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Feb 6, 2020 at 23:14 answer added Mike Earnest timeline score: 1
Feb 6, 2020 at 7:25 comment added Lee David Chung Lin How about the answer by @GEdgar in this old post?
Feb 5, 2020 at 17:41 comment added StubbornAtom If $a_{(1)}\le a_{(2)}\le \cdots\le a_{(n)}$ are the ordered observations, then median of $(a_1,a_2,\ldots,a_n)$ for odd $n$ is $a_{\left(\frac{n+1}{2}\right)}$.
Feb 5, 2020 at 12:50 answer added Yannik timeline score: 0
Feb 5, 2020 at 9:48 answer added aryan bansal timeline score: -1
Feb 5, 2020 at 9:42 comment added pH 74 @Idnoknow see this answer
Feb 5, 2020 at 9:25 comment added Idonknow @PeterSheldrick May I have a proof of the formula or reference that you provide above?
Feb 5, 2020 at 9:17 comment added user11260 For the median pick one of the $a_1,a_2,...$ to find the minimum of the sums $|a_1-a_1|+|a_2-a_1|+|a_3-a_1|...$ and $|a_1-a_2|+|a_2-a_2|+|a_3-a_2|...$ and so on
Feb 5, 2020 at 9:12 history edited Idonknow CC BY-SA 4.0
edited title
Feb 5, 2020 at 9:03 history asked Idonknow CC BY-SA 4.0