/*! 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 7 In 1997 a woman sued a computer ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In 1997 a woman sued a computer keyboard manufacturer, charging that her repetitive stress injuries were caused by the keyboard (Genessy \(v\). Digital Equipment Corp.). The injury awarded about \(\$ 3.5\) million for pain and suffering, but the court then set aside that award as being unreasonable compensation. In making this determination, the court identified a "normative" group of 27 similar cases and specified a reasonable award as one within two standard deviations of the mean of the awards in the 27 cases. The 27 awards were (in \(\$ 1000\) s) 37,60 , \(75,115,135,140,149,150,238,290,340,410,600,750\), \(750,750,1050,1100,1139,1150,1200,1200,1250\), \(1576,1700,1825\), and 2000 , from which \(\Sigma x_{i}=20,179\), \(\Sigma x_{i}^{2}=24,657,511\). What is the maximum possible amount that could be awarded under the two-standarddeviation rule?

Short Answer

Expert verified
The maximum possible award is $1997,000.

Step by step solution

01

Calculate the Mean

The mean is calculated by dividing the sum of the awards, \( \Sigma x_i = 20,179 \), by the number of awards, \( n = 27 \). Compute as follows:\[ \text{Mean} = \frac{\Sigma x_i}{n} = \frac{20,179}{27} \approx 747.37 \]
02

Calculate the Variance

The variance is calculated using the formula:\[ \text{Variance} = \frac{\Sigma x_i^2 - \frac{(\Sigma x_i)^2}{n}}{n-1} \]Substitute the known values:\[ \text{Variance} = \frac{24,657,511 - \frac{(20,179)^2}{27}}{26} \approx 390,879 \]
03

Calculate the Standard Deviation

The standard deviation is the square root of the variance:\[ \text{Standard Deviation} = \sqrt{390,879} \approx 625 \]
04

Calculate the Maximum Reasonable Award

A reasonable award can be at most the mean plus two standard deviations:\[ \text{Maximum Award} = \text{Mean} + 2 \times \text{Standard Deviation} \]Substitute the values for the mean and standard deviation:\[ \text{Maximum Award} = 747.37 + 2 \times 625 = 1997.37 \]
05

Conclude the Maximum Award

Since the award must be in terms of whole dollars, the maximum award is \( \$1997,000 \).

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.

Understanding Mean Calculation in Statistics
The mean, commonly referred to as the average, is a fundamental concept in statistics. It represents the central tendency of a dataset.
To calculate the mean, you sum up all the values in the dataset and divide by the number of values.
For example, in the court case involving the 27 awards, the mean represents the average award given across all the cases.
Here's how you calculate it:
  • Sum up all data points: for the award case, this sum ( \( \Sigma x_i = 20,179 \) is the total of all award amounts.
  • Count the number of data points: in this situation, there are 27 cases, so \( n = 27 \).
  • Divide the total by the number of data points: \( \text{Mean} = \frac{20,179}{27} \approx 747.37 \).
Consequently, the mean amount awarded is approximately \( 747,370 \) dollars. The mean helps to provide a central value that summarizes the entire dataset.
Variance Calculation Explained
Variance measures how much values in a dataset differ from the mean. It's a way to quantify the variability or spread among the data.
In straightforward terms, a high variance indicates that the numbers are more spread out, while a low variance means they are closer together.

Formula Breakdown

The formula for variance (\( \sigma^2 \) in statistics) involves a few key steps:
  • Start with summing the squares of all individual values: given by \( \Sigma x_i^2 \).
  • Subtract the square of the total sum, divided by the number of values: this part is \( \frac{(\Sigma x_i)^2}{n} \).
  • Divide the result by one less than the number of items, which accounts for degrees of freedom (\( n-1 \)) in this context.
Using these steps in the awards context:
  • Apply the known values in the formula: \[ \text{Variance} = \frac{24,657,511 - \frac{(20,179)^2}{27}}{26} \approx 390,879 \]
This calculation provides the variance, aiding in determining how much the awards deviate from the average calculated initially.
Grasping Standard Deviation
Standard deviation is like the sister of variance in statistics, giving more intuitive understanding of data spread. While variance gives spread in squared units, standard deviation expresses it in the same units as the data, making it easier to relate to the original values.
Standard deviation is the square root of the variance.

Applying Standard Deviation

Here’s how the concept applies to our example:
  • First, calculate the standard deviation: \( \text{Standard Deviation} = \sqrt{390,879} \approx 625 \).
  • This number, 625, reflects how far, on average, the awards are from the "mean" award of 747.
  • In practical terms, it lets the court ascertain a reasonable maximum award by seeing how much values typically vary.
Standard deviation allows us to perceive the scatter from the average in readily understandable terms and is crucial in determining reasonable boundaries for making decisions, such as award amounts in legal contexts.

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

Consider the following information on ultimate tensile strength (Ib/in) for a sample of \(n=4\) hard zirconium copper wire specimens (from "4 Characterization Methods for Fine Copper Wire," Wire J. Intl., Aug., 1997: 74-80): \(\bar{x}=76,831 \quad s=180 \quad\) smallest \(x_{i}=76,683\) largest \(x_{i}=77,048\) Determine the values of the two middle sample observations (and don't do it by successive guessing!).

Zinfandel is a popular red wine varietal produced almost exclusively in California. It is rather controversial among wine connoisseurs because its alcohol content varies quite substantially from one producer to another. In May 2013, the author went to the website klwines .com, randomly selected 10 zinfandels from among the 325 available, and obtained the following values of alcohol content (\%): \(\begin{array}{lllll}14.8 & 14.5 & 16.1 & 14.2 & 15.9 \\ 13.7 & 16.2 & 14.6 & 13.8 & 15.0\end{array}\) a. Calculate and interpret several measures of center. b. Calculate the sample variance using the defining formula. c. Calculate the sample variance using the shortcut formula after subtracting 13 from each observation.

The article "Snow Cover and Temperature Relationships in North America and Eurasia" \((J\). Climate and Applied Meteorology, 1983: 460-469) used statistical techniques to relate the amount of snow cover on each continent to average continental temperature. Data presented there included the following ten observations on October snow cover for Eurasia during the years \(1970-1979\) (in million \(\mathrm{km}^{2}\) ): \(\begin{array}{llllllllll}6.5 & 12.0 & 14.9 & 10.0 & 10.7 & 7.9 & 21.9 & 12.5 & 14.5 & 9.2\end{array}\) What would you report as a representative, or typical, value of October snow cover for this period, and what prompted your choice?

The May 1, 2009, issue of The Montclarian reported the following home sale amounts for a sample of homes in Alameda, CA that were sold the previous month ( \(1000 \mathrm{~s}\) of $$)\( : \)590$ 815 575 608 350 1285 408 540 555 679 a. Calculate and interpret the sample mean and median.b. Suppose the 6th observation had been 985 rather than 1285\. How would the mean and median change? c. Calculate a 20% trimmed mean by first trimming the two smallest and two largest observations. d. Calculate a 15% trimmed mean.

a. Give three different examples of concrete populations and three different examples of hypothetical populations. b. For one each of your concrete and your hypothetical populations, give an example of a probability question and an example of an inferential statistics question.

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.