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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) \( \lambda = 1.8 \); (b) \( 0.1653 \); (c) \( 0.5371 \); (d) \( 0.2693 \); (e) \( 0.1066 \).

Step by step solution

01

Understanding when to use the Poisson Approximation

The Poisson approximation is a suitable choice for a binomial distribution when the number of trials \( n \) is large, and the probability of success \( p \) is small. In this case, we have \( n = 1000 \) and \( p = \frac{1}{569} \), which fits these criteria.
02

Calculate the rate parameter \( \lambda \)

The parameter \( \lambda \) for a Poisson distribution is given by \( \lambda = np \). Substituting the values, \( \lambda = 1000 \times \frac{1}{569} \approx 1.8 \). So, \( \lambda = 1.8 \) to the nearest tenth.
03

Calculate Probability for \( r = 0 \)

Using the Poisson probability formula \( P(r) = \frac{\lambda^r e^{-\lambda}}{r!} \), for \( r = 0 \): \[ P(0) = \frac{1.8^0 e^{-1.8}}{0!} = e^{-1.8} \approx 0.1653 \].
04

Calculate Probability for \( r > 1 \)

First, calculate \( P(r \leq 1) \): 1. \[ P(1) = \frac{1.8^1 e^{-1.8}}{1!} = 1.8 e^{-1.8} \approx 0.2976 \].2. \[ P(0) + P(1) = 0.1653 + 0.2976 = 0.4629 \].Therefore, \( P(r > 1) = 1 - P(r \leq 1) = 1 - 0.4629 = 0.5371 \).
05

Calculate Probability for \( r > 2 \)

First, calculate \( P(r \leq 2) \): 1. \[ P(2) = \frac{1.8^2 e^{-1.8}}{2!} = \frac{3.24 e^{-1.8}}{2} \approx 0.2678 \].2. \[ P(0) + P(1) + P(2) = 0.1653 + 0.2976 + 0.2678 = 0.7307 \].Therefore, \( P(r > 2) = 1 - P(r \leq 2) = 1 - 0.7307 = 0.2693 \).
06

Calculate Probability for \( r > 3 \)

First, calculate \( P(r \leq 3) \): 1. \[ P(3) = \frac{1.8^3 e^{-1.8}}{3!} = \frac{5.832 e^{-1.8}}{6} \approx 0.1607 \].2. \[ P(0) + P(1) + P(2) + P(3) = 0.1653 + 0.2976 + 0.2678 + 0.1607 = 0.8934 \].Therefore, \( P(r > 3) = 1 - P(r \leq 3) = 1 - 0.8934 = 0.1066 \).

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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 common probability distribution used to model the number of successes in a series of independent and identical trials. In each trial, there are only two possible outcomes: success or failure. For example, when flipping a coin, each flip could be considered a trial with 'heads' representing success and 'tails' representing failure.

The binomial distribution is defined by two parameters: the number of trials \(n\) and the probability of success in a single trial \(p\). The probability mass function for a binomial distribution is given by:

\[ P(k; n, p) = \binom{n}{k} p^k (1-p)^{n-k} \]

where \(k\) represents the number of successes. In the context of the exercise, we have 1000 trials (the number of convicted drunk drivers) and a very small probability of any one of them being a repeat offender \(p = \frac{1}{569}\).

When the number of trials is large and the probability \(p\) is small, calculating binomial probabilities directly can be challenging due to large computational demands. This is where using a Poisson approximation can be very useful.
Probability Calculations
Probability calculations are crucial in statistical analysis to predict the likelihood of various outcomes. In many real-world cases, exact probability calculations using distributions like binomial can be tedious, especially with a large number of trials and very small probabilities.

To simplify these calculations, approximations and estimates are often used. One useful method involves the Poisson approximation, which is helpful for calculating probabilities when dealing with rare events in large populations. This helps in overcoming computational complexities.

For instance, in this exercise, we are interested in finding the probabilities that a specific event (like having more than one repeat offender) occurs. We use the converted Poisson probabilities to easily determine the chances of observing different numbers of repeat offenders.
Poisson Distribution
The Poisson distribution is a powerful tool used to model the number of times an event occurs within a fixed interval of time or space. It is ideal for situations where events happen independently and at a constant average rate over time.

The key parameter of the Poisson distribution is \( \lambda \), which represents the average number of occurrences within the designated interval. The probability of observing \( r \) occurrences is given by the formula:

\[ P(r; \lambda) = \frac{\lambda^r e^{-\lambda}}{r!} \]

In cases where the binomial distribution is challenging to compute directly due to high \( n \) and low \( p \), the Poisson distribution provides a simpler approximation. The rate parameter \( \lambda \) for the Poisson approximation is given by \( \lambda = np \).

In our exercise, with \( n = 1000 \) and \( p = \frac{1}{569} \), we find \( \lambda \approx 1.8 \). This simplifies the calculations and allows us to easily compute the probabilities of having 0, 1, 2, or more repeat offenders.

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Most popular questions from this chapter

Henry Petroski is a professor of civil engineering at Duke University. In his book To Engineer Is Human: The Role of Failure in Successful Design, Professor Petroski says that up to \(95 \%\) of all structural failures, including those of bridges, airplanes, and other commonplace products of technology, are believed to be the result of crack growth. In most cases, the cracks grow slowly. It is only when the cracks reach intolerable proportions and still go undetected that catastrophe can occur. In a cement retaining wall, occasional hairline cracks are normal and nothing to worry about. If these cracks are spread out and not too close together, the wall is considered safe. However, if a number of cracks group together in a small region, there may be real trouble. Suppose a given cement retaining wall is considered safe if hairline cracks are evenly spread out and occur on the average of 4.2 cracks per 30 -foot section of wall. (a) Explain why a Poisson probability distribution would be a good choice for the random variable \(r=\) number of hairline cracks for a given length of retaining wall. (b) In a 50 -foot section of safe wall, what is the probability of three (evenly spread-out) hairline cracks? What is the probability of three or more (evenly spread-out) hairline cracks? (c) Answer part (b) for a 20 -foot section of wall. (d) Answer part (b) for a 2 -foot section of wall. Round \(\lambda\) to the nearest tenth. (e) Consider your answers to parts (b), (c), and (d). If you had three hairline cracks evenly spread out over a 50 -foot section of wall, should this be cause for concern? The probability is low. Could this mean that you are lucky to have so few cracks? On a 20-foot section of wall [part (c)], the probability of three cracks is higher. Does this mean that this distribution of cracks is closer to what we should expect? For part (d), the probability is very small. Could this mean you are not so lucky and have something to worry about? Explain your answers.

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