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Some previous studies have shown a relationship between emergency-room admissions per day and level of pollution on a given day. A small local hospital finds that the number of admissions to the emergency ward on a single day ordinarily (unless there is unusually high pollution) follows a Poisson distribution with mean \(=2.0\) admissions per day. Suppose each admitted person to the emergency ward stays there for exactly 1 day and is then discharged. The hospital is planning a new emergency-room facility. It wants enough beds in the emergency ward so that for at least \(95 \%\) of normal-pollution days it will not need to turn anyone away. What is the smallest number of beds it should have to satisfy this criterion?

Short Answer

Expert verified
The hospital should have 5 beds.

Step by step solution

01

Understand the Poisson Distribution

A Poisson distribution is used for counting the number of events (admissions, in this case) that occur within a fixed interval of time or space, given a constant mean rate of occurrence. Here, the mean number of admissions is given as 2 per day, and admissions follow a Poisson distribution.
02

Define the Problem

We need to find the smallest number of beds, denoted as "k", such that the hospital will not need to turn anyone away on at least 95% of normal-pollution days. This implies that the cumulative probability of having up to "k" admissions should be at least 0.95.
03

Use the Poisson Cumulative Distribution Function (CDF)

The cumulative distribution function for a Poisson random variable with mean \( \lambda \) is given by the sum: \[ P(X \leq k) = \sum_{x=0}^{k} \frac{e^{-\lambda} \lambda^x}{x!} \], where \( \lambda = 2 \). We need to find the smallest \( k \) such that \( P(X \leq k) \geq 0.95 \).
04

Calculate Cumulative Probabilities

Calculate \( P(X \leq k) \) for successive values of \( k \) starting from 2, until the cumulative probability reaches or exceeds 0.95. Use the values: - For \( k = 2 \): \( P(X \leq 2) = e^{-2} \left(1 + 2 + \frac{2^2}{2}\right) = 0.6767 \)- For \( k = 3 \): \( P(X \leq 3) = 0.8571 \)- For \( k = 4 \): \( P(X \leq 4) = 0.9473 \)- For \( k = 5 \): \( P(X \leq 5) = 0.9834 \)
05

Determine the Minimum Number of Beds

Looking at the calculations, \( P(X \leq 4) \) is 0.9473, which is slightly below 0.95, and \( P(X \leq 5) \) is 0.9834, which is above 0.95. Hence, the smallest integer \( k \) that satisfies the condition \( P(X \leq k) \geq 0.95 \) is 5.

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Key Concepts

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

Cumulative Distribution Function
The cumulative distribution function (CDF) is an essential tool in statistics that helps us understand the probability of a random variable being less than or equal to a specific value.
In the context of a Poisson distribution, which we are dealing with, the CDF gives us the probability that the number of events (like emergency room admissions) will be a certain count or less within a specific period.
For example, if each day, the average number of admissions is 2, and we want to ensure we have enough beds for 95% of the days, we'd use the CDF to find this probability for different bed numbers.

### Calculating CDF for Poisson
The formula used is: \[ P(X \leq k) = \sum_{x=0}^{k} \frac{e^{-\lambda} \lambda^x}{x!} \]Where:- \( \lambda \) is the average number of events per time period (in our case, 2 admissions per day)- \( k \) is the maximum number of events we can accommodate (the number of beds)
By calculating this for successive values of \( k \), we can determine the minimum \( k \) where the probability meets or exceeds our desired threshold, here being 0.95 for 95% coverage.
Hospital Capacity Planning
Hospital capacity planning is a crucial process that healthcare facilities use to ensure they have the necessary resources to meet patient demand.
In an emergency room setting, this involves determining the right number of beds, considering factors like expected patient inflow and average treatment periods.
By forecasting these numbers using tools such as the Poisson distribution, hospitals can avoid turning patients away due to lack of capacity, especially on 95% of the days which is often the target for service reliability.

### Application of Poisson in Planning
The use of statistical models helps hospitals manage uncertainties in patient admissions. With an average admission rate (like 2 per day in our situation), they calculate the required resources – such as beds – to handle typical demand without overextending resources.
  • This keeps operational costs manageable.
  • Improves patient satisfaction by reducing wait times.
  • Aims to achieve service goals without excessive empty space or staff idle time when admissions are below average.
Having a planned approach allows hospitals to maintain readiness for emergencies while efficiently using resources.
Emergency Room Admissions
Emergency room admissions can be unpredictable, with numbers fluctuating based on various factors, including pollution levels, as noted in the exercise.
Adopting statistical models like the Poisson distribution helps anticipate these variations, allowing hospitals to prepare adequately.
The goal is to develop a system where hospitals are equipped to handle typical daily variations while mitigating the risk of overcrowding.

### Practical Implications
A model showing an average of 2 daily admissions introduces predictable patterns despite some inherent variability.
By using this data, hospitals can optimize their operations:
  • They align staffing levels with patient flow expectations, ensuring sufficient medical attention without undue delays.
  • Resource allocation becomes more efficient, minimizing waste and enhancing the quality of service provided.
In conclusion, understanding emergency room admissions through statistical interpretations ensures better service delivery and optimizes hospital operations, preparing for both typical days and potential emergency surges.

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