Talk:Median absolute deviation
|WikiProject Statistics||(Rated Start-class, Low-importance)|
- It seems almost obvious that for a continous distribution which is symmetric about 0, half the distribution is further from the centre/median than the 3rd quartile (i.e. above the 3rd quartile or below the 1st quatile) and half is closer.
- In other words, if F(x)=1-F(-x) and F(0)=1/2, then both F-1(3/4)-F-1(1/2) and F-1(1/2)-F-1(1/4) are equal to F-1(3/4), and Pr(|X-F-1(1/2)|>F-1(3/4)) is 1/2, making F-1(3/4) the median absolute deviation --Rumping (talk) 01:53, 11 December 2009 (UTC)
- That much is obvious, but I don't think that's what was being referred to. Take a look at the article. Michael Hardy (talk) 19:40, 15 December 2009 (UTC)
Relationship with absolute deviation?
Is it really worth two articles?. Does MAD stand for Median absolute deviation, or Mean absolute deviation, or both/either? See http://www.google.com/search?q=MAD+%22absolute+deviation%22+-wikipedia --Rumping (talk) 00:03, 11 December 2009 (UTC)
I have seen some textbooks refer to "Median absolute deviation", but using this formula:
Is there a mistake?
Is this statement correct? "For a symmetric distribution with zero mean, the population MAD is the 75th percentile of the distribution." Wouldn't it be the 75th percentile minus the 25th percentile, not merely the 75th percentile? — Preceding unsigned comment added by 188.8.131.52 (talk) 17:44, 7 August 2015 (UTC)