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Chapter 2: Problem 62

Fiber in the Diet The number of grams of fiber eaten in one day for a sample of ten people are \(\begin{array}{ll}10 & 11\end{array}\) \(\begin{array}{ll}11 & 14\end{array}\) \(\begin{array}{llllll}15 & 17 & 21 & 24 & 28 & 115\end{array}\) (a) Find the mean and the median for these data. (b) The value of 115 appears to be an obvious outlier. Compute the mean and the median for the nine numbers with the outlier excluded. (c) Comment on the effect of the outlier on the mean and on the median.

Short Answer

Expert verified
With the outlier, the mean is 25.6 and the median is 16. Without the outlier, the mean is 15.67 and the median is 15. This shows that the mean is more affected by the outlier than the median, making the median a more resistant measure of central tendency in this case.

Step by step solution

01

Calculate the Mean (with outlier)

First, find the sum of all the data points: \(10 + 11 + 11 + 14 + 15 + 17 + 21 + 24 + 28 + 115 = 256\). Then divide this by the total number of data points (10 in this case) to get the mean: \(256 / 10 = 25.6\)
02

Calculate the Median (with outlier)

First, list the numbers in ascending order: \(10, 11, 11, 14, 15, 17, 21, 24, 28, 115\). Since there are 10 data points, the median is the average of the 5th and 6th data points, which is \( (15 + 17) / 2 = 16\).
03

Calculate the Mean (without outlier)

Now, exclude the outlier (115) and repeat the process for finding the mean: \(10 + 11 + 11 + 14 + 15 + 17 + 21 + 24 + 28 = 141\). Then divide this by the number of data points minus the outlier (9 in this case) to get the new mean: \(141 / 9 = 15.67\)
04

Calculate the Median (without outlier)

Again, exclude the outlier and list the numbers in ascending order: \(10, 11, 11, 14, 15, 17, 21, 24, 28\). There are now 9 data points and the median is the 5th, which is 15.
05

Discuss the Effect of the Outlier

Notice how the mean got smaller when the outlier was excluded, from 25.6 to 15.67. This shows that the mean is sensitive to extreme values. However, the median only decreased slightly, from 16 to 15. This demonstrates that the median is resistant to outliers, and hence it is a better representation of the central value when outliers are present.

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