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

The formula for mean is stated as:

$\stackrel{炉}{x}=\frac{鈭�x}{n}$, 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.

Substitute the values in the formula stated in step 2.

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

Thus, the mean value is 14.975 Mbps.

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.

Substitute the values in the formula stated in step 2.

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

Thus, the mean value is 40.38 Mbps.

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.

The summarized values are:

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.