/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 The article "Doctors Misdiagnose... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article "Doctors Misdiagnose More Women, Blacks" (San Luis Obispo Tribune, April 20, 2000) gave the following information, which is based on a large study of more than 10,000 patients treated in emergency rooms in the eastern and midwestern United States: 1\. Doctors misdiagnosed heart attacks in \(2.1 \%\) of all patients. 2\. Doctors misdiagnosed heart attacks in \(4.3 \%\) of black patients. 3\. Doctors misdiagnosed heart attacks in \(7 \%\) of women under 55 years old. Use the following event definitions: \(M=\) event that a heart attack is misdiagnosed, \(B=\) event that a patient is black, and \(W=\) event that a patient is a woman under 55 years old. Translate each of the three statements into probability notation.

Short Answer

Expert verified
The probabilities translated into probability notation are: \(P(M) = 0.021\), \(P(M|B) = 0.043\), and \(P(M|W) = 0.07\).

Step by step solution

01

Translate the first statement

The first statement informs us that doctors misdiagnosed heart attacks in 2.1% of all patients. In probability notation, this can be written as \(P(M) = 0.021\) because M represents the event of a heart attack being misdiagnosed.
02

Translate the second statement

The second statement indicates that doctors misdiagnosed heart attacks in 4.3% of black patients. This means that given a patient is black, the chance of misdiagnosis is 4.3%. Therefore, this can be written as \(P(M|B) = 0.043\), where M|B represents the event of a heart attack being misdiagnosed given the patient is black.
03

Translate the third statement

The third statement states that doctors misdiagnosed heart attacks in 7% of women under 55 years old. When translated into probability notation, this can be represented as \(P(M|W) = 0.07\), which signifies the probability of a misdiagnosed heart attack given that the patient is a woman under 55.

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.

Conditional Probability
Understanding conditional probability is crucial when dealing with scenarios where occurrences are interconnected. It's simply the likelihood of an event happening, given the occurrence of another event. Imagine having two interlocking gears; the motion of one influences the other. This relationship is what conditional probability measures.

For example, in a medical context, the chance of misdiagnosis can depend on a patient's demographics, such as age and race. From our exercise, if we let the event of being a black patient be 'B' and the event of a heart attack being misdiagnosed be 'M', then the conditional probability of M given B, denoted as \(P(M|B)\), is the probability that a heart attack is misdiagnosed for the subgroup of black patients. Conversely, \(P(M|W)\) would represent the misdiagnosis rate among women under 55.

These probabilities are important in identifying potential patterns of misdiagnosis within specific patient categories and can lead to improved healthcare strategies tailored to these groups. When learning about conditional probability, remember that it allows for more nuanced predictions and conclusions based on subsets of a larger population.
Statistical Inference
Statistical inference is a mighty tool in the world of statistics, as it allows us to make decisions and draw conclusions about a population based on a sample of data. Think of it as a detective carefully piecing together evidence to make a broader claim. It encompasses various methods including hypothesis testing, determining confidence intervals, and regression analysis.

In the context of our healthcare exercise, statistical inference could be used to analyze the data from the 10,000 patients to make more general claims about misdiagnosis rates in hospitals. By comparing the observed rates of misdiagnosis for heart attacks—an event 'M'—in the entire sample and in subgroups like black patients 'B' or women under 55 'W', we apply statistical inference to identify if the differences are significant or if they might be due to random chance.

The two key types of errors in statistical inference, Type I and Type II errors, are crucial in medical research, as they refer to the incorrect rejection or acceptance of a null hypothesis, respectively. Misclassification of these errors could have serious implications, like mistaking a pattern of misdiagnosis for a random occurrence.
Misdiagnosis Rates
In medical terms, the rate of misdiagnosis is a measure of how often medical conditions are incorrectly identified. Sadly, these statistics are not rare birds but rather common occurrences that affect patient care. High misdiagnosis rates can lead to delayed treatments or inappropriate care, worsening patient outcomes.

From the exercise, it's clear that misdiagnosis rates vary across different demographics, with patients who are black or women under 55 having higher rates than the general population. This information can effect transformative changes in strategies for medical diagnosis and treatment, ensuring more accurate and equitable healthcare services. Moreover, by studying misdiagnosis rates through the lens of probability, we can better prepare healthcare providers to identify and address the factors contributing to these disparities. This knowledge can be a powerful catalyst for change and improvement in healthcare systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

6.12 Consider a Venn diagram picturing two events \(A\) and \(B\) that are not disjoint. a. Shade the event \((A \cup B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cup B^{C} .\) How are these two events related? b. Shade the event \((A \cap B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cup B^{C}\). How are these two events related? (Note: These two relationships together are called DeMorgan's laws.)

The Australian newspaper The Mercury (May 30 , 1995) reported that, based on a survey of 600 reformed and current smokers, \(11.3 \%\) of those who had attempted to quit smoking in the previous 2 years had used a nicotine aid (such as a nicotine patch). It also reported that \(62 \%\) of those who quit smoking without a nicotine aid be gan smoking again within 2 weeks and \(60 \%\) of those who used a nicotine aid began smoking again within 2 weeks. If a smoker who is trying to quit smoking is selected at random, are the events selected smoker who is trying to quit uses a nicotine aid and selected smoker who has attempted to quit begins smoking again within 2 weeks inde pendent or dependent events? Justify your answer using the given information.

Refer to Exercise 6.18. Adding probabilities in the first row of the given table yields \(P(\) midsize \()=.45\), whereas from the first column, \(\mathrm{P}\left(4 \frac{3}{8}\right.\) in. grip) \(=.30\). Is the following true? $$ P\left(\text { midsize } \text { or } 4 \frac{3}{8} \text { in. grip }\right)=.45+.30=.75 $$ Explain.

A deck of 52 cards is mixed well, and 5 cards are dealt. a. It can be shown that (disregarding the order in which the cards are dealt) there are \(2,598,960\) possible hands, of which only 1287 are hands consisting entirely of spades. What is the probability that a hand will consist entirely of spades? What is the probability that a hand will consist entirely of a single suit? b. It can be shown that exactly 63,206 hands contain only spades and clubs, with both suits represented. What is the probability that a hand consists entirely of spades and clubs with both suits represented? c. Using the result of Part (b), what is the probability that a hand contains cards from exactly two suits?

In a small city, approximately \(15 \%\) of those eligible are called for jury duty in any one calendar year. People are selected for jury duty at random from those eligible, and the same individual cannot be called more than once in the same year. What is the probability that a particular eligible person in this city is selected two years in a row? three years in a row?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.