Exercise 30 Page P3

We want to find the mean, median, mode, range, and standard deviation of the given data set, as well as whether there are any outliers.

Mean

The mean of a data set is the sum of the values divided by the total number of values in the set. Let's start by calculating the sum of the given values. There are $17$ values in our set, so we have to divide the sum by $17.$

Median

When the data are arranged in numerical order, the median is the middle value — or the mean of the two middle values — in a set of data. Let's arrange the given values and find the median. The median of this set is $10.$

Mode

The mode is the value or values that appear most often in a set of data. Let's find the mode of the given values.

$$\begin{array}{c}10+15+9+8+11+10+9+8+\\ 13+9+12+10+13+11+9+12+10\end{array}$$

Since the data set has two values that appear more often than the other values but equally as often as each other, there are two modes.

$$\begin{array}{c}\text{Mode:}9\text{ and }10\end{array}$$

Range

The range is the difference between the greatest and least values in a set of data. For our exercise, the greatest value is $15$ and the least value is $8.$

$$\begin{array}{c}\text{Range:}15-8=7\end{array}$$

Standard Deviation

The standard deviation in a set of data is the average amount by which each individual value deviates or differs from the mean.

$$\begin{array}{c}\underline{\text{Standard Deviation}}\\ \sqrt{\frac{(x_1-\mu {)}^2+(x_2-\mu {)}^2+\cdots +(x_n-\mu {)}^2}{n}}\end{array}$$

In the above formula, $x_1,\dots ,x_n$ are the values of the set of data, $\mu$ is the mean, and $n$ is the number of values. For this exercise, the mean is the sum of the values $179$ divided by the number of values $17.$

$$\begin{array}{c}\mu =\frac{179}{17}\approx 10.529\end{array}$$

Let's use this value and apply the formula to each value in the set.

$x_n$	$x_n-\mu$	$(x_n-\mu {)}^2$
$10$	$10-10.529=\text{-}0.529$	$(\text{-}0.529{)}^2\approx 0.280$
$15$	$15-10.529=4.471$	$4.471^2\approx 19.990$
$9$	$9-10.529=\text{-}1.529$	$(\text{-}1.529{)}^2\approx 2.338$
$8$	$8-10.529=\text{-}2.529$	$(\text{-}2.529{)}^2\approx 6.396$
$11$	$11-10.529=0.471$	$0.471^2\approx 0.222$
$10$	$10-10.529=\text{-}0.529$	$(\text{-}0.529{)}^2\approx 0.280$
$9$	$9-10.529=\text{-}1.529$	$(\text{-}1.529{)}^2\approx 2.338$
$8$	$8-10.529=\text{-}2.529$	$(\text{-}2.529{)}^2\approx 6.396$
$13$	$13-10.529=2.471$	$2.471^2\approx 6.106$
$9$	$9-10.529=\text{-}1.529$	$(\text{-}1.529{)}^2\approx 2.338$
$12$	$12-10.529=1.471$	$1.471^2\approx 2.164$
$10$	$10-10.529=\text{-}0.529$	$(\text{-}0.529{)}^2\approx 0.280$
$13$	$13-10.529=2.471$	$2.471^2\approx 6.106$
$11$	$11-10.529=0.471$	$0.471^2\approx 0.222$
$9$	$9-10.529=\text{-}1.529$	$(\text{-}1.529{)}^2\approx 2.338$
$12$	$12-10.529=1.471$	$1.471^2\approx 2.164$
$10$	$10-10.529=\text{-}0.529$	$(\text{-}0.529{)}^2\approx 0.280$
Sum of Values		$\approx 60.238$

Finally, we need to divide by $17$ and then calculate the square root.

$$\begin{array}{c}\text{Standard Deviation:}\sqrt{\frac{60.238}{17}}\approx 1.882\end{array}$$

Outliers

To find the outliers, let's start by defining some important characteristics of data sets.

• Lower Quartile (Q$1$) is the median of the lower half of the data set.
• Upper Quartile (Q$3$) is the median of the upper half of the data set.
• Interquartile Range is the difference between the upper and lower quartiles $(Q3-Q1).$

Let's find the upper quartile, the lower quartile, and the interquartile range for the given data set. Do not forget to write the values in numerical order!

$$\begin{array}{c}8,8,9,9,9,9,10,10,\\ |10|\\ 10,11,11,12,12,13,13,15\end{array}$$

We have two middle values for each half. Thus, we need to calculate the mean of those middle values.

$$\begin{array}{rl}\text{Upper Quartile:}& \frac{12+12}{2}=12\\ \text{Lower Quartile:}& \frac{9+9}{2}=9\\ \text{Interquartile Range:}& 12-9=3\end{array}$$

An outlier is an extremely high or extremely low value when compared with the rest of the values in the set. To check for them, we look for data values that are beyond the upper or lower quartiles by more than $1.5$ times the interquartile range.

$$\begin{array}{rl}\text{Minimum}=& \text{Q}1-\text{IQR}\times 1.5\\ \text{Maximum}=& \text{Q}2+\text{IQR}\times 1.5\end{array}$$

Let's find the minimum and maximum for a value not to be an outlier.

$$\begin{array}{ccc}\text{Minimum}& & \text{Maximum}\\ 9-1.5\times 3=4.5& & 12+1.5\times 3=16.5\end{array}$$

Any value between $4.5$ and $16.5$ is not an outlier. Notice that the data set does not have any outliers.