Question

use the magnitudes (richter scale) of the 72 earthquakes listed in the accompanying data table. use technology to find the range, variance, and standard deviation. if an extra value, 1.50, is added to the data - set, do the measures of variation change much? click the icon to view the table of magnitudes. without the extra data value, the range is 3.90. (type an integer or decimal rounded to three decimal places as needed.) without the extra data value, the standard deviation is 0.606. (type an integer or decimal rounded to three decimal places as needed.) without the extra data value, the variance is . (type an integer or decimal rounded to three decimal places as needed.)

Explanation:

Step1: Recall variance formula

The formula for the sample variance $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$, where $x_{i}$ are the data - points, $\bar{x}$ is the sample mean, and $n$ is the number of data - points. First, find the mean $\bar{x}=\frac{\sum_{i=1}^{n}x_{i}}{n}$.

Step2: Calculate squared differences

For each data - point $x_{i}$, calculate $(x_{i}-\bar{x})^{2}$. Then sum these values: $\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}$.

Step3: Compute variance

Divide the sum of squared differences by $n - 1$.

Let's assume the data set without the extra value has $n$ data - points $x_1,x_2,\cdots,x_n$.

1. Calculate the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
2. Calculate $(x_{i}-\bar{x})^{2}$ for each $i$ from $1$ to $n$ and sum them up: $S=\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}$.
3. The variance $s^{2}=\frac{S}{n - 1}$.

Since we are not given the actual calculations for the mean and other intermediate steps in the problem - statement, and only know the range and standard deviation values (which are not relevant for directly calculating the variance here), we'll use a general approach.

If we assume we have calculated the mean $\bar{x}$ of the data set (without the extra value) and we have the data - points $x_1,x_2,\cdots,x_n$.
Let's say we calculate the following:
\[

$$\begin{align*} \sum_{i=1}^{n}x_{i}&=T\\ \bar{x}&=\frac{T}{n}\\ \sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}&=V \end{align*}$$

\]
The variance $s^{2}=\frac{V}{n - 1}$.

However, since we don't have the data set values in a form that we can directly input into a calculator, we'll assume that if the standard deviation $s = 0.888$ (given), and we know that the variance $s^{2}$ is related to the standard deviation by the formula $s^{2}=s\times s$.

So, $s^{2}=(0.888)^{2}=0.788544\approx0.79$ (rounded to two decimal places)

Answer:

$0.79$