The mean deviation of the data 3, 10, 10, 4,7 , 10, 5 from the mean is
2
2.57
3
3.75
Given, observations are 3, 10, I0, 4, J , I0 and 5
∴x¯=3+10+10+4+7+10+57=497=7
Now, M.D. =ΣdiN=187=2.57