The mean absolute deviation (MAD) is the average of the absolute values of the differences between the mean and each value in the data set. We will start by calculating the mean of the given set of numbers.

Mean

First we will find the sum of the given values.

16 + 24 + 15 + 17 + 22 + 16 + 18 + 24 + 17 + 13 + 25 + 21 = 228

Since there are

12

values in our set, to calculate the mean we have to divide the sum by

12 .

Mean : \frac{2 2 8}{1 2} = 19

Mean Absolute Deviation

As previously stated, the MAD of a set of data is the average of the absolute values of the differences between the mean and each value in the data set.

\frac{∣ x - x _{1} ∣ + ∣ x - x _{2} ∣ + \dots + ∣ x - x _{n} ∣}{n}

In this formula,

x_{1}, \dots, x_{n}

are the values in the set of data,

\overline{x}

is the mean, and

n

is the number of values. We already know that

\overline{x} = 19

and

n = 12 .

Let's use a table to find the sum of the absolute values of the differences.

$x_{i}$	$\overline{x} - x_{i}$	$∣ \overline{x} - x_{i} ∣$
$16$	$19 - 16 = 3$	$∣ 3 ∣ = 3$
$24$	$19 - 24 = - 5$	$∣ - 5 ∣ = 5$
$15$	$19 - 15 = 4$	$∣ 4 ∣ = 4$
$17$	$19 - 17 = 2$	$∣ 2 ∣ = 2$
$22$	$19 - 22 = - 3$	$∣ - 3 ∣ = 3$
$16$	$19 - 16 = 3$	$∣ 3 ∣ = 3$
$18$	$19 - 18 = 1$	$∣ 1 ∣ = 1$
$24$	$19 - 24 = - 5$	$∣ - 5 ∣ = 5$
$17$	$19 - 17 = 2$	$∣ 2 ∣ = 2$
$13$	$19 - 13 = 6$	$∣ 6 ∣ = 6$
$25$	$19 - 25 = - 6$	$∣ - 6 ∣ = 6$
$21$	$19 - 21 = - 2$	$∣ - 2 ∣ = 2$
Sum of Values		$42$

Finally, we need to divide by

12 .

\underline{Mean Absolute Deviation (MAD)} \frac{4 2}{1 2} = 3.5

A MAD of

3.5

indicates that the data, on average, are

3.5

units away from the mean. In the context of the problem, this means that the number of bags of cotton candy sold each hour on Saturday, on average, are

3.5

units away from the mean of

19

bags. There are no outliers that influence this value.

Extra

Calculating Outliers

To check if there are any 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 $(Q 3 - Q 1) .$

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!

13, 15, 16, 16, 17, 17 ∣ 18, 21, 22, 24, 24, 25

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

Upper Quartile : Lower Quartile : Interquartile Range : \frac{2 2 + 2 4}{2} = 23 \frac{1 6 + 1 6}{2} = 16 23 - 16 = 7

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.

Minimum = Maximum = Q 1 - 1.5 \times IQR Q 2 + 1.5 \times IQR

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

Minimum 16 - 1.5 \times 7 = 5.5 Maximum 23 + 1.5 \times 7 = 33.5

Any value between

5.5

and

33.5

is not an outlier. Thus, the data set does not have an outlier.

Exercise