Question

How does the mean absolute deviation (MAD) of the data in set 1 compare to the mean absolute deviation of the data in set 2? Set 1: 5, 10, 7, 2 Set 2: 5, 10, 7, 42

Answer 1

Set 2 is 10.5 more than set 1.

Answer 2

Answer:

The MAD of the set 2 is greater than the MAD of the set 1.

Step-by-step explanation:

The given data set is:

Set 1.  5, 10, 7, 2

Let us first find the mean of the given data set, that is

$M=\frac{5+10+7+2}{4}=\frac{24}{2}=12$

Now, find the distance between each data point and the mean.

Data point                                                     Distance from mean

5                                                                        |12-5|=7

10                                                                      |12-10|=2

7                                                                        |12-7|=5

2                                                                       |12-2|=10

Now, Add the distances together and Divide the sum by the number of data points, thus

$MAD=\frac{7+2+5+10}{4}=12$

Thus, the Mean absolute deviation of set 1 is 12.

The given set is:

Set 2. 5, 10, 7, 42

Let us first find the mean of the given data set, that is

$M=\frac{5+10+7+42}{4}=\frac{64}{2}=32$

Now, find the distance between each data point and the mean.

Data point                                                     Distance from mean

5                                                                        |32-5|=27

10                                                                      |32-10|=22

7                                                                        |32-7|=25

42                                                                      |32-42|=10

Now, Add the distances together and Divide the sum by the number of data points, thus

$MAD=\frac{27+22+25+10}{4}=42$

Thus, the Mean absolute deviation of set 1 is 42.

Therefore, the MAD of the set 2 is greater than the MAD of the set 1.