.S. households that buy photo gear & \(80 \% 6\) & \(50 \%\) \\\ \hline \end{tabular} Let us say you are a market research person who interviews a random sample of 10 households. (a) Suppose you interview 10 households with children under the age of 2 years. Let \(r\) represent the number of such households that buy photo gear. Make a histogram showing the probability distribution of \(r\) for \(r=0\) through \(r=10 .\) Find the mean and standard deviation of this probability distribution. (b) Suppose that the 10 households are chosen to have no children under 21 years old. Let \(r\) represent the number of such households that buy photo gear. Make a histogram showing the probability distribution of \(r\) for \(r=0\) through \(r=10 .\) Find the mean and standard deviation of this probability distribution. (c) Interpretation Compare the distributions in parts (a) and (b). You are designing TV ads to sell photo gear. Could you justify featuring ads of parents taking pictures of toddlers? Explain your answer.

Statistical Literacy Consider each distribution. Determine if it is a valid probability distribution or not, and explain your answer. (a) \begin{tabular}{l|ccc} \hline\(x\) & 0 & 1 & 2 \\ \hline\(P(x)\) & \(0.25\) & \(0.60\) & \(0.15\) \\ \hline \end{tabular} (b) \begin{tabular}{c|ccc} \hline \(\mathbf{x}\) & 0 & 1 & 2 \\ \hline\(P(x)\) & \(0.25\) & \(0.60\) & \(0.20\) \\ \hline \end{tabular}

Statistical Literacy Which of the following are continuous variables, and which are discrete? (a) Speed of an airplane (b) Age of a college professor chosen at random (c) Number of books in the college bookstore (d) Weight of a football player chosen at random (e) Number of lightning strikes in Rocky Mountain National Park on a given day

Binomial Probabilities: Multiple-Cboice Quiz Richard has just been given a 10 -question multiple-choice quiz in his history class. Each question has five answers, of which only one is correct. Since Richard has not attended class recently, he doesn't know any of the answers. Assuming that Richard guesses on all 10 questions, find the indicated probabilities. (a) What is the probability that he will answer all questions correctly? (b) What is the probability that he will answer all questions incorrectly? (c) What is the probability that he will answer at least one of the questions correctly? Compute this probability two ways. First, use the rule for mutually exclusive events and the probabilities shown in Table 3 of Appendix II. Then use the fact that \(P(r \geq 1)=1-P(r=0)\). Compare the two results. Should they be equal? Are they equal? If not, how do you account for the difference? (d) What is the probability that Richard will answer at least half the questions correctly?

Combination of Random Variables: Insurance Risk Insurance companies know the risk of insurance is greatly reduced if the company insures not just one person, but many people. How does this work? Let \(x\) be a random variable representing the expectation of life in years for a 25 -year-old male (i.e., number of years until death). Then the mean and standard deviation of \(x\) are \(\mu=50.2\) years and \(\sigma=11.5\) years (Vital Statistics Section of the Statistical Abstract of the United States, 116 th Edition). Suppose Big Rock Insurance Company has sold life insurance policies to Joel and David. Both are 25 years old, unrelated, live in different states, and have about the same health record. Let \(x_{1}\) and \(x_{2}\) be random variables representing Joel's and David's life expectancies. It is reasonable to assume \(x_{1}\) and \(x_{2}\) are independent. Joel, \(x_{1}: \mu_{1}=50.2 ; \sigma_{1}=11.5\) David, \(x_{2}: \mu_{2}=50.2 ; \sigma_{2}=11.5\) If life expectancy can be predicted with more accuracy, Big Rock will have less risk in its insurance business. Risk in this case is measured by \(\sigma\) (larger \(\sigma\) means more risk). (a) The average life expectancy for Joel and David is \(W=0.5 x_{1}+0.5 x_{2}\). Compute the mean, variance, and standard deviation of \(W\). (b) Compare the mean life expectancy for a single policy \(\left(x_{1}\right)\) with that for two policies \((W)\) (c) Compare the standard deviation of the life expectancy for a single policy \(\left(x_{1}\right)\) with that for two policies \((W)\). (d) The mean life expectancy is the same for a single policy \(\left(x_{1}\right)\) as it is for two policies (W), but the standard deviation is smaller for two policies. What happens to the mean life expectancy and the standard deviation when we include more policies issued to people whose life expectancies have the same mean and standard deviation (i.e., 25 -year-old males)? For instance, for three policies, \(W=(\mu+\mu+\mu) / 3=\mu\) and \(\sigma_{W}^{2}=(1 / 3)^{2} \sigma^{2}+(1 / 3)^{2} \sigma^{2}+(1 / 3)^{2} \sigma^{2}=\) \((1 / 3)^{2}\left(3 \sigma^{2}\right)=(1 / 3) \sigma^{2}\) and \(\sigma_{W}=\frac{1}{\sqrt{3}} \sigma .\) Likewise, for \(n\) such policies, \(W=\mu\) and \(\sigma_{W}^{2}=(1 / n) \sigma^{2}\) and \(\sigma_{W}=\frac{1}{\sqrt{n}} \sigma .\) Looking at the general result, is it appropriate to say that when we increase the number of policies to \(n\), the risk decreases by a factor of \(\sigma_{w}=\frac{1}{\sqrt{n}} ?\)
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