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 \(\sum x_{i}=\) \(20,179, \sum x_{i}^{2}=24,657,511\). What is the maximum possible amount that could be awarded under the two-standarddeviation rule?

Step by step solution

01

Calculate the Mean

To calculate the mean of the awards, use the formula for the mean: \( \bar{x} = \frac{\sum x_{i}}{n} \), where \( \sum x_{i} = 20179 \) and \( n = 27 \). Thus, \( \bar{x} = \frac{20179}{27} = 747.37 \).

02

Calculate the Variance

The variance \( \sigma^2 \) is calculated using the formula: \( \sigma^2 = \frac{\sum x_{i}^2}{n} - \left(\frac{\sum x_{i}}{n}\right)^2 \). First, compute \( \frac{\sum x_{i}^2}{n} \): \( \frac{24657511}{27} = 912500.41 \). Then, calculate \( \bar{x}^2 = 747.37^2 = 558570.81 \). Thus, \( \sigma^2 = 912500.41 - 558570.81 = 353929.6 \).

03

Calculate the Standard Deviation

The standard deviation \( \sigma \) is the square root of the variance: \( \sigma = \sqrt{353929.6} \approx 595 \).

04

Determine the Award Range

To determine the range within two standard deviations, calculate the upper bound as \( \mu + 2\sigma \). With \( \mu = 747.37 \) and \( \sigma = 595 \), calculate the upper bound: \( 747.37 + 2(595) = 1937.37 \).

05

Find the Maximum Possible Award

The maximum possible award is the largest award from the 27 awards that does not exceed the calculated upper bound. Review the list of awards: the largest award below 1937.37 is \( 1825 \).

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Key Concepts

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

Mean Calculation

The mean, often called the average, is the starting point of many statistical analyses. It provides a measure of the central tendency of a dataset. To find the mean of a set of values, add all the values together and divide by the number of values. This formula is represented as: \[ \bar{x} = \frac{\sum x_{i}}{n} \]where \( \sum x_{i} \) is the sum of all the data points and \( n \) is the number of observations in the dataset.

• In the given case study, the sum of all awards \((\$)\) was 20179, and there were 27 awards in total.
• Using the formula above, the mean comes out to be \( 747.37 \), which represents the average award among the cases considered.

Understanding the mean is crucial because it serves as a baseline for further statistical analysis, such as calculating variance or determining ranges.

Variance and Standard Deviation

Variance and standard deviation are measures of data dispersion, showing how much the values differ from the mean.**Variance**Variance is defined by the formula: \[ \sigma^2 = \frac{\sum x_{i}^2}{n} - \left(\frac{\sum x_{i}}{n}\right)^2 \] It essentially measures the average of the squared differences from the mean.

• First, calculate \( \frac{\sum x_{i}^2}{n} \), which was found to be 912500.41 in this problem.
• Deduct the square of the mean (\( \bar{x}^2 = 558570.81 \)) from this value, resulting in a variance \( \sigma^2 \) of 353929.6.

**Standard Deviation**Standard deviation is simply the square root of the variance and measures the average distance of each data point from the mean. The formula for standard deviation is: \[ \sigma = \sqrt{\sigma^2} \]

• In our problem, the standard deviation \( \sigma \) is approximately 595, indicating how much the awards typically vary from the mean.

These metrics are pivotal in understanding the spread of the data, helping us assess the variability across the dataset.

Two-Standard Deviation Rule

The Two-Standard Deviation Rule is a quick way of determining which values are likely to be considered typical or normal.This statistical rule of thumb suggests that most data points (approximately 95%) in a normal distribution lie within two standard deviations from the mean.

• To apply this rule, compute the range as \( \mu \pm 2\sigma \), where \( \mu \) is the mean and \( \sigma \) is the standard deviation.
• In the context of the exercise, the calculated upper bound (mean plus two times the standard deviation) was 1937.37.
• Only awards below this threshold are considered reasonable as per the court's standard.
• The largest award within this range from the given dataset was 1825.

This approach helps identify any unusually high or low data points and is useful in decision-making processes where determining typical cases is necessary.