Question

Resistant Measures Here are four of the Verizon data speeds (Mbps) from Figure 3-1: 13.5, 10.2, 21.1, 15.1. Find the mean and median of these four values. Then find the mean and median after including a fifth value of 142, which is an outlier. (One of the Verizon data speeds is 14.2 Mbps, but 142 is used here as an error resulting from an entry with a missing decimal point.) Compare the two sets of results. How much was the mean affected by the inclusion of the outlier? How much is the median affected by the inclusion of the outlier?

Short answer

The mean and median values with and without including the outliers:

Case	Mean (in Mbps)	Median (in Mbps)
Without outlier	14.975	14.3
With outlier	40.38	15.1

The mean changes drastically while the median changes by a small amount.

Step-by-step solution

Given information

Four observations of data speeds for Verizon (in Mbps): 13.5, 10.2, 21.1, 15.1

Formula of mean and median

The formula for mean is stated as:

$\overline{x}=\frac{\sum x}{n}$, where x are the observations and n are the count of the observations.

The formula for median is stated as:

• If n is odd, the middlemost observation is the median.
• If n is even, the average of the two observations in the middle is the median.

Compute mean

Substitute the values in the formula stated in step 2.

$$\begin{gathered} \overline{x}=\frac{13.5+10.2+...+15.1}{4} \\ =\frac{59.9}{4} \\ =14.975 \end{gathered}$$

Thus, the mean value is 14.975 Mbps.

Compute median

The number of observations is 4.

Arrange the observations in ascending order.

10.2,13.5,15.1,21.1

The middlemost observations are 13.5 and 15.1.

The median is given as:

$$\begin{gathered} M=\frac{13.5+15.1}{2} \\ =\frac{28.6}{2} \\ =14.3 \end{gathered}$$

Thus, the median is 14.3 Mbps.

Compute mean including the outlier value 142

Substitute the values in the formula stated in step 2.

$$\begin{gathered} \overline{x}=\frac{13.5+10.2+...+15.1+142}{5} \\ =\frac{201.9}{5} \\ =40.38 \end{gathered}$$

Thus, the mean value is 40.38 Mbps.

Compute median including the outlier value 142

The number of observations is 5.

Arrange the observations in ascending order.

10.2,13.5,15.1,21.1,142

The middlemost observation is 15.1.

The median is given as:

$M=15.1$

Thus, the median is 15.1 Mbps.

Summarized values

The summarized values are:

Case	Mean (in Mbps)	Median (in Mbps)
Without outlier	14.975	14.3
With outlier	40.38	15.1

The mean changes drastically from 14.975 Mbps to 40.38 Mbps when the outlier is included in the set of observations.

The median changes by a small margin from 14.3 Mbps to 15.1 Mbps when the outlier is included in the set of observations.

Thus, it can be said that the median value does not get much affected by outliers, whereas the median is affected by outliers.