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The article "An Association Between Fine Particles and Asthma Emergency Department Visits for Children in Seattle" [Environmental Health Perspectives June, 1999107(6)\(]\) used Poisson models for the number of asthma emergency department (ED) visits per day. For the zip codes studied, the mean ED visits were 1.8 per day. Determine the following: (a) Probability of more than five visits in a day. (b) Probability of fewer than five visits in a week. (c) Number of days such that the probability of at least one visit is \(0.99 .\) (d) Instead of a mean of 1.8 per day, determine the mean visits per day such that the probability of more than five visits in a day is 0.1 .

Short Answer

Expert verified
(a) Use the cumulative Poisson formula: \( P(X>5) = 0.0138 \). (b) Calculate \( P(X<5) \) for weekly, \( P(X<5) = 0.0316 \). (c) Solve \( n = 3 \) days. (d) \( \lambda \approx 3.57 \).

Step by step solution

01

Understand the Poisson Distribution

The number of ED visits per day follows a Poisson distribution. This distribution is defined by a single parameter, \( \lambda \), which represents the average number of occurrences in a fixed interval (in this case, 1.8 visits per day). Define the probability mass function for a Poisson distribution as \( P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} \).
02

Calculate Probability for More Than Five Visits in a Day

To find the probability of more than five visits in a day, compute \( P(X > 5) = 1 - P(X \leq 5) \). Use the cumulative probability formula for \( X \leq 5 \) and sum the probabilities from 0 to 5 using \( P(X=k) = \frac{1.8^k e^{-1.8}}{k!} \) for \( k = 0, 1, 2, 3, 4, 5 \). Subtract this from 1 to get \( P(X > 5) \).
03

Calculate the Probability of Fewer Than Five Visits in a Week

First, calculate the mean visits per week as \( \lambda_{week} = 1.8 \times 7 = 12.6 \). Now, calculate \( P(X < 5) = \sum_{k=0}^{4} \frac{12.6^k e^{-12.6}}{k!} \).
04

Calculate Days for Probability of At Least One Visit Being 0.99

The probability of zero visits on a given day can be found as \( P(X=0) = e^{-\lambda} \). If we need the probability of at least one visit to be 0.99, we solve \( 1 - (e^{-1.8})^n = 0.99 \) where \( n \) is the number of days. This can be rearranged to \( (e^{-1.8})^n = 0.01 \). Find \( n \) by solving \( n = \frac{\ln(0.01)}{-1.8} \).
05

Determine Mean Visits for Probability of More Than Five Being 0.1

We want to find \( \lambda \) such that \( P(X > 5) = 0.1 \). Start by expressing this as \( 1 - P(X \leq 5) = 0.1 \), giving \( P(X \leq 5) = 0.9 \). Calculate \( P(X \leq 5) \) using different \( \lambda \) values until the sum from 0 to 5 meets 0.9.

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Key Concepts

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

Probability Calculations
When dealing with Poisson distribution, one of the most essential skills is being able to calculate probabilities. This involves understanding and applying the probability mass function (PMF) of the Poisson distribution, which is defined as \( P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} \). Here, \( \lambda \) represents the average number of occurrences, and \( k \) is the number of occurrences you are interested in.

For example, to find the probability of more than five asthma emergency department visits per day, you need to calculate \( P(X > 5) \). This requires summing the probabilities of \( X \) from 0 to 5 using the PMF, and then subtracting the result from 1 to get the cumulative probability of the desired range. Similarly, calculating probabilities over a week or for a specific requirement involves adjusting \( \lambda \) to reflect the time period or conditions. Knowing how to manipulate and combine these probabilities is crucial for accurate calculations.
Asthma Emergency Department Visits
In the study discussed, the focus was on emergency department visits for asthma among children. This context is crucial since healthcare data often follow a Poisson distribution due to their nature of involving counts of events over time. For example, the average number of emergency visits per day was found to be 1.8 for the studied zip codes.

Understanding this average is critical, as it serves as the mean (\( \lambda \)) for setting up the Poisson model. This average helps in predicting the likelihood of any given number of visits on a particular day or over a week. It’s also important to note how external factors like environmental conditions could impact this number, potentially altering the probability calculations.
Cumulative Distribution
Cumulative distribution functions (CDF) play a key role when working with probabilities in a Poisson distribution. Specifically, when we want the probability of \( X \) being less than or equal to a certain value, we use the cumulative distribution.

For instance, calculating \( P(X \leq 5) \) as part of determining the probability of having more than five visits involved summing the probabilities of events from 0 to 5. This cumulative approach simplifies finding intervals as it allows you to determine how likely a range of outcomes is, rather than individual occurrences. CDFs enable a broader view of probabilities in context, which is particularly useful for decision-making in healthcare scenarios.
Mean Visits
The concept of mean visits is crucial in the context of Poisson distribution, as it directly influences the probability of different event counts. The mean, denoted as \( \lambda \), represents the expected number of occurrences within a fixed timeframe, like a day or a week.

In the case study mentioned, the mean was 1.8 visits per day. This becomes the base parameter for calculating probabilities. If the healthcare center aimed for a certain likelihood of visits, such as ensuring the probability of more than five visits is 0.1, the task would be to find a different \( \lambda \) that meets these modified conditions. Adjusting this mean helps adapt to new scenarios or goals, providing flexibility and precision in planning and resource allocation.

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