The Mean Absolute Deviation is:
[tex]\begin{gathered} \text{MAD}=\frac{1}{N}\sum ^N_{i\mathop=1}\mleft|x_i-\bar{x}\mright| \\ \text{where }\bar{\text{x}}\text{ is the mean} \end{gathered}[/tex]For set1:
[tex]\begin{gathered} \operatorname{mean}_1=\frac{12+8+10+50}{4}=\frac{80}{4}=20 \\ \text{MAD}_1=\frac{1}{4}\lbrack\mleft|12-20\mright|+\mleft|8-20\mright|+\mleft|10-20\mright|+\mleft|50-20\mright|\rbrack \\ \text{MAD}_1=\frac{1}{4}(8+12+10+30)=\frac{60}{4}=15 \end{gathered}[/tex]For set2:
[tex]\begin{gathered} \operatorname{mean}_2=\frac{13+9+8}{3}=\frac{30}{3}=10 \\ \text{MAD}_2=\frac{1}{3}\lbrack\mleft|13-10\mright|+\mleft|9-10\mright|+\mleft|8-10\mright|\rbrack \\ \text{MAD}_2=\frac{1}{3}(3+1+2)=\frac{6}{3}=2 \end{gathered}[/tex]So the option B is the correct answer. MAD1 is 13 more than MAD2
