What does IQR/Med tell you about the data? What is the purpose of dividing the IQR by the Median?


IQR = 3.0  
Med = 8.1  
IQR/Med = .37

IQR = 2.1  
Med = 7.1  
IQR/Med = .29

I got the numbers from this site. I understand IQR and Median, and I was told that IQR/Med is an indicator of data quality (<0.3 means good data). But I don't understand why or how. I need to explain IQR/Med to a general audience, but I need to understand its utility first.


2 Answers 2


I can't reach the site you link. It's hard to know exactly what was meant by this number without more context. I can make an educated guess, though.

Many numbers will have a greater or lesser amount of variability depending on the level of their center. Starting with stats 101, we often think in terms of the normal distribution, in which case we can think of the standard deviation becoming larger with larger means. In such cases, it is often true that, although there is heteroscedasticity, there is a constant coefficient of variation ($SD/\bar X$). However, people may prefer to work outside of parameters like the mean and standard deviation, which have a strong connection to the normal distribution and are sensitive to outliers. The interquartile range can be considered a more robust measure of spread, and the median can likewise be considered a more robust measure of central tendency. Thus, the IQR/median can fill a similar role as the coefficient of variation while remaining more resistant to outliers. (For what it's worth, for this purpose I might prefer the median absolute deviation from the median, MADM/median, personally.)

I have never heard of the rule of thumb that a CoV (or in this case, IQR/median) <.30 means "good data". In line with the interpretation above, I would guess the idea is that the variability isn't that large relative to the location of the middle of the distribution, so the data will tend to be more stable. That is, small changes to the median won't create large swings in the data.

  • $\begingroup$ In my work, the numbers people use are ultrasound elastography samples. <br/> $\endgroup$
    – NAK
    Commented Oct 31, 2016 at 19:46
  • $\begingroup$ I suspect people are worried that the data are non-normal, possibly contaminated w/ outliers or other phenomena, & possibly unstable. Checking the IQR/median allows them to assess the stability of the data w/o having to make assumptions they don't want to rely on. $\endgroup$ Commented Oct 31, 2016 at 19:51
  • $\begingroup$ To clarify: by stability of the data, you mean reliability, consistency -- is that correct? $\endgroup$
    – NAK
    Commented Oct 31, 2016 at 19:53
  • 2
    $\begingroup$ Much of what can be said about the coefficient of variation (e.g. stats.stackexchange.com/questions/118497/…) carries over simply to this measure. In particularly the measure is most useful and easiest to think about when loosely IQR is proportional to Med, so that their ratio is natural for the data concerned. The biggest pitfall, regardless of the robustness correctly advertised here, is that the measure is either unstable or meaningless if the median is very small (especially if zero!) or negative. $\endgroup$
    – Nick Cox
    Commented Oct 31, 2016 at 19:54
  • $\begingroup$ @NAK, yes, & also that it would be possible to have a reasonably strong signal in the inevitable noise. $\endgroup$ Commented Oct 31, 2016 at 19:55

Yes, the "Quartile based CV" [QCV in short] for data that have a skewed distribution may be defined and estimated as

QCV = [(Q3 - Q1) / Q2] x 100

wherein Q1, Q2, Q3 are the first, second and third quartiles of the distribution. The second quartile Q2 is the median of the distribution. The interpretation of QCV will be similar to CV = (SD/Mean) x 100 commonly used for data that have symmetric (not necessarily normal) distribution. The term QCV appears not to have been used in published literature as yet.


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