91Ó°ÊÓ

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?

Step by step solution

01

Understand the Problem

We have a neighborhood with 316 homes, and we want to know the likelihood of burglaries using the given average rate per 100,000 homes.

02

Identify Parameters for Poisson Approximation

The binomial distribution is approximated by a Poisson distribution when the number of trials is large, and the probability of success is small. Here, the number of trials is the number of homes, so \(n = 316\). The probability of burglary for one home is \(p = \frac{661}{100,000}\), which is quite small.

03

Calculate Rate \( \lambda \) for Poisson Distribution

For the Poisson approximation, the parameter \( \lambda \) is calculated as \( \lambda = n \times p = 316 \times \frac{661}{100,000} = 2.08876 \). Rounding to the nearest tenth, we have \( \lambda \approx 2.1 \).

04

Compute Probability for Zero Burglaries

The probability of zero burglaries, given a Poisson distribution with \( \lambda = 2.1 \), is calculated using \( P(X=0) = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-2.1} \approx 0.1225 \).

05

Compute Probability for No More Than One Burglary

To find the probability of no more than one burglary, we calculate \( P(X \leq 1) = P(X=0) + P(X=1) \). We've already found \( P(X=0) \). Now, \( P(X=1) = \frac{\lambda^1 e^{-\lambda}}{1!} = 2.1 e^{-2.1} \approx 0.2573 \). So, \( P(X \leq 1) = 0.1225 + 0.2573 = 0.3798 \).

06

Compute Probability for Two or More Burglaries

The probability that there will be two or more burglaries is the complement of having no more than one burglary: \( P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.3798 = 0.6202 \).

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Key Concepts

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

Understanding the Binomial Distribution

Understanding the Binomial Distribution is essential when we have a specific number of independent trials, each with two possible outcomes: success or failure. In the context of our burglary problem, we can think of each house having either the outcome of being burglarized (success) or not (failure).
However, to make the situation clear, the Binomial Distribution is characterized by:

• **Number of Trials (n):** This is the fixed number of independent trials or events. In our problem, it refers to the 316 homes in the neighborhood.
• **Probability of Success (p):** Each trial, such as a house being burglarized, has the same probability of success. Here, the probability of a single house being burglarized is 661 out of 100,000 homes.
• **Discrete Possible Outcomes:** We can calculate the probability of having exactly 0, 1, 2, ... up to 316 successful outcomes (s), but this becomes cumbersome when dealing with events having low probabilities and a large number of trials.

Because in this scenario, we have a large number of trials and low probability of success, the Poisson approximation is often more practical to use.

Making Probability Calculations Simple

Probability Calculations involve determining how likely an event is to happen. For our example, we need to calculate different probabilities concerning burglaries in the neighborhood using the Poisson Distribution.
The Poisson Distribution formula is: \[P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} \]Where:

• \(k\) represents the number of successes that result from the experiment (e.g., burglars).
• \(\lambda\) is the average rate of success (in this situation, the lambda is 2.1, calculated as the number of homes times the small probability of them being burglarized). The value of \(\lambda\) is crucial as it sets the average number of incidents.

For zero burglaries, we use \(k = 0\). For one, it is \(k = 1\) and so forth. Notice that probabilities for different \(k\) can be added to find cumulative probabilities such as the possibility of no more than one burglary.These calculations help us easily figure out the chances of different numbers of burglaries occurring, which is particularly helpful for understanding scenarios where events are rare but possible in large populations.

Why Poisson Approximation Is Useful

The Poisson Approximation is a powerful tool, especially when dealing with events that happen with low probability over a high number of occurrences. Here's why it works well for our burglary problem:

• **Many Trials, Small Probability:** For the approximation to be valid, you need a large number of trials (316 homes) with each having a small probability of the event happening (a burglary).
• **Simplifies Complex Calculations:** The Poisson approach simplifies probability computations when \(n\) is large and \(p\) is small, which would be painful to compute with the traditional Binomial model.
• **Characteristic Lambda:** It uses \(\lambda = np\) to determine average occurrences (in this case, approximately 2.1 burglaries in one year), making it easier to compute probabilities for any \(k\) number of burglaries.

Thus, using the Poisson Approximation in scenarios like these doesn't just make calculations more manageable. It helps us provide a clear picture of the probability landscape in real-world problems with numerous potential events but limited actual occurrences.