Question

how does the mean absolute deviation (mad) of the data in set 2 compare to the mean absolute deviation of the data in set 1?
set 1: 16, 15, 10, 12
set 2: 16, 62, 15, 10, 12
the mad of set 2 is 10 less than the mad of set 1.
the mad of set 2 is 13.35 more than the mad of set 1.
the mad of set 2 is 10 more than the mad of set 1.
the mad of set 2 is 13.35 less than the mad of set 1.

Explanation:

Step1: Calculate mean of Set 1

The formula for the mean $\bar{x}$ of a data set is $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the data points and $n$ is the number of data points.
For Set 1: $x_1 = 16$, $x_2=15$, $x_3 = 10$, $x_4=12$, $n = 4$.
$\sum_{i=1}^{4}x_{i}=16 + 15+10 + 12=53$.
Mean of Set 1, $\bar{x}_1=\frac{53}{4}=13.25$.

Step2: Calculate MAD of Set 1

The formula for Mean Absolute Deviation (MAD) is $MAD=\frac{\sum_{i = 1}^{n}|x_{i}-\bar{x}|}{n}$.
For each data point in Set 1:

• $|16 - 13.25|=2.75$
• $|15 - 13.25| = 1.75$
• $|10 - 13.25|=3.25$
• $|12 - 13.25| = 1.25$

$\sum_{i = 1}^{4}|x_{i}-\bar{x}_1|=2.75+1.75 + 3.25+1.25=9$.
$MAD_1=\frac{9}{4}=2.25$.

Step3: Calculate mean of Set 2

For Set 2: $x_1 = 16$, $x_2=62$, $x_3 = 15$, $x_4=10$, $x_5=12$, $n = 5$.
$\sum_{i=1}^{5}x_{i}=16+62 + 15+10+12=115$.
Mean of Set 2, $\bar{x}_2=\frac{115}{5}=23$.

Step4: Calculate MAD of Set 2

For each data point in Set 2:

• $|16 - 23|=7$
• $|62 - 23| = 39$
• $|15 - 23|=8$
• $|10 - 23|=13$
• $|12 - 23| = 11$

$\sum_{i=1}^{5}|x_{i}-\bar{x}_2|=7 + 39+8+13+11=78$.
$MAD_2=\frac{78}{5}=15.6$.

Step5: Compare MAD of Set 2 and Set 1

Find the difference: $MAD_2-MAD_1=15.6 - 2.25 = 13.35$.
So the MAD of set 2 is 13.35 more than the MAD of set 1.

Answer:

The MAD of set 2 is 13.35 more than the MAD of set 1. (Corresponding option: The MAD of set 2 is 13.35 more than the MAD of set 1.)