Question

because the mean is very sensitive to extreme values, it is not a resistant measure of center. by deleting some low values and high values, the trimmed mean is more resistant. to find the 10% trimmed mean for a data set, first arrange the data in order, then delete the bottom 10% of the values and delete the top 10% of the values, then calculate the mean of the remaining values. use the axial loads (pounds) of aluminum cans listed below for cans that are 0.0111 in thick. identify any outliers, then compare the median, mean, 10% trimmed mean, and 20% trimmed mean. 248 261 267 274 276 279 280 283 283 285 285 287 289 291 294 295 295 299 309 506 a. the outlier(s) is/are 506 pounds (type a whole number. use a comma to separate answers as needed) b. there are no outliers. the median is 285 pounds (type an integer or decimal rounded to one decimal place as needed) the untrimmed mean is 294.3 pounds (type an integer or decimal rounded to one decimal place as needed) the 10% trimmed mean is 285.1 pounds (type an integer or decimal rounded to one decimal place as needed) the 20% trimmed mean is pounds (type an integer or decimal rounded to one decimal place as needed)

Explanation:

Step1: Calculate the 20% trimmed mean

First, find 20% of the number of data points. There are \(n = 20\) data points. \(20\%\) of \(20\) is \(0.2\times20=4\). So we delete \(4\) lowest - value and \(4\) highest - value data points.
The ordered data set is: \(248,261,267,274,276,279,280,283,283,285,285,287,289,291,294,295,295,299,309,506\)
After deleting \(4\) lowest (\(248,261,267,274\)) and \(4\) highest (\(299,309,506\)) values, the remaining data set is: \(276,279,280,283,283,285,285,287,289,291,294,295,295\)
The sum of the remaining data is \(\sum x=276 + 279+280+283+283+285+285+287+289+291+294+295+295\)

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The number of remaining data points \(n'=20-(4 + 4)=12\)
The mean formula is \(\bar{x}=\frac{\sum x}{n'}\), so \(\bar{x}=\frac{3722}{12}\approx310.2\)

Answer:

The \(20\%\) trimmed mean is \(310.2\) pounds.