In order to answer this question correctly, we need to solve for the mean, median, and mode of this data set. 

To begin, let's sort the data from least to greatest:

\(\displaystyle 1,2,2,3,3,4,4,4,4,4,7,8,8,9,9\)

Now that our data is ordered from least to greatest, we can solve for the median:

\(\displaystyle 1,2,2,3,3,4,4,{\color{Red} 4},4,4,7,8,8,9,9\)

Remember, the median is the middle most number when a data set is ordered from least to greatest.

The median for this data set is \(\displaystyle 4\)

Next, we can look at our data set to determine the mode:

The mode for this data set is \(\displaystyle 4\)

Remember, the mode is the number in a set that appears most often. 

Finally, we can solve for the mean:

\(\displaystyle 2+7+2+4+1+8+4+9+3+4+9+8+4+3+4=72\) 

\(\displaystyle \frac{72}{15}=4.8\)

Remember, the mean of a data set is the average of the numbers in a data set. 

The mean for this data set is \(\displaystyle 4.8\)

Now that we've done our calculation we should have:

Median: \(\displaystyle 4\)

Mode: \(\displaystyle 4\) 

Mean: \(\displaystyle 4.8\)

We are looking for the value that is representative of the center of the data; thus the mode or median would be the best measurement to use. 

In order to answer this question correctly, we need to solve for the range and the interquartile range. 

To begin, let's sort the data from least to greatest:

\(\displaystyle 1,2,2,3,3,4,4,4,4,4,7,8,8,9,9\)

Next, we can solve for the range. Remember, the range of a data set is the difference between the highest value and the lowest value in the set. 

\(\displaystyle \frac{\begin{array}[b]{r}11\\ -\ 1\end{array}}{ \ \ \ \space 10}\)

The range for this data set is \(\displaystyle 10\)

Now, we can solve for the interquartile range. Remember, the interquartile range is the difference between the upper quartile median and the lower quartile median. This means that we need to first calculate these two values. In order to do this, we need to split the data set into quartiles. 

First, we will find the median:

\(\displaystyle 1,2,2,3,3,4,4,{\color{Red} 4},4,4,7,8,8,9,9\)

We will then use the median to split the data in half. Next,  we must find the median of the first half—or lower quartile—and then the median of the second half—or upper quartile:

\(\displaystyle 1,2,2,{\color{Teal} 3},3,4,4,{\color{Red} 4},4,4,7,{\color{Teal} 8},8,9,9\)

Now we can solve for the difference between the upper quartile median and the lower quartile median:

\(\displaystyle \frac{\begin{array}[b]{r}8\\ -\ 3\end{array}}{ \ \ \ \space 5}\)

Now that we have completed these operations, we should have calculated the following values:

Range: \(\displaystyle 10\)

Interquartile range: \(\displaystyle 5\)

As you can see, solving for the interquartile range requires more steps because it takes into account more of the data points; thus, given our options, interquartile range is best to use when solving for variability. 

In order to answer this question correctly, we need to solve for the mean, median, and mode of this data set. 

To begin, let's sort the data from least to greatest:

\(\displaystyle 50,58,64,78,79,79,82\)

Now that our data is ordered from least to greatest, we can solve for the median:

\(\displaystyle 50,58,64,{\color{Red} 78},79,79,82\)

Remember, the median is the middle most number when a data set is ordered from least to greatest.

The median for this data set is \(\displaystyle 78\)

Next, we can look at our data set to determine the mode:

The mode for this data set is \(\displaystyle 79\)

Remember, the mode is the number in a set that appears most often. 

Finally, we can solve for the mean:

\(\displaystyle 50+58+64+78+79+79+82=490\) 

\(\displaystyle \frac{490}{7}=70\)

Remember, the mean of a data set is the average of the numbers in a data set. 

The mean for this data set is \(\displaystyle 70\)

Now that we've done our calculation we should have:

Median: \(\displaystyle 78\)

Mode: \(\displaystyle 79\) 

Mean: \(\displaystyle 70\)

We are looking for the value that is representative of the center of the data; thus the mean would be the best measure because \(\displaystyle 78\) and \(\displaystyle 79\) represent the greatest values of our data set, but \(\displaystyle 70\) is more reflective of the center of all of the values. Normally, when a data set is varied the mean is normally the best measure of center. 

In order to answer this question correctly, we need to solve for the range and the interquartile range. 

To begin, let's sort the data from least to greatest:

\(\displaystyle 50,58,64,78,79,79,82\)

Next, we can solve for the range. Remember, the range of a data set is the difference between the highest value and the lowest value in the set. 

\(\displaystyle \frac{\begin{array}[b]{r}82\\ -\ 50\end{array}}{ \ \ \ \space 32}\)

The range for this data set is \(\displaystyle 32\)

Now, we can solve for the interquartile range. Remember, the interquartile range is the difference between the upper quartile median and the lower quartile median. This means that we need to first calculate these two values. In order to do this, we need to split the data set into quartiles. 

First, we will find the median:

\(\displaystyle 50,58,64,{\color{Red} 78},79,79,82\)

We will then use the median to split the data in half. Next,  we must find the median of the first half—or lower quartile—and then the median of the second half—or upper quartile:

\(\displaystyle 50,{\color{Teal} 58},64,{\color{Red} 78},79,{\color{Teal} 79},82\)

Now we can solve for the difference between the upper quartile median and the lower quartile median:

\(\displaystyle \frac{\begin{array}[b]{r}79\\ -\ 58\end{array}}{ \ \ \ \space 21}\)

Now that we have completed these operations, we should have calculated the following values:

Range: \(\displaystyle 32\)

Interquartile range: \(\displaystyle 21\)

As you can see, solving for the interquartile range requires more steps because it takes into account more of the data points; thus, given our options, interquartile range is best to use when solving for variability. 

In order to answer this question correctly, we need to solve for the range and the interquartile range. 

To begin, let's sort the data from least to greatest:

\(\displaystyle 0,13,21,25,35,43,52\)

Next, we can solve for the range. Remember, the range of a data set is the difference between the highest value and the lowest value in the set. 

\(\displaystyle \frac{\begin{array}[b]{r}52\\ -\ 0\end{array}}{ \ \ \ \space 52}\)

The range for this data set is \(\displaystyle 52\)

Now, we can solve for the interquartile range. Remember, the interquartile range is the difference between the upper quartile median and the lower quartile median. This means that we need to first calculate these two values. In order to do this, we need to split the data set into quartiles. 

First, we will find the median:

\(\displaystyle 0,13,21,{\color{Red} 25},35,43,52\)

We will then use the median to split the data in half. Next,  we must find the median of the first half—or lower quartile—and then the median of the second half—or upper quartile:

\(\displaystyle 0,{\color{Teal} 13},21,{\color{Red} 25},35,{\color{Teal} 43},52\)

Now we can solve for the difference between the upper quartile median and the lower quartile median:

\(\displaystyle \frac{\begin{array}[b]{r}43\\ -\ 13\end{array}}{ \ \ \ \space 20}\)

Now that we have completed these operations, we should have calculated the following values:

Range: \(\displaystyle 52\)

Interquartile range: \(\displaystyle 20\)

As you can see, solving for the interquartile range requires more steps because it takes into account more of the data points; thus, given our options, interquartile range is best to use when solving for variability. 

In order to answer this question correctly, we need to solve for the mean, median, and mode of this data set. 

To begin, let's sort the data from least to greatest:

\(\displaystyle 0,13,21,25,35,43,52\)

Now that our data is ordered from least to greatest, we can solve for the median:

\(\displaystyle 0,13,21,{\color{Red} 25},35,43,52\)

Remember, the median is the middle most number when a data set is ordered from least to greatest.

The median for this data set is \(\displaystyle 25\)

Next, we can look at our data set to determine the mode:

There is no mode for this set because each value only appears once.

Remember, the mode is the number in a set that appears most often. 

Finally, we can solve for the mean:

\(\displaystyle 0+13+21+25+35+43+52=189\) 

\(\displaystyle \frac{189}{7}=27\)

Remember, the mean of a data set is the average of the numbers in a data set. 

The mean for this data set is \(\displaystyle 27\)

Now that we've done our calculation we should have:

Median: \(\displaystyle 25\)

Mode: None 

Mean: \(\displaystyle 27\)

We are looking for the value that is representative of the center of the data; thus the mean would be the best measure because of how varied the set is. Normally, when a data set is varied the mean is normally the best measure of center.

In order to answer this question correctly, we need to solve for the range and the interquartile range. 

To begin, let's sort the data from least to greatest:

\(\displaystyle 20,25,30,50,50,50,50\)

Next, we can solve for the range. Remember, the range of a data set is the difference between the highest value and the lowest value in the set. 

\(\displaystyle \frac{\begin{array}[b]{r}50\\ -\ 20\end{array}}{ \ \ \ \space 30}\)

The range for this data set is \(\displaystyle 30\)

Now, we can solve for the interquartile range. Remember, the interquartile range is the difference between the upper quartile median and the lower quartile median. This means that we need to first calculate these two values. In order to do this, we need to split the data set into quartiles. 

First, we will find the median:

\(\displaystyle 20,25,30,{\color{Red} 50},50,50,50\)

We will then use the median to split the data in half. Next,  we must find the median of the first half—or lower quartile—and then the median of the second half—or upper quartile:

\(\displaystyle 20,{\color{Teal} 25},30,{\color{Red} 50},50,{\color{Teal} 50},50\)

Now we can solve for the difference between the upper quartile median and the lower quartile median:

\(\displaystyle \frac{\begin{array}[b]{r}50\\ -\ 25\end{array}}{ \ \ \ \space 25}\)

Now that we have completed these operations, we should have calculated the following values:

Range: \(\displaystyle 30\)

Interquartile range: \(\displaystyle 25\)

As you can see, solving for the interquartile range requires more steps because it takes into account more of the data points; thus, given our options, interquartile range is best to use when solving for variability. 

In order to answer this question correctly, we need to solve for the mean, median, and mode of this data set. 

To begin, let's sort the data from least to greatest:

\(\displaystyle 20,25,30,50,50,50,50\)

Now that our data is ordered from least to greatest, we can solve for the median:

\(\displaystyle 20,25,30,{\color{Red} 50},50,50,50\)

Remember, the median is the middle most number when a data set is ordered from least to greatest.

The median for this data set is \(\displaystyle 50\)

Next, we can look at our data set to determine the mode:

The mode for this set is \(\displaystyle 50\)

Remember, the mode is the number in a set that appears most often. 

Finally, we can solve for the mean:

\(\displaystyle 20+25+30+50+50+50+50=275\) 

\(\displaystyle \frac{275}{7}=39.3\)

Remember, the mean of a data set is the average of the numbers in a data set. 

The mean for this data set is \(\displaystyle 39.3\)

Now that we've done our calculation we should have:

Median: \(\displaystyle 50\)

Mode: \(\displaystyle 50\)

Mean: \(\displaystyle 39.3\)

We are looking for the value that is representative of the center of the data; thus the mean would be the best measure because of how varied the set is. Normally, when a data set is varied the mean is normally the best measure of center.

In order to answer this question correctly, we need to solve for the mean, median, and mode of this data set. 

To begin, let's sort the data from least to greatest:

\(\displaystyle 2,2,2,3,5,7,29\)

Now that our data is ordered from least to greatest, we can solve for the median:

\(\displaystyle 2,2,2,{\color{Red} 3},5,7,29\)

Remember, the median is the middle most number when a data set is ordered from least to greatest.

The median for this data set is \(\displaystyle 3\)

Next, we can look at our data set to determine the mode:

The mode for this set is \(\displaystyle 2\)

Remember, the mode is the number in a set that appears most often. 

Finally, we can solve for the mean:

\(\displaystyle 2+2+2+3+5+7+29=50\) 

\(\displaystyle \frac{50}{7}=7.14\)

Remember, the mean of a data set is the average of the numbers in a data set. 

The mean for this data set is \(\displaystyle 7.17\)

Now that we've done our calculation we should have:

Median: \(\displaystyle 3\)

Mode: \(\displaystyle 2\)

Mean: \(\displaystyle 7.17\)

We are looking for the value that is representative of the center of the data. In this data set we have an outlier, which means the mean is not going to be the best measure of center. Also, the mode is the lowest value in our set; thus, the median is the best measure of center. 

In order to answer this question correctly, we need to solve for the range and the interquartile range. 

To begin, let's sort the data from least to greatest:

\(\displaystyle 2,2,2,3,5,7,29\)

Next, we can solve for the range. Remember, the range of a data set is the difference between the highest value and the lowest value in the set. 

\(\displaystyle \frac{\begin{array}[b]{r}29\\ -\ 2\end{array}}{ \ \ \ \space 27}\)

The range for this data set is \(\displaystyle 27\)

Now, we can solve for the interquartile range. Remember, the interquartile range is the difference between the upper quartile median and the lower quartile median. This means that we need to first calculate these two values. In order to do this, we need to split the data set into quartiles. 

First, we will find the median:

\(\displaystyle 2,2,2,{\color{Red} 3},5,7,29\)

We will then use the median to split the data in half. Next,  we must find the median of the first half—or lower quartile—and then the median of the second half—or upper quartile:

\(\displaystyle 2,{\color{Teal} 2},2,{\color{Red} 3},5,{\color{Teal} 7},29\)

Now we can solve for the difference between the upper quartile median and the lower quartile median:

\(\displaystyle \frac{\begin{array}[b]{r}7\\ -\ 2\end{array}}{ \ \ \ \space 5}\)

Now that we have completed these operations, we should have calculated the following values:

Range: \(\displaystyle 27\)

Interquartile range: \(\displaystyle 5\)

As you can see, solving for the interquartile range requires more steps because it takes into account more of the data points; thus, given our options, interquartile range is best to use when solving for variability.