Question

use the magnitudes (richter scale) of the 120 earthquakes listed in the accompanying data table. use technology to find the range, variance, and standard deviation. if another value, 7.00, is added to the data - set, do the measures of variation change much? (type an integer or decimal rounded to three decimal places as needed.) without the extra data value, the standard deviation is 0.054 (type an integer or decimal rounded to three decimal places as needed.) without the extra data value, the variance is 0.426 (type an integer or decimal rounded to three decimal places as needed.) with the extra data value, the range is 5.000 (type an integer or decimal rounded to three decimal places as needed.) with the extra d (type an integer or decimal rounded to three decimal places as needed.) with the extra d within 5 percentage points of each other, (type an integer or decimal rounded to three decimal places as needed.) do the measure more than 5 percentage points apart,? choose the correct answer below the ranges are the variances are and the standard deviations are

Explanation:

Step1: Recall range formula

Range = Maximum - Minimum

Step2: Recall variance and standard - deviation formulas

Variance $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$, Standard deviation $s=\sqrt{s^{2}}$
We are not given the full data - set to calculate from scratch. But we know that adding a new data point (7.00) can change the measures of variation.
If the original range was say $R_1$ and new range is $R_2 = 5.800$ (given with extra data value). We need to consider how much the variance and standard deviation change.
Let the original variance be $V_1=0.426$ and original standard deviation be $S_1 = 0.054$.
When a new data point is added, we recalculate the mean $\bar{x}$ (not given the data so we can't calculate exactly), then the squared - differences $(x_{i}-\bar{x})^{2}$, sum them up and divide by $n$ (or $n - 1$ depending on sample/population) to get the new variance $V_2$ and then the new standard deviation $S_2=\sqrt{V_2}$.
To check if they change significantly, we can calculate the percentage change. For example, for variance, percentage change $=\frac{V_2 - V_1}{V_1}\times100\%$.
However, since we are not given the full data - set to recalculate the variance and standard deviation exactly after adding the new point, we assume based on the nature of these measures.
The range is highly affected by extreme values. If 7.00 is an extreme value compared to the original data set, the range will change.
The variance and standard deviation are also affected by extreme values. A large value like 7.00 can increase the spread of the data, thus increasing the variance and standard deviation.
Since we are not given the original range, we can't comment on its percentage change. But we know that the new range is 5.800.
For variance and standard deviation, we can say that adding a value like 7.00 (assuming it is far from the original mean) will increase them.
If we assume the original range was much smaller than 5.800, the range has changed significantly.
For variance and standard deviation, since 7.00 is likely an outlier, the variance and standard deviation will also change significantly.

Answer:

The ranges are more than 5 percentage points apart, the variances are more than 5 percentage points apart, and the standard deviations are more than 5 percentage points apart.