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I have a set of temperature datasets all from different products. I wish to calculate the variation in the products. I have tried using coefficient of variation. But because the temperature data varies from say 20-40 deg C, the mean would be high and standard deviation low, the coefficient of variation is very low. I wish to know an alternative in this case to give a proper representation.

It would be great if you can help.

Ex. 27.2 30.4 30.4 31.0 31.9 35.2
Mean:31.0;
SD=2.6; CV=8.4

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  • $\begingroup$ Dear Keerthi, maybe you can explain why you think the fact that the coefficient of variation is low is a problem? Also, usually, the coefficient of variation is defined as $\dfrac{\sigma}{\mu}$, so in your example would be $\dfrac{2.6}{31} = 0.083$. For some reason, you seem to use that value multiplied by 100. Also, the wikipedia page on coefficient of variation cautions using them with interval data, the example use temperatures. $\endgroup$
    – Antoine Vernet
    Commented Aug 19, 2016 at 8:09
  • $\begingroup$ Thank you Vernet! yes i am multiplying it by 100 to just give it as a percentage. I have used the same method to calculate the CV for precipitation as well.. and in my case the CV for precipitation should be lower than that for temperature. Which I am unable to achieve. $\endgroup$
    – Keerthi
    Commented Aug 20, 2016 at 3:47

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Please refer to CV wiki - https://en.wikipedia.org/wiki/Coefficient_of_variation#Alternative

First, CV may not be relevant for Celsius scale as it takes both positive & negative values. You can convert it to Kelvin & compute.

There are alternatives provided on the same page

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