Is it valid to use mean length ($h$) and mean weight ($w$)) from a given population to calculate mean Body Mass Index ($BMI = \frac{w}{h^2}$) for that population?

  • $\begingroup$ The question is not that abstract (per @JoeTaxpayer's comment): typical real-world BMIs (15.0-30.0) are already going to be rounded anyway to 1dp, so if the error in mean BMI due to using the ratio of means is ~ 0.05-0.1 it's mathematically negligible (<1%); for most real-world distributions of (height, weight). We're not talking about Laurel and Hardy... $\endgroup$
    – smci
    Commented May 8, 2019 at 12:17

3 Answers 3


Mathematically, it's not the case that these are necessarily close. It would work if it was the case that $E(Y/X^2) = E(Y)/E(X)^2$ but this is false in general and in some particular situations it might be quite far out.

However, for a fairly realistic set of bivariate height and weight data it looks like the impact will be small.

For example, consider the model for US adult male height and weight in Brainard and Burmaster (1992) [1]; this model is a bivariate normal in height and log(weight), which fits the height-weight data pretty well and makes it easy to get realistic simulations. A good model for females is a little more complicated, but I don't expect it would make all that much difference to the quality of the BMI approximation; I'm just going to do the males because a very simple model is quite good.

Converting the model there for male height and weight to metric and simulating 100,000 bivariate points in R before calculating individual BMIs and hence mean BMI, as well as calculating mean height on (mean weight)-squared, it turns out the result was that mean BMI was (to four figures) 25.21 and $\bar{h}/\bar{w}^2$ was 25.22, which looks pretty close.

Looking at the effect of varying the parameters, it looks like the impact of using the biased means-of-variables estimator for the women would probably be slightly larger but still not substantial enough that it's likely to be much of an issue.

Ideally something closer to whatever situation you want to use it for should be checked, but it's probably going to be pretty good.

So for a typical situation, it would seem it's unlikely to be much of a problem in practice.

[1]: Brainard, J. and Burmaster, D.E. (1992),
"Bivariate Distributions for Height and Weight of Men and Women in the United States",
Risk Analysis, Vol. 12, No. 2, p 267-275


It's not completely correct, but it will usually not make a huge difference.

For example, suppose your population has weights 80, 90 and 100kg, and is 1.7, 1.8 and 1.9m tall. Then the BMIs are 27.68, 27.78 and 27.70. The mean of the BMIs is 27.72. If you calculate the BMI from the means of the weights and heights, you get 27.78, which is slightly different, but should usually not make all that much of a difference.

  • $\begingroup$ Thanks a bunch for your answer! So this means that this method of computation would probably not be appropriate for any type of statistical analysis, correct? $\endgroup$ Commented May 7, 2019 at 10:39
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    $\begingroup$ Quite honestly, I would not go quite as far. Statistics is dealing with noisy data, anyway, and a slight imprecision in measuring heights or weights would dwarf the difference we have here. I'd recommend you simulate something along these lines with the data you are thinking about using, and then think about whether these tiny differences will truly have an impact on your statistical analysis. $\endgroup$ Commented May 7, 2019 at 10:45
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    $\begingroup$ "If you calculate the BMI from the means of the weights and heights, you get 27.78" But that's equal to the maximum of the BMIs! Getting the maximum instead of the mean seems like a big difference to me. $\endgroup$ Commented May 7, 2019 at 15:49
  • $\begingroup$ @Acccumulation - Stephan's point was dead on, but the choice of numbers didn't illustrate the situation well. BMI for "normal" is 18.5 to 24.9. The 3 BMIs in the example have a low to high range of .1. A 1/2% difference on this data is noise. A 5'8" man's BMI goes up from 27.4 to 27.5 as his weight rises from 180 to 181. 1lb is the difference between drinking a tall glass of water before or after getting on the scale. $\endgroup$ Commented May 7, 2019 at 21:01
  • $\begingroup$ @JoeTaxpayer: I think your point is that typical BMIs (15.0-30.0) are already going to be rounded anyway to 1dp, so if the error in mean BMI due to using the ratio of means is < 0.05 it's mathematically negligible. Correct? $\endgroup$
    – smci
    Commented May 8, 2019 at 9:06

Although I agree with the other answers that it is likely that this method will approximate the mean BMI, I would like to point out this is only an approximation.

I'm actually inclined to say you should not use the method you describe, as it is simply less accurate. It is trivial to calculate BMI's for each individual and then take the mean of that, giving you the real mean BMI.

Here I illustrate two extremes, where the means of weight and length remain the same, but the average BMI is actually different:

Using the following (matlab) code:

weight = [60, 61, 62, 100, 101, 102]; % OUR DATA
length = [1.5, 1.5, 1.5, 1.8, 1.8, 1.8;]; % OUR DATA
length = length.^2;
bmi = weight./length;
scatter(1:size(weight,2), bmi, 'filled');
legend('BMI', 'mean(bmi)', 'mean(weight)/mean(length)', 'Location','northwest');

We get: mean_bmi2

If we simply re-order the lengths, we get a different mean BMI while mean(weight)/mean(length^2) remains the same:

weight = [60, 61, 62, 100, 101, 102]; % OUR DATA
length = [1.8, 1.8, 1.8, 1.5, 1.5, 1.5;]; % OUR DATA (REORDERED)
... % rest is the same

mean bmi

Again, using real data it is likely that your method will approximate the real mean BMI, but why would you use a less accurate method?

Outside the scope of the question: It's always a good idea to visualise your data so you can actually see the distributions. If you notice certain clusters for example, you can also consider getting separate means for those clusters (e.g. separately for the first 3 and last 3 people in my example)

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    $\begingroup$ " It is trivial to calculate BMI's for each individual and then take the mean of that, giving you the real mean BMI." Yes, if you have the raw data. If all you have are summary statistics for a population, like the average height and weight, then deriving a "population BMI" from these is the best you can do, and the question of how bad an approximation that is is quite valid. $\endgroup$ Commented May 8, 2019 at 9:25
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    $\begingroup$ +1 Don't aggregate aggregates if you can get the same kind of answer aggregating raw data. There are answers and comments that in this case it makes little/no/slim/negligible difference, but don't do it. Learn and use healthy practices about data analysis, do it the right way. $\endgroup$
    – Stian
    Commented May 8, 2019 at 10:00

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