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Mar 6, 2023 at 15:50 history edited LSpice CC BY-SA 4.0
Link to answer, while this is on the front page
Nov 5, 2009 at 7:57 comment added Yoo If we define the standard deviation with absolute value instead of squares, we would still have Chebyshev's theorem except with 1/k in place of 1/k^2
Oct 18, 2009 at 22:02 history answered Anna Varvak CC BY-SA 2.5