/*! 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 40 In a survey conducted in the fal... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a survey conducted in the fall 2006, 800 homeowners were asked about their expectations regarding the value of their home in the next few years; the results of the survey are summarized below: $$\begin{array}{lc} \hline \text { Expectations } & \text { Homeowners } \\ \hline \text { Decrease } & 48 \\ \hline \text { Stay the same } & 152 \\ \hline \text { Increase less than } 5 \% & 232 \\ \hline \text { Increase 5-10\% } & 240 \\ \hline \text { Increase more than 10\% } & 128 \\ \hline \end{array}$$ If a homeowner in the survey is chosen at random, what is the probability that he or she expected his or her home to a. Stay the same or decrease in value in the next few years? b. Increase \(5 \%\) or more in value in the next few years?

Short Answer

Expert verified
a. \(P(\text{Stay the same or Decrease}) = 0.25\) b. \(P(\text{Increase } 5\% \text{ or more}) = 0.46\)

Step by step solution

01

Understand the data

The data is given in a table format showing homeowners' expectations and the number of homeowners with each expectation. The total number of homeowners is 800, and we are given the number of homeowners who have each of the five possible expectations.
02

Calculate the probability of each expectation

To calculate the probability of selecting a homeowner with each expectation at random, we need to divide the number of homeowners with that specific expectation by the total number of homeowners (800). 1. Probability of Decrease: \[P(\text{Decrease}) = \frac{48}{800} = 0.06\] 2. Probability of Stay the same: \[P(\text{Stay the same}) = \frac{152}{800} = 0.19\] 3. Probability of Increase less than 5%: \[P(\text{Increase less than } 5\%) = \frac{232}{800} = 0.29\] 4. Probability of Increase 5-10%: \[P(\text{Increase } 5-10\%) = \frac{240}{800} = 0.30\] 5. Probability of Increase more than 10%: \[P(\text{Increase more than } 10\%) = \frac{128}{800} = 0.16\]
03

Calculate the probability for question (a)

For question (a), we want to find the probability that a randomly chosen homeowner expects their home's value to stay the same or decrease. We can find this probability by summing the probabilities of these two expectations: \[P(\text{Stay the same or Decrease}) = P(\text{Stay the same}) + P(\text{Decrease})\] \[P(\text{Stay the same or Decrease}) = 0.19 + 0.06 = 0.25\]
04

Calculate the probability for question (b)

For question (b), we want to find the probability that a randomly chosen homeowner expects their home's value to increase by 5% or more. We can find this probability by summing the probabilities of the two expectations that meet this condition: \[P(\text{Increase } 5\% \text{ or more}) = P(\text{Increase } 5-10\%) + P(\text{Increase more than } 10\%)\] \[P(\text{Increase } 5\% \text{ or more}) = 0.30 + 0.16 = 0.46\] #Answer#: a. The probability that a homeowner expects their home's value to stay the same or decrease is 0.25. b. The probability that a homeowner expects their home's value to increase by 5% or more is 0.46.

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 Calculation
Probability is a branch of mathematics that deals with the likelihood of different outcomes. In the context of our survey, probability tells us how likely it is for a randomly selected homeowner to have certain expectations about their home's value. To calculate probability, you divide the number of favorable outcomes by the total number of possible outcomes. For example, if 48 homeowners expect a decrease, the probability of selecting one of these is \( \frac{48}{800} \). These basic probability calculations allow us to analyze data more efficiently.
Survey Data
Survey data is collected from a group of people to understand their opinions or behaviors. In this exercise, the survey data involves 800 homeowners, each expressing expectations about their home values. The data's reliability depends on the random selection of participants and their honest responses. Surveys help in gathering a wide array of viewpoints, which can be used for statistical analysis and to draw meaningful conclusions.
Statistical Analysis
Statistical analysis involves using mathematical techniques to interpret data. This includes calculating probabilities, averages, or other statistical measures. It helps in extracting important information from raw data. In our case, the analysis is used to determine the likelihood of different homeowner expectations regarding house value changes. It transforms complex data sets into understandable results, providing insights into homeowner trends and preferences.
Expectation in Mathematics
Expectation in mathematics refers to the average outcome you can anticipate from a random event. In surveying homeowners, the expectation would involve estimating the average consensus on future home values. By understanding the different probabilities, one can predict potential economic trends. These expectations help stakeholders make informed decisions. In this context, knowing that a certain percentage of homeowners expect an increase might influence market strategies or policies.

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

Before being allowed to enter a maximum-security area at a military installation, a person must pass three independent identification tests: a voice-pattern test, a fingerprint test, and a handwriting test. If the reliability of the first test is \(97 \%\), the reliability of the second test is \(98.5 \%\), and that of the third is \(98.5 \%\), what is the probability that this security system will allow an improperly identified person to enter the maximumsecurity area?enter the maximumsecurity area?

A nationwide survey conducted by the National Cancer Society revealed the following information. Of 10,000 people surveyed, 3200 were "heavy coffee drinkers" and 160 had cancer of the pancreas. Of those who had cancer of the pancreas, 132 were heavy coffee drinkers. Using the data in this survey, determine whether the events "being a heavy coffee drinker" and "having cancer of the pancreas" are independent events.

Refer to the following experiment: Urn A contains four white and six black balls. Urn B contains three white and five black balls. A ball is drawn from urn A and then transferred to urn B. A ball is then drawn from urn B. What is the probability that the transferred ball was white given that the second ball drawn was white?

Copykwik has four photocopy machines: \(A, B, C\), and \(D .\) The probability that a given machine will break down on a particular day is \(P(A)=\frac{1}{50} \quad P(B)=\frac{1}{60} \quad P(C)=\frac{1}{75} \quad P(D)=\frac{1}{40}\) Assuming independence, what is the probability on a particular day that a. All four machines will break down? b. None of the machines will break down?

A study was conducted among a certain group of union members whose health insurance policies required second opinions prior to surgery. Of those members whose doctors advised them to have surgery, \(20 \%\) were informed by a second doctor that no surgery was needed. Of these, \(70 \%\) took the second doctor's opinion and did not go through with the surgery. Of the members who were advised to have surgery by both doctors, \(95 \%\) went through with the surgery. What is the probability that a union member who had surgery was advised to do so by a second doctor?

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.