Question 1

How does the mean absolute deviation (MAD) of the data in set 1. compare to the mean absolute deviation of the data inSet 1: 12, 8, 10, 50Set 2: 13, 9, 8 A- The MAD of set 1 is 13 less than the MAD of set 2B- The MAD of set 1 is 13 more than the MAD of set 2C- The MAD of set 1 is 2 more than the MAD of set 2D- The MAD of set 1 is 2 less than the MAD of set 2

Question 2

The Mean Absolute Deviation is:

$\begin{gathered} \text{MAD}=(1)/(N)\sum ^N_(i\mathop=1)\mleft|x_i-\bar{x}\mright| \\ \text{where }\bar{\text{x}}\text{ is the mean} \end{gathered}$

For set1:

$\begin{gathered} \operatorname{mean}_1=(12+8+10+50)/(4)=(80)/(4)=20 \\ \text{MAD}_1=(1)/(4)\lbrack\mleft|12-20\mright|+\mleft|8-20\mright|+\mleft|10-20\mright|+\mleft|50-20\mright|\rbrack \\ \text{MAD}_1=(1)/(4)(8+12+10+30)=(60)/(4)=15 \end{gathered}$

For set2:

$\begin{gathered} \operatorname{mean}_2=(13+9+8)/(3)=(30)/(3)=10 \\ \text{MAD}_2=(1)/(3)\lbrack\mleft|13-10\mright|+\mleft|9-10\mright|+\mleft|8-10\mright|\rbrack \\ \text{MAD}_2=(1)/(3)(3+1+2)=(6)/(3)=2 \end{gathered}$

So the option B is the correct answer. MAD1 is 13 more than MAD2

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