/*! 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 32 According to the U.S. Bureau of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 32

According to the U.S. Bureau of Labor Statistics, there is a \(4.9 \%\) probability that a randomly selected employed individual has more than one job (a multiple-job holder). Also, there is a \(46.6 \%\) probability that a randomly selected employed individual is male, given that he has more than one job. What is the probability that a randomly selected employed individual is a multiple-job holder and male? Would it be unusual to randomly select an employed individual who is a multiple-job holder and male?

Short Answer

Expert verified
The probability is 0.022834. It would be unusual to randomly select an employed individual who is both a multiple-job holder and male.

Step by step solution

01

Understand the Given Probabilities

Identify the probabilities given in the problem. - The probability that an individual is a multiple-job holder \(P(A) = 0.049\).- The probability that an individual is male given that he is a multiple-job holder \(P(B|A) = 0.466\).
02

Use Conditional Probability for Joint Probability

To find the probability that a randomly selected employed individual is both a multiple-job holder and male, use the conditional probability formula.The formula to find this joint probability is:\[P(A \cap B) = P(B|A) \cdot P(A)\]Substitute the given values into the formula and calculate the result.
03

Perform the Calculation

Use the values from Step 1 in Step 2's formula:\[P(A \cap B) = 0.466 \cdot 0.049\]Calculate the product to find the probability:\[P(A \cap B) = 0.022834\]
04

Interpret the Result

To determine if it would be unusual to select an individual who is a multiple-job holder and male, compare the calculated probability to a commonly used threshold for unusual events (such as 0.05).Since \(P(A \cap B) = 0.022834\) is less than 0.05, it would indeed be considered unusual.

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!

Key Concepts

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

Probability
Probability measures how likely an event is to happen. It ranges from 0 (impossible) to 1 (certain). For example, the probability that a randomly selected employed individual has more than one job is 0.049. This means that there is a 4.9% chance that any given employed person you pick will hold multiple jobs. Understanding probability helps us predict outcomes and make sense of uncertainty in everyday scenarios. In our example, knowing that the probability of being a multiple-job holder is 0.049 helps us understand how common or rare this situation is in the employed population.
Unusual Events
Unusual events are those that have a low probability of occurring, commonly below a threshold of 0.05 (5%). To decide if an event is unusual, we compare its probability to this benchmark. In our problem, the joint probability of selecting a multiple-job holder who is also male is calculated to be 0.022834 (2.28%). Since this probability is less than 0.05, we classify it as unusual. This means that finding an individual fitting both criteria (male and multiple-job holder) is relatively rare in the employed population. Recognizing unusual events helps us make informed decisions and understand anomalies in data.

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

Four members from a 50-person committee are to be selected randomly to serve as chairperson, vice-chairperson, secretary, and treasurer. The first person selected is the chairperson; the second, the vice-chairperson; the third, the secretary; and the fourth, the treasurer. How many different leadership structures are possible?

A family has six children. If this family has exactly two boys, how many different birth and gender orders are possible?

Suppose that a computer chip company has just shipped 10,000 computer chips to a computer company. Unfortunately, 50 of the chips are defective. (a) Compute the probability that two randomly selected chips are defective using conditional probability. (b) There are 50 defective chips out of 10,000 shipped. The probability that the first chip randomly selected is defective is \(\frac{50}{10,000}=0.005 .\) Compute the probability that two randomly selected chips are defective under the assumption of independent events. Compare your results to part (a). Conclude that, when small samples are taken from large populations without replacement, the assumption of independence does not significantly affect the probability.

A combination lock has 50 numbers on it. To open it, you turn counterclockwise to a number, then rotate clockwise to a second number, and then counterclockwise to the third number. Repetitions are allowed.

Go to www.pearsonhighered.com/sullivanstats to obtain the data file SullivanStatsSurveyI using the file format of your choice for the version of the text you are using. The data represent the results of a survey conducted by the author. The variable "Text while Driving" represents the response to the question, "Have you ever texted while driving?" The variable "Tickets" represents the response to the question, "How many speeding tickets have you received in the past 12 months?" Treat the individuals in the survey as a random sample of all U.S. drivers. (a) Build a contingency table treating "Text while Driving" as the row variable and "Tickets" as the column variable. (b) Determine the marginal relative frequency distribution for both the row and column variable. (c) What is the probability a randomly selected U.S. driver texts while driving? (d) What is the probability a randomly selected U.S. driver received three speeding tickets in the past 12 months? (e) What is the probability a randomly selected U.S. driver texts while driving or received three speeding tickets in the past 12 months?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

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.