MCQ
Sum of the absolute deviations remains minimum with respect to
- Amean
- ✓median
- Cmode
- Dgeometric mean
MD from median $=\frac{\sum \mid X-\text { median } \mid}{n}$
MD from mode $=\frac{\sum \mid X-\text { mode } \mid}{n}$
Since median $>$ mean $(\bar{X})$ and median $>$ mode.
So, It is clear that the mean deviation from median has the least value.
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