The mean absolute deviation is the average of the absolute values of the differences between the mean and each value in the data set. Start by calculating the mean of the given set of numbers.
Mean Absolute Deviation: 11.4 Interpretation: The amount of milligrams of caffeine in certain types of tea is, on average, 11.4 units away from the mean.
The mean of a data set or x, is calculated by finding the sum of all of the values in the set and then dividing by the number of values in the set. In this case, there are 10 values. Let's now look at the table with given values.
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.
|x-x_1|+|x-x_2|+...+|x-x_n|/n
In this formula, x_1,...,x_n are the values in the set of data, x is the mean, and n is the number of values. We already know that x=30 and n= 10. Let's use a table to find the sum of the absolute values of the differences.
x_n
x-x_n
|x-x_n|
9
30- 9=21
|21|=21
46
30- 46=- 16
|- 16|=16
18
30- 18=12
|12|=12
35
30- 35=- 5
|- 5|=5
30
30- 30=0
|0|=0
12
30- 12=18
|18|=18
56
30- 56=- 26
|- 26|=26
24
30- 24=6
|6|=6
38
30- 38=- 8
|- 8|=8
32
30- 32=- 2
|- 2|=2
Sum of Values
114
Finally, we need to divide by 10.
Mean Absolute Deviation (MAD)
114/10=11.4
A MAD of 11.4 indicates that the data, on average, are 11.4 units away from the mean. In the context of the problem, this means that the amount of caffeine in milligrams in certain types of tea is, on average, 11.4 units away from the mean of 30 milligrams. 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 (Q1): is the median of the lower half of the data set.
Upper Quartile (Q3): 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!
9, 12, 18, 24, 30 | 32, 35, 38, 48, 56
We can use the upper and lower quartiles to find the interquartile range, or IQR.
Upper Quartile:& 38
Lower Quartile:& 18
IQR:& 38- 18= 20
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=& Q1-1.5* IQR
Maximum=& Q2+1.5* IQR
Let's find the minimum and maximum for a value not to be an outlier.
Minimum
18-1.5* 20=- 12 [0.8em]
Maximum
38+1.5* 20=68
Any value between 14.5 and 41.5 is not an outlier. Thus, the data set has no outliers.
