/*! 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 47 A Gallup survey of 2002 adults f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 47

A Gallup survey of 2002 adults found that \(46 \%\) of women and \(37 \%\) of men experience pain daily (San Luis Obispo Tribune, April 6,2000 ). Suppose that this information is representative of adult Americans. If an adult American is selected at random, are the events selected adult is male and selected adult experiences pain daily independent or dependent? Explain.

Short Answer

Expert verified
The events 'selected adult is male' and 'selected adult experiences pain daily' are not independent.

Step by step solution

01

Defining the problem and understanding the provided information

The problem provides that \(46 \%\) of women and \(37 \%\) of men experience pain daily. In addition, it asks whether the events: 'selected adult is male' and 'selected adult experiences pain daily' are independent. To know this, first, we need to understand the probabilities of these events.
02

Calculating individual probabilities

Assuming that the distribution of men and women in the adult population is roughly equal, we can use the provided information to get the following probabilities: \[ P(\text{Male}) = P(\text{Female}) = 0.5 \]and\[ P(\text{Experiences Pain} | \text{Male}) = 0.37 \]\[ P(\text{Experiences Pain} | \text{Female}) = 0.46 \]
03

Checking for independence

Two events A and B are independent if and only if the probability of A given B, denoted by \( P(A|B) \), is equal to the probability of A, denoted by \( P(A) \), i.e., \( P(A|B) = P(A) \). Therefore, the events 'selected adult is male' and 'selected adult experiences pain daily' would be independent only if the probability of experience daily pain in general, \( P(\text{Experiences Pain}) \), was equal to the probability of a male adult experiencing daily pain, \( P(\text{Experiences Pain} | \text{Male}) = 0.37 \). However, since \( P(\text{Experiences Pain}) = 0.5 * 0.37 + 0.5 * 0.46 = 0.415 \), implying that \( P(\text{Experiences Pain}) \neq P(\text{Experiences Pain}|\text{Male}) \), the events are not independent.

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 91影视!

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

In an article that appears on the web site of the American Statistical Association (www.amstat.org), Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in 100 . And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given- say 99 in \(100-\) are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty, \(T C=\) event that the test result is clean, \(D=\) event that the person tested is actually dirty, and \(C=\) event that the person tested is actually clean. a. Using the information in the quote, what are the values of \mathbf{i} . ~ \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C)\) iv. \(P(D)\) b. Use the law of total probability to find \(P(T D)\). c. Use Bayes' rule to evaluate \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

A construction firm bids on two different contracts. Let \(E_{1}\) be the event that the bid on the first contract is successful, and define \(E_{2}\) analogously for the second contract. Suppose that \(P\left(E_{1}\right)=.4\) and \(P\left(E_{2}\right)=.3\) and that \(E_{1}\) and \(E_{2}\) are independent events. a. Calculate the probability that both bids are successful (the probability of the event \(E_{1}\) and \(E_{2}\) ). b. Calculate the probability that neither bid is successful (the probability of the event \(\left(\right.\) not \(\left.E_{1}\right)\) and \(\left(\right.\) not \(\left.E_{2}\right)\). c. What is the probability that the firm is successful in at least one of the two bids?

The report "TV Drama/Comedy Viewers and Health Information" (www.cdc.gov/Healthmarketing) describes the results of a large survey involving approximately 3500 people that was conducted for the Center for Disease Control. The sample was selected in a way that the Center for Disease Control believed would result in a sample that was representative of adult Americans. One question on the survey asked respondents if they had learned something new about a health issue or disease from a TV show in the previous 6 months. Data from the survey was used to estimate the following probabilities, where \(L=\) event that a randomly selected adult American reports learning something new about a health issue or disease from a TV show in the previous 6 months and \(F=\) event that a randomly selected adult American is female $$ P(L)=.58 \quad P(L \cap F)=.31 $$ Assume that \(P(F)=.5\). Are the events \(L\) and \(F\) independent events? Use probabilities to justify your answer.

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 \(A^{C} \cap B^{C} .\) 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.)

Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Let \(E\) denote the event that the first airline's flight is fully booked on a particular day, and let \(F\) denote the event that the second airline's flight is fully booked on that same day. Suppose that \(P(E)=.7\), \(P(F)=.6\), and \(P(E \cap F)=.54 .\) a. Calculate \(P(E \mid F)\), the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked. b. Calculate \(P(F \mid E)\).
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation
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.