Given electric bills of family A and family C.
To find mean absolute deviation of family A,
A).
First find mean of family A,
Mean
Now, mean absolute deviation
![\begin{gathered} =(\lvert88.82-78.585\rvert+\lvert84.80-78.585\rvert+\lvert62.35-78.585\rvert+\lvert65.17-78.585\rvert+\lvert60.20-78.585\rvert+\lvert74.86-78.585\rvert+\lvert92.15-78.585\rvert+\lvert99.89-78.585\rvert+\lvert96.45-78.585\rvert+\lvert81.63-78.585\rvert+\lvert64.12-78.585\rvert+\lvert72.58+78.585\rvert)/(12) \\ =(10.235+6.215+16.235+13.415+18.385+3.725+13.565+21.305+17.865+3.045+14.465+6.005)/(12) \\ =(144.458)/(12) \\ =12.038 \end{gathered}]()
12.038 is the mean absolute deviation for family A.
Now. mean absolute deviation for family C,
Mean
![\begin{gathered} =(134.51+118.06+109.28+90.15+91.20+96.36+130.71+129.16+107.77+100.95+107.64+123.82)/(12) \\ =(1339.61)/(12) \\ =111.63 \end{gathered}]()
Now, Mean absolute deviation
![\begin{gathered} =(\lvert134.51-111.63\rvert+\lvert118.06-111.63\rvert+\lvert109.28-111.63\rvert+\lvert90.15-111.63\rvert+\lvert91.20-111.63\rvert+\lvert96.36-111.63\rvert+\lvert130.71-111.63\rvert+\lvert129.16-111.63\rvert+\lvert107.77-111.63\rvert+\lvert100.95-111.63\rvert+\lvert107.64-111.63\rvert+\lvert123.82-111.63\rvert)/(12) \\ =(22.88+6.43+2.35+21.48+20.43+15.27+19.08+17.53+3.86+10.68+3.99+12.19)/(12) \\ =(156.17)/(12) \\ =13.04 \end{gathered}]()
Hence. 13.04 is the required mean absolute deviation for family B.