/*! 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 79 The 2011 gross sales of all comp... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 79

The 2011 gross sales of all companies in a large city have a mean of \(\$ 2.3\) million and a standard deviation of \(\$ .6\) million. Using Chebyshev's theorem, find at least what percentage of companies in this city had 2011 gross sales of a). \(\$ 1.1\) to \(\$ 3.5\) million b. \(\$ .8\) to \(\$ 3.8\) million c. \(\$ .5\) to \(\$ 4.1\) million

Short Answer

Expert verified
a) 75\% of the companies have gross sales between \(\$ 1.1\) - \(\$ 3.5\) million. b) 84\% of the companies have gross sales between \(\$ .8\) - \(\$ 3.8\) million. c) 89\% of the companies have gross sales between \(\$ .5\) - \(\$ 4.1\) million.

Step by step solution

01

Identify the Mean and Standard Deviation

The mean is given as \(\$ 2.3\) million and the standard deviation is \(\$ .6\) million.
02

Apply Chebyshev's Theorem for each case

a) \(\$ 1.1\) to \(\$ 3.5\) million: To calculate the number of standard deviations that \(\$ 1.1\) million to \(\$ 3.5\) million range covers, subtract \(\$ 1.1\) million from the mean (\(\$ 2.3\) million) and divide by the standard deviation (\(\$ .6\) million). Then, find the maximum value obtained from comparing this value to the value obtained from subtracting the mean from \(\$ 3.5\) million and dividing by the standard deviation. In this case, the maximum value is approximately 2. So, according to Chebyshev's theorem, at least \(1 - (1/ 2^2)) = 75\% of the companies have their gross sales between \(\$ 1.1\) and \(\$ 3.5\) million. b) Repeat the same process for \(\$ .8\) to \(\$ 3.8\) million range. The maximum value is approximately 2.5, yielding at least \(1 - (1/ 2.5^2)) = 84\% of companies within this range. c) Repeat the same process for \(\$ .5\) to \(\$ 4.1\) million. The maximum value is approximately 3, yielding at least \(1 - (1/ 3^2)) = 89\% of companies within this range.

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.

Gross Sales Distribution
Gross sales distribution provides an overview of how revenue is spread across various companies in a defined area. Think of it as a roadmap showing the financial achievements of a group of businesses. In statistics, understanding the distribution of any data set helps businesses assess performance and set future targets. The gross sales of companies in a city reflect both high-performing companies and those that may need improvement. By analyzing this distribution, companies can benchmark their achievements against industry standards or local competitors. When this data is analyzed, it usually follows various patterns or distributions, such as normal distribution. However, not all data fits perfectly into these patterns, which is where Chebyshev's theorem becomes valuable.
Mean and Standard Deviation
The terms 'mean' and 'standard deviation' are crucial in summarizing a data set. The mean, or average, is calculated by adding all values and dividing by the number of values. It gives a central value of the dataset鈥攊n this case, the average gross sales amount, \(\$2.3\) million.
Standard deviation measures the dispersion or spread of the data from the mean. A lower standard deviation means data points are close to the mean, indicating consistency. The given standard deviation here is \(.6\) million, indicating how much the company's sales typically vary from the average. This measure helps investors understand the stability and predictability of a company's performance and is foundational in invoking Chebyshev's theorem to estimate the percentage of data within specific ranges.
Statistics Problem Solving
Problem-solving in statistics involves breaking down questions using various analytical tools. For instance, our exercise begins by identifying key statistics like the mean and standard deviation. Knowing these, one approaches the problem methodically to analyze the data spread. Considering different ranges (e.g., \(\\(1.1\) to \(\\)3.5\) million) allows us to apply statistical theorems effectively. The exercise's goal is to understand what percentage of companies fall within certain sales brackets.
  • First, always identify known values, like mean and standard deviation.
  • Next, calculate how many standard deviations lie between given sales figures and the mean.
  • Finally, use statistical principles or theorems such as Chebyshev's theorem to find percentages.
In this step-by-step process, statistical problem-solving becomes less daunting and more intuitive for students.
Percentage Calculation Using Chebyshev's Theorem
Chebyshev's theorem is incredibly useful because it applies to any data distribution. It helps calculate the minimum percentage of observations that fall within a k standard deviation range from the mean for a set of data. The theorem states that at least \(1 - \frac{1}{k^2}\) of data falls within \(k\) standard deviations.
Applying Chebyshev's theorem involves a few steps:
  • Identify the range to analyze, such as \(\\(1.1\) million to \(\\)3.5\) million.
  • Calculate how many standard deviations these endpoints are from the mean.
  • Use the larger of these two standard deviation calculations to derive the percentage with Chebyshev's formula.
This method allowed us to determine that at least 75% of companies had gross sales between \(\\(1.1\) and \(\\)3.5\) million, a useful insight into the distribution of corporate financial performance in the city.

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

Although the standard workweek is 40 hours a week, many people work a lot more than 40 hours a week. The following data give the numbers of hours worked last week by 50 people. \(\begin{array}{llllllllll}40.5 & 41.3 & 41.4 & 41.5 & 42.0 & 42.2 & 42.4 & 42.4 & 42.6 & 43.3 \\ 43.7 & 43.9 & 45.0 & 45.0 & 45.2 & 45.8 & 45.9 & 46.2 & 47.2 & 47.5 \\ 47.8 & 48.2 & 48.3 & 48.8 & 49.0 & 49.2 & 49.9 & 50.1 & 50.6 & 50.6 \\ 50.8 & 51.5 & 51.5 & 52.3 & 52.3 & 52.6 & 52.7 & 52.7 & 53.4 & 53.9 \\\ 54.4 & 54.8 & 55.0 & 55.4 & 55.4 & 55.4 & 56.2 & 56.3 & 57.8 & 58.7\end{array}\) a. The sample mean and sample standard deviation for this data set are \(49.012\) and \(5.080\), respectively. Using Chebyshev's theorem, calculate the intervals that contain at least \(75 \%, 88.89 \%\) and \(93.75 \%\) of the data. b. Determine the actual percentages of the given data values that fall in each of the intervals that you calculated in part a. Also calculate the percentage of the data values that fall within one standard deviation of the mean. c. Do you think the lower endpoints provided by Chebyshev's theorem in part a are useful for this problem? Explain your answer. d. Suppose that the individual with the first number \((54.4)\) in the fifth row of the data is a workaholic who actually worked \(84.4\) hours last week and not \(54.4\) hours. With this change now \(\bar{x}=49.61\) and \(s=7.10\). Recalculate the intervals for part a and the actual percentages for part b. Did your percentages change a lot or a little? e. How many standard deviations above the mean would you have to go to capture all 50 data values? What is the lower bound for the percentage of the data that should fall in the interval, according to the Chebyshev theorem.

The following table gives the frequency distribution of the times (in minutes) that 50 commuter students at a large university spent looking for parking spaces on the first day of classes in the Fall semester of 2012 . \begin{tabular}{lc} \hline \multicolumn{1}{c} { Time } & Number of Students \\ \hline 0 to less than 4 & 1 \\ 4 to less than 8 & 7 \\ 8 to less than 12 & 15 \\ 12 to less than 16 & 18 \\ 16 to less than 20 & 6 \\ 20 to less than 24 & 3 \\ \hline \end{tabular} Find the mean, variance, and standard deviation. Are the values of these summary measures population parameters or sample statistics?

Consider the following two data sets. \(\begin{array}{llllrl}\text { Data Set I: } & 12 & 25 & 37 & 8 & 41 \\ \text { Data Set II: } & 19 & 32 & 44 & 15 & 48\end{array}\) Note that each value of the second data set is obtained by adding 7 to the corresponding value of the first data set. Calculate the standard deviation for each of these two data sets using the formula for sample data. Comment on the relationship between the two standard deviations.

The following stem-and-leaf diagram gives the distances (in thousands of miles) driven during the past year by a sample of drivers in a city. $$ \begin{array}{l|llllll} 0 & 3 & 6 & 9 & & & \\ 1 & 2 & 8 & 5 & 1 & 0 & 5 \\ 2 & 5 & 1 & 6 & & & \\ 3 & 8 & & & & & \\ 4 & 1 & & & & & \\ 5 & & & & & & \\ 6 & 2 & & & & \end{array} $$ a. Compute the sample mean, median, and mode for the data on distances driven. b. Compute the range, variance, and standard deviation for these data. c. Compute the first and third quartiles. d. Compute the interquartile range. Describe what properties the interquartile range has. When would the IQR be preferable to using the standard deviation when measuring variation?

According to the Kaiser Family Foundation, U.S. workers who had employer- provided health insur. ance paid an average premium of \(\$ 4129\) for family coverage during 2011 (USA TODAY, October 10,2011 ). Suppose that the premiums for such family coverage paid this year by all such workers have a bell-shaped distribution with a mean of \(\$ 4129\) and a standard deviation of \(\$ 600\). Using the empirical rule, find the ap: proximate percentage of such workers who pay premiums for such family coverage between a. \(\$ 2329\) and \(\$ 5929\) b. \(\$ 3529\) and \(\$ 4729\) c. \(\$ 2929\) and \(\$ 5329\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Statistics

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.