# Talk:Median absolute deviation

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We need a reference for the 3/4 in the formula relating MAD to standard deviation. —Preceding unsigned comment added by 99.251.254.165 (talkcontribs)

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)
....OK,.... indeed, that is not what is referred to, but what is referred to is similarly obvious. Michael Hardy (talk) 19:47, 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)

## Alternative formula

I have seen some textbooks refer to "Median absolute deviation", but using this formula:

${\displaystyle \operatorname {MAD} ={\frac {1}{N}}\sum _{i=1}^{n}|X_{i}-\operatorname {median} |}$

Since I have not much knowledge about statistics I don't know if it is worth mentioning it. --Tgor (talk) 19:05, 17 October 2013 (UTC)

## 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 98.116.5.5 (talk) 17:44, 7 August 2015 (UTC)