91Ó°ÊÓ

Heart failure is due to either natural occurrences \((87 \%)\) or outside factors \((13 \%) .\) Outside factors are related to induced substances or foreign objects. Natural occurrences are caused by arterial blockage, disease, and infection. Suppose that 20 patients will visit an emergency room with heart failure. Assume that causes of heart failure between individuals are independent. (a) What is the probability that three individuals have conditions caused by outside factors? (b) What is the probability that three or more individuals have conditions caused by outside factors? (c) What are the mean and standard deviation of the number of individuals with conditions caused by outside factors?

Step by step solution

01

Identify the Type of Distribution

Since we're dealing with a fixed number of trials (20 patients) and each patient has two possible outcomes (heart failure due to natural occurrences or outside factors), we use a Binomial Distribution. The probability of success (heart failure due to outside factors) is \( p = 0.13 \), and the probability of failure (heart failure due to natural occurrences) is \( 1 - p = 0.87 \).

02

Calculate Probability of Exactly Three Individuals with Conditions Caused by Outside Factors

For question (a), we need the probability that exactly 3 individuals have heart failure caused by outside factors. Using the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \( n = 20 \), \( k = 3 \), and \( p = 0.13 \). So, \[ P(X = 3) = \binom{20}{3} (0.13)^3 (0.87)^{17} \] Calculating this gives approximately \( P(X = 3) \approx 0.2314 \).

03

Calculate Probability of Three or More Individuals with Conditions Caused by Outside Factors

For question (b), we want the probability that three or more individuals have heart failure due to outside factors, i.e., \( P(X \geq 3) \). This is calculated by finding \( 1 - P(X \leq 2) \). We find \( P(X \leq 2) \) by calculating \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \) using the binomial formula and sum them:\[ P(X = 0) = \binom{20}{0} (0.13)^0 (0.87)^{20} \]\[ P(X = 1) = \binom{20}{1} (0.13)^1 (0.87)^{19} \]\[ P(X = 2) = \binom{20}{2} (0.13)^2 (0.87)^{18} \]Calculating these and adding them gives \( P(X \leq 2) \approx 0.5432 \). So, \( P(X \geq 3) = 1 - 0.5432 = 0.4568 \).

04

Calculate the Mean and Standard Deviation

For question (c), the mean \( \mu \) and standard deviation \( \sigma \) of a Binomial Distribution are given by:\[ \mu = np = 20 \times 0.13 = 2.6 \]\[ \sigma = \sqrt{np(1-p)} = \sqrt{20 \times 0.13 \times 0.87} = \sqrt{2.262} \approx 1.504 \]So, the mean number of individuals with conditions caused by outside factors is 2.6, and the standard deviation is approximately 1.504.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability

Probability is a way to measure the likelihood of an event happening. It is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. In our scenario with heart failure patients:

• The probability of heart failure due to outside factors is given as 0.13.
• This is contrasted with the 0.87 probability of natural causes leading to heart failure.

The formula used to calculate these probabilities for different numbers of patients is called the binomial probability formula. It considers:

• The number of trials (\(n\)), which is 20 in this case because 20 patients are considered.
• The probability of success (\(p\)), or failure due to outside factors, which is 0.13.
• The specific number of successes (\(k\)), like having 3 patients affected by outside factors.

The overall concept of probability allows us to quantify uncertainty and make predictions based on available data.

Heart Failure Causes

Heart failure can be triggered by several factors. In this exercise, these causes are categorized into natural occurrences and outside factors:
Natural occurrences include:

• Arterial blockage, which restricts blood flow.
• Diseases that affect heart tissue.
• Infections that can weaken heart function.

Outside factors, which are what this exercise focuses on, relate to:

• Induced substances like alcohol and drugs.
• Foreign objects or interventions.

By distinguishing between these categories, we gain insights into different health outcomes and the likelihood of intervention in emergency settings. This understanding helps healthcare providers advise on preventive strategies and manage risks.

Mean and Standard Deviation

The mean and standard deviation are essential statistics used to describe a data set's central tendency and variability.
In the context of our problem:

• The mean refers to the average, expected number of patients having conditions caused by outside factors. It is calculated as \( \mu = np \), yielding 2.6 patients.

• The standard deviation tells us about the variability around this mean. Calculated as \( \sigma = \sqrt{np(1-p)} \), it amounts to approximately 1.504.

These values are crucial in understanding both the expected outcome and the typical dispersion around it. This understanding is valuable in analyzing patterns and preparing for possible variations in medical scenarios.

Independent Events

In probability, independent events mean that the occurrence of one event doesn't affect the occurrence of another. Our exercise assumes that each patient's heart failure cause is independent of others.

• This means that whether a patient's heart failure is due to outside factors or natural causes does not influence the likelihood of another patient's failure cause.

This assumption simplifies calculations and separates individual risk profiles. It allows usage of the Binomial Distribution to model the scenario accurately.
Applying the concept of independent events is vital in many fields, especially in statistical modeling, as it often means treating complex systems as arrays of simpler units for analysis.