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91Ó°ÊÓ

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) The Poisson approximation applies because \(n=1000\) is large and \(p=1/569\) is small. \(\lambda=1.8\). (b) \(P(r=0) \approx 0.1653\). (c) \(P(r>1) \approx 0.5372\). (d) \(P(r>2) \approx 0.2695\). (e) \(P(r>3) \approx 0.1089\).

Step by step solution

01

Understanding Why to Use Poisson Approximation

The Poisson approximation to the binomial distribution is appropriate when the probability of success \( p \) is small, and the number of trials \( n \) is large. Given that \( p = \frac{1}{569} \approx 0.00176 \) is indeed small, and \( n = 1000 \) is relatively large, the conditions for Poisson approximation are satisfied.
02

Calculate Lambda for Poisson Distribution

In a Poisson distribution, the parameter \( \lambda \) is calculated as \( \lambda = np \). Substitute \( n = 1000 \) and \( p = \frac{1}{569} \) to get \( \lambda = 1000 \times \frac{1}{569} \approx 1.757 \). Round this to the nearest tenth, \( \lambda \approx 1.8 \).
03

Find Probability for r=0

For a Poisson distribution, the probability that \( r = 0 \) is given by \( P(r=0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-\lambda} \). So \( P(r=0) = e^{-1.8} \approx 0.1653 \).
04

Find Probability for r>1

To find \( P(r>1) \), one can subtract the probabilities of \( r = 0 \) and \( r = 1 \) from 1. First, calculate \( P(r=1) = \frac{e^{-1.8} \times 1.8^1}{1!} = 1.8e^{-1.8} \approx 0.2975 \). Then, \( P(r>1) = 1 - (P(r=0) + P(r=1)) = 1 - (0.1653 + 0.2975) \approx 0.5372 \).
05

Find Probability for r>2

Next, calculate \( P(r=2) = \frac{e^{-1.8} \times 1.8^2}{2!} = \frac{1.8^2}{2} \times e^{-1.8} \approx 0.2677 \). Therefore, \( P(r>2) = 1 - (P(r=0) + P(r=1) + P(r=2)) = 1 - (0.1653 + 0.2975 + 0.2677) \approx 0.2695 \).
06

Find Probability for r>3

Calculate \( P(r=3) = \frac{e^{-1.8} \times 1.8^3}{3!} = \frac{1.8^3}{6} \times e^{-1.8} \approx 0.1606 \). Thus, \( P(r>3) = 1 - (P(r=0) + P(r=1) + P(r=2) + P(r=3)) = 1 - (0.1653 + 0.2975 + 0.2677 + 0.1606) \approx 0.1089 \).

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Key Concepts

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

Binomial Approximation
The Binomial Approximation is a technique used in statistics to simplify calculations involving the binomial distribution. It is particularly useful when certain conditions are met:
  • The number of trials, denoted as \( n \) in the experiment, is large.
  • The probability of success, \( p \), in each trial is small.
In the context of the exercise, we are looking at 1000 newly convicted drunk drivers with a probability \( p = \frac{1}{569} \) of becoming repeat offenders. Given that \( n = 1000 \) and \( p \approx 0.00176 \) which is small, this is ideal for using the Poisson approximation.

A Binomial Distribution models the total number of successes when each trial is independent. When trials grow large, but each has a tiny probability of success, calculating probabilities directly through binomial formulas becomes cumbersome. Here, entering Poisson Distribution simplifies the process, estimating those very low-probability events efficiently. This is why Poisson is applied in the given problem, making it manageable to work with a large number of trials and achieving approximations quickly.
Probability Theory
Probability Theory is the mathematical framework for quantifying the chances of various outcomes. Central to many statistical analyses, it provides tools for understanding random events and phenomena.

In the exercise, the probability of a repeat offender event plays a crucial role. By assuming that each driver has a small chance (\( p = \frac{1}{569} \)) to reoffend, we focus on extracting probabilities for different numbers of repeat offenders (e.g., \( r = 0, r > 1 \)) using Poisson's approximation.
  • The probability of exactly zero repeat offenders \( r = 0 \) is given by the Poisson formula \( P(r=0) = e^{-\lambda} \), providing an efficient way to determine improbable events.
  • Probability calculations rely on fundamental properties such as independence, mutual exclusivity, and the cumulative probability which considers various scenarios revealing real-world implications.
    • Each unique configuration can be understood and managed by assigning probability values that make sense in the context of the data and constraints provided.
Statistical Analysis
Statistical Analysis involves examining and evaluating data using various statistical methods and models. It allows us to make informed conclusions based on data patterns rather than mere conjecture.

In the context of the problem, we conducted a statistical analysis by employing Poisson Distribution to approximate the probabilities for different numbers of repeat offenders:
  • The analysis starts with calculating \( \lambda = np \) leading to \( \lambda = 1.8 \), representing the average rate of repeat offenses assumed across the given trials.
  • Using Poisson, we then ascertain probabilities for specific cases like \( r=0 \) to \( r>3 \) through stepwise calculations. These provide actionable insights into expected outcomes, informing possible policy decisions.
The analysis allows us to understand the effectiveness of measures like morgue tours in preventing repeat offenses. It combines theoretical savor with empirical data to offer compelling predictions and conclusions beneficial for informed decision-making.

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Most popular questions from this chapter

Poisson Approximation to Binomial: Comparisons (a) For \(n=100, p=0.02\), and \(r=2\), compute \(P(r)\) using the formula for the binomial distribution and your calculator: $$ P(r)=C_{n, t} p^{r}(1-p)^{n-r} $$ (b) For \(n=100, p=0.02\), and \(r=2\), estimate \(P(r)\) using the Poisson approximation to the binomial. (c) Compare the results of parts (a) and (b). Does it appear that the Poisson distribution with \(\lambda=n p\) provides a good approximation for \(P(r=2) ?\) (d) Repeat parts (a) to \((c)\) for \(r=3\)

What does the expected value of a binomial distribution with \(n\) trials tell you?

Approximately \(3.6 \%\) of all (untreated) Jonathan apples had bitter pit in a study conducted by the botanists Ratkowsky and Martin (Source: Australian Journal of Agricultural Research, Vol. 25, pp. \(783-790) .\) (Bitter pit is a disease of apples resulting in a soggy core, which can be caused either by overwatering the apple tree or by a calcium deficiency in the soil.) Let \(n\) be a random variable that represents the first Jonathan apple chosen at random that has bitter pit. (a) Write out a formula for the probability distribution of the random variable \(n .\) (b) Find the probabilities that \(n=3, n=5\), and \(n=12\). (c) Find the probability that \(n \geq 5\). (d) What is the expected number of apples that must be examined to find the first one with bitter pit? Hint: Use \(\mu\) for the geometric distribution and round.

Consider two binomial distributions, with \(n\) trials each. The first distribution has a higher probability of success on each trial than the second. How does the expected value of the first distribution compare to that of the second?

Consider a binomial distribution of 200 trials with expected value 80 and standard deviation of about \(6.9 .\) Use the criterion that it is unusual to have data values more than \(2.5\) standard deviations above the mean or \(2.5\) standard deviations below the mean to answer the following questions. (a) Would it be unusual to have more than 120 successes out of 200 trials? Explain. (b) Would it be unusual to have fewer than 40 successes out of 200 trials? Explain. (c) Would it be unusual to have from 70 to 90 successes out of 200 trials? Explain.

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