/*! 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 23 According to a survey of 176 ret... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

According to a survey of 176 retailers, \(46 \%\) of them use electronic tags as protection against shoplifting and employee theft. If one of these retailers is selected at random, what is the probability that he or she uses electronic tags as antitheft devices?

Short Answer

Expert verified
The probability that a randomly selected retailer uses electronic tags as antitheft devices is approximately 0.46 or 46%.

Step by step solution

01

Determine the number of retailers using electronic tags

First, we have to find the actual number of retailers using electronic tags by multiplying the given percentage by the total number of retailers surveyed. In this case, the percentage is \(46\%\) and the total number of retailers is 176. Number of retailers using electronic tags: \(\frac{46}{100}\) * 176
02

Calculate the number of retailers using electronic tags

Now, we will calculate the actual number of retailers using electronic tags: Number of retailers using electronic tags: \(\frac{46}{100}\) * 176 = 80.96 Since we cannot have a fraction of a retailer, we can round this number to the nearest whole number. In this case, it would be 81.
03

Calculate the probability of a randomly selected retailer using electronic tags

To find the probability, we will divide the number of retailers using electronic tags by the total number of retailers surveyed. Probability: \(\frac{\text{Number of retailers using electronic tags}}{\text{Total number of retailers surveyed}}\) Probability: \(\frac{81}{176}\)
04

Simplify the probability

Now we will simplify the probability: Probability: \(\frac{81}{176}\) ≈ 0.46 The probability that a randomly selected retailer uses electronic tags as antitheft devices is approximately 0.46 or 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.

Survey Analysis
Survey analysis is a method used to gather and interpret data from a specific group or sample to understand trends, behavior, and opinions. In the example of retailers surveyed about the use of electronic tags as antitheft devices, we gather data from 176 retailers.
- A survey consists of structured questions aimed at extracting quantitative data, like percentages, to infer conclusions. - Surveys can be conducted through various methods such as online questionnaires, face-to-face interviews, or phone calls. - Important considerations in survey analysis include sample size, question clarity, and respondent honesty.
In our exercise, the survey's outcome shows that 46% of retailers have adopted electronic tags. This data is critical for understanding industry trends and making informed business decisions, like adopting similar security measures or analyzing the effectiveness of these devices.
Percentage Calculation
Percentage calculation is a fundamental math skill used to express a number as a fraction of 100. In numerous real-life scenarios, such as surveys, percentages help present data efficiently. To find out how many retailers out of 176 use electronic tags: Multiply the total number of retailers (176) by the percentage using tags (46%). The formula is: \[ \text{Number of retailers using tags} = \left( \frac{46}{100} \right) \times 176 \]
This yields approximately 81 retailers using electronic tags. Knowing how to perform percentage calculations enables you to quickly convert survey data into actionable insights. It's a valuable skill for analyzing probability and making everyday decisions.
Antitheft Devices
Antitheft devices are tools designed to prevent or reduce theft in businesses and retail stores. Among these, electronic tags are common due to their effectiveness and ease of use. Here's why electronic tags are preferred: - They deter potential shoplifters by alerting store personnel when an item is being removed without authorization. - These devices enable stores to secure merchandise without encasing items in bulky security boxes. - Electronic tags are economical for large retailers, who can tag thousands of products efficiently.
Understanding the use of antitheft devices like electronic tags can inform retailers in making decisions about inventory protection and loss prevention strategies, which are vital for maintaining profitability.
Simplifying Fractions
Simplifying fractions is a method to reduce a fraction to its simplest form where the numerator and denominator have no common factors other than 1. In the given exercise, simplifying the fraction representing the probability involves: The probability that a randomly selected retailer uses electronic tags is given by the fraction: \[ \frac{81}{176} \] To simplify, find the greatest common divisor (GCD) of 81 and 176, and divide both the numerator and the denominator by this GCD. This fraction simplifies the representation of probabilities, making them easier to understand and communicate, especially in reviewing survey results or making predictions based on data.
Simplifying fractions is essential not only in probability but also in general mathematics, ensuring clarity and efficiency in calculations and decision-making.

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

A pair of fair dice is rolled. What is the probability that a. The sum of the numbers shown uppermost is less than 5 ? b. At least one 6 is rolled?

A poll was conducted among 250 residents of a certain city regarding tougher gun-control laws. The results of the poll are shown in the table: $$ \begin{array}{lccccc} \hline & \begin{array}{c} \text { Own } \\ \text { Only a } \\ \text { Handgun } \end{array} & \begin{array}{c} \text { Own } \\ \text { Only a } \\ \text { Rifle } \end{array} & \begin{array}{c} \text { Own a } \\ \text { Handgun } \\ \text { and a Rifle } \end{array} & \begin{array}{c} \text { Own } \\ \text { Neither } \end{array} & \text { Total } \\ \hline \text { Favor } & & & & & \\ \text { Tougher Laws } & 0 & 12 & 0 & 138 & 150 \\ \hline \begin{array}{l} \text { Oppose } \\ \text { Tougher Laws } \end{array} & 58 & 5 & 25 & 0 & 88 \\ \hline \text { No } & & & & & \\ \text { Opinion } & 0 & 0 & 0 & 12 & 12 \\ \hline \text { Total } & 58 & 17 & 25 & 150 & 250 \\ \hline \end{array} $$ If one of the participants in this poll is selected at random, what is the probability that he or she a. Favors tougher gun-control laws? b. Owns a handgun? c. Owns a handgun but not a rifle? d. Favors tougher gun-control laws and does not own a handgun?

An experiment consists of selecting a card at random from a 52-card deck. Refer to this experiment and find the probability of the event. A face card (i.e., a jack, queen, or king) is drawn.

CoURSE ENROLLMENTS Among 500 freshmen pursuing a business degree at a university, 320 are enrolled in an economics course, 225 are enrolled in a mathematics course, and 140 are enrolled in both an economics and a mathematics course. What is the probability that a freshman selected at random from this group is enrolled in a. An economics and/or a mathematics course? b. Exactly one of these two courses? c. Neither an economics course nor a mathematics course?

STAYING IN ToucH In a poll conducted in 2007, 2000 adults ages 18 yr and older were asked how frequently they are in touch with their parents by phone. The results of the poll are as follows: $$ \begin{array}{lccccc} \hline \text { Answer } & \text { Monthly } & \text { Weekly } & \text { Daily } & \text { Don't know } & \text { Less } \\ \hline \text { Respondents, \% } & 11 & 47 & 32 & 2 & 8 \\ \hline \end{array} $$ If a person who participated in the poll is selected at random, what is the probability that the person said he or she kept in touch with his or her parents a. Once a week? b. At least once a week?

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.