Question

use the magnitudes (richter scale) of the earthquakes listed in the data set below. find the mean and median of the data set. is the magnitude of an earthquake measuring 7.0 an outlier (data value that is very far away from the others) when considered in the context of the sample data given in the data set? click the icon to view the earthquake richter scale data. find the mean and median of the data set using a calculator or similar data - analysis technology. the mean of the data set is . (type an integer or decimal rounded to three decimal places as needed.) more info 0.72 1.62 2.94 2.75 1.93 1.26 1.20 2.02 2.09 1.08 0.06 1.61 2.35 0.77 2.11 0.00 2.53 0.03 1.15 2.31 0.02 0.33 1.07 1.06 0.71 0.57 0.31 2.22 0.11 2.39 2.33 2.62 2.03 1.41 2.00 0.59 1.13 1.24 1.31 2.10 0.70 1.31 2.38 2.56 2.81 2.34 0.27 0.08 0.01 2.32

Explanation:

Step1: Recall mean formula

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are data - points and $n$ is the number of data - points. First, count the number of data - points $n$. Counting the values in the data set, $n = 50$.

Step2: Calculate the sum of data - points

$\sum_{i=1}^{50}x_{i}=0.72 + 1.62+2.94+2.75+1.93+1.26+1.20+2.02+2.09+1.00+0.06+1.61+2.35+0.77+2.11+0.00+2.53+0.03+1.16+2.31+0.02+0.33+1.07+1.06+0.71+0.57+0.31+2.22+0.11+2.39+2.33+2.62+2.03+1.41+2.00+0.59+1.13+1.24+1.31+2.10+0.70+1.31+2.38+2.56+2.81+2.34+0.27+0.08+0.01+3.02$
$\sum_{i = 1}^{50}x_{i}=74.94$

Step3: Calculate the mean

$\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{74.94}{50}=1.499$

Step4: Recall median formula for $n = 50$ (even number of data - points)

When $n$ is even, the median $M=\frac{x_{\frac{n}{2}}+x_{\frac{n}{2}+1}}{2}$ after arranging the data in ascending order. Arranging the 50 data - points in ascending order: $0.00,0.01,0.02,0.03,0.06,0.08,0.11,0.27,0.31,0.33,0.57,0.59,0.70,0.71,0.72,0.77,1.00,1.06,1.07,1.13,1.16,1.20,1.24,1.26,1.31,1.31,1.41,1.61,1.62,1.93,2.00,2.03,2.09,2.10,2.11,2.22,2.31,2.33,2.34,2.35,2.38,2.53,2.56,2.62,2.75,2.81,2.94,3.02$
$x_{25}=1.26$ and $x_{26}=1.31$
$M=\frac{1.26 + 1.31}{2}=\frac{2.57}{2}=1.285$

Answer:

The mean of the data set is $1.499$. The median of the data set is $1.285$.