I have data that includes ~20 groups with descriptive stats of median, count, and mean for each of the groups. I know that the means are heavily skewed, and I don't have access to the underlying data to remove outliers.

What I want to do essentially is find a weighted average of the medians, so I can visualize the relative importance of each of the categories, while using only median and count.

What I had done is just take the median * count for each category, then divide by the sum product of median * count for all categories to get a percentage weighting.

Is this a "legit" statistical approach? Is there a more direct, or at least easier to explain approach to this.

The second step would be to try to apply the weights to the median of all the data points. So if the median is lets say 8, could I apply the weighting from the above approach to this figure of 8. e.g. group 1 has a 20% weight. I can somewhat say then that group 1's "share" of the grand total median is 1.6. Is this feasible?

I understand I am really just trying to find a weighted average, but I can't use means and I only have the summary data points, not raw data.

  • 2
    $\begingroup$ You might get better answers if you can give descriptive statistics for three or four of your 20 groups, to give some idea of sample sizes, degree of skewness, etc. What is your purpose in trying to get the overall median of the combined groups? Do you have any idea from what family of distributions the 20 groups were randomly sampled? // My Answer below shows some of the fundamental difficulties by looking at three groups of equal sample sizes, so that weighting is not an issue. $\endgroup$
    – BruceET
    Jan 14, 2021 at 5:44

1 Answer 1


Consider three samples (groups) a,b, and c, each of size $n=20$ from exponential distributions with different rates $\lambda_i,$ where means are reciprocal rates $\mu_i = 1/\lambda_i$

We can find the mean of the grand sample of all sixty observations by averaging the the three group means (weighted average if sample sizes differed), but it is not possible to find the median of the grand sample from the medians of the three groups.

Three groups of equal size:

a = rexp(20, 1/2); length(a); mean(a); median(a)
[1] 20
[1] 1.825926
[1] 1.322669

b = rexp(20, 1/10); length(b); mean(b); median(b)
[1] 20
[1] 9.402898
[1] 4.726509

c = rexp(20, 1/50); length(c); mean(c); median(c)
[1] 20
[1] 43.62316
[1] 38.04221

Grand sample x (possible because all 60 observations are available).

x = c(a,b,c); length(x); mean(x); median(x)
[1] 60
[1] 18.284    # mean of grand sample
[1] 4.40221   $ median of grand sample

Finding mean of x from group means:

[1] 18.284    # matches mean of grand sample above

Failure to find median of x from group medians:

[1] 14.69713  # not equal to 4.40221
[1] 4.726509  # median of (b); not equal to 4.40221

Of the failed attempts, the median of group medians comes closest to the median of the grand sample. If samples were of different sizes you might try to find the 'median' group by some criterion and use its median. [Here it seems obvious to use b because it has 20 observations "above" and 20 observations "below" (ignoring some overlaps).]

boxplot(a,b,c,x, col=c("red","green3", "skyblue2", "grey"), 
        horizontal=T, varwidth=T)

enter image description here

Notes: (1) If you know the distributional form of the populations from which groups were sampled, you might be able to simulate each group. That is feasible here because the three exponential rates can be estimated as the reciprocals of group sample means. So y below might be a reasonable 'reconstruction' of the grand sample x and the sample median of y might be near to the sample median of x:

y = c(rexp(20, 1/mean(a)), rexp(20, 1/mean(b)), rexp(20, 1/mean(c)))
[1] 5.08311   # roughly the same as median of `x`: 4.402
boxplot(x,y, horizontal=T)

enter image description here

This is hardly a high-precision procedure, its values are usually in the interval $(2.4, 8.5),$ which does include 4.4.

 r1 = 1/mean(a); r2 = 1/mean(b);  r3 = 1/mean(c)
 h = replicate(10^5, median( c(rexp(20,r1), rexp(20,r2), rexp(20,r3)) ))
 quantile(h, c(.025,.975))
    2.5%    97.5% 
3.406778 8.586979 

(2) If group sample sizes are very large, the sample means do not differ much from group to group, and the ratio between sample means and sample variances is nearly the same in each group, then you may be able to approximate the sample median as a multiple of the weighted average.

Example: For exponential data the population median $\eta = \ln(2)\mu$ and this relationship between medians and means is approximately true for large samples.

w = rexp(10000, 1/5)
mean(w);  median(w);  mean(w)*log(2)
[1] 5.026372
[1] 3.515211
[1] 3.484016

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