Question

Use the magnitudes​ (Richter scale) of the earthquakes listed in the data set below. Find the mean and median of this data set. Is the magnitude of an earthquake measuring 7.0 on the Richter scale an outlier​ (data value that is very far away from the​ others) when considered in the context of the sample data given in this data​ set? Explain.

Find the mean and median of the data set using a calculator or similar data analysis technology.

2.53 1.33 0.64 2.55 0.67 0.57 1.61 0.69 2.06 1.84

2.01 2.74 1.98 0.99 1.01 2.59 1.52 1.41 1.53 0.92

0.14 1.15 0.01 2.02 0.81 1.48 0.08 1.16 0.74 0.24

1.77 2.74 0.35 0.65 2.77 1.04 0.01 2.31 1.62 2.91

2.77 2.41 1.45 2.06 2.66 1.73 1.58 0.82 2.03 1.78

Answer 1

`\large \boxed{\text{Mean = 1.49; Mode= 1.52; 7.0 is an outlier}}`

Explanation:

When sorted, your data are:

0.01, 0.01, 0.08, 0.14, 0.24, 0.35, 0.57, 0.64, 0.65, 0.67, 0.69, 0.74, 0.81

0.82, 0.92, 0.99, 1.01, 1.04, 1.15, 1.16, 1.33, 1.41, 1.45, 1.48, 1.52

1.53, 1.58, 1.61, 1.62, 1.73, 1.77, 1.78, 1.84, 1.98, 2.01, 2.02, 2.03, 2.06

2.06, 2.31, 2.41, 2.53, 2.55, 2.59, 2.66, 2.74, 2.74, 2.77, 2.77, 2.91

I entered the data into a statistical and box plot generator.

The results were:

mean = 1.49

Median = 1.52

Q1 = 0.81

Q2 = 1.525

Q3 = 2.06

IQR = 1.25

One test for an outlier is if the value is greater than the upper fence: Q3 + 1.5IQR.

`\begin{array}{rcl}v & > & Q3 + 1.5IQR\\7.0 & > &2.06 + 1.5*1.25\\7 .0& > & 2.06 + 1.875\\7.0 & > & 3.935\\\end{array}\\\textbf{TRUE}\\\\\large \boxed{\textbf{Mean = 1.49; Mode= 1.52; 7.0 is an outlier}}`

The box-and-whisker plot shows that a value of 7.0 is well above the upper fence (in red).

Answer 2

mean = 1.49

median = 1.53

Explanation:

Using Excel, the procedure is the following

1. Insert data set in a sheet. In my case I used cells A1 to AX1
2. Compute mean with AVERAGE formula. In my case, AVERAGE(A1:AX1)
3. Compute median with MEDIAN formula. In my case, MEDIAN(A1:AX1)

7.0 is an outlier because all data points are below 2.91