I'm working with wage data and would like to compute total aggregate wages earned for individual job classifications. The survey data I'm working from does not report aggregate wages, but it does include each job classification's total employment, 10th, 25th, 50th, 75th and 90th percentile wage estimates.

While I wouldn't be able to compute a precise total, I'd like to see if it's possible to multiply the employment estimates by the percentile values to approximate the total wages earned for each classification. I'd greatly appreciate any ideas or best practices for an approach to take.


The standard datasets which quote wage percentiles also provide the mean, and you can calculate the aggregate from the mean. E.g. see this example from the Bureau of Labor Statistics. So I'll restrict to the case where the mean is not also quoted.

If most of the profession’s income is earned by the people above the 90th percentile, you may not be able to estimate those incomes well, and then you won't have any good way to aggregate.

On the other hand, if you assume a form for the distribution, like the lognormal, you can make an educated guess. E.g.: Suppose the given quantiles are $q_{10}, \ldots, q_{90}$. You can fit the distribution by finding the $\mu$ and $\sigma$ which minimize

$$(F(q_{10})-.10)^2 + \cdots + (F(q_{90})-.90)^2$$

where $F$ is the cumulative distribution function for the lognormal distribution parameterized by $\mu$ and $\sigma$. This will have to be done numerically; it's the same as finding the $\mu$ and $\sigma$ which minimize

$$\left(\Phi\left(\frac{\ln(q_{10})-\mu}{\sigma}\right)-.10\right)^2 + \cdots + \left(\Phi\left(\frac{\ln(q_{90})-\mu}{\sigma}\right)-.90\right)^2$$

where $\Phi$ is the cdf for the standard normal. Once you have the minimizing $\mu$ and $\sigma$, you can estimate the mean wage as the mean of that lognormal distribution, and get the aggregate wages from there.

Taking the example of computer systems analysts at insurance carriers, from the BLS data, the percentiles are 55,390, 68,820, 86,790, 104,490, 125,690. This procedure suggests a lognormal distribution with $\mu$=11.35, $\sigma$=.315, and mean 89,691, which is not far from the actual mean reported.

  • $\begingroup$ This sum of squared deviations isn't the right one to minimize. A maximum likelihood estimate will work better (see stats.stackexchange.com/questions/34882) or, if you don't want to go to that effort, use a weighted sum of squares, recognizing that the variance of $F(q_p)$ ought to be proportional to $(p/100)(1-p/100)$ (using your notation). $\endgroup$ – whuber Apr 25 '20 at 0:44
  • $\begingroup$ @whuber, I appreciate MLE as an option, but if you’re going to use normative words like “right” and “better”, then I’d also appreciate some justification for them. $\endgroup$ – Matt F. Apr 25 '20 at 20:23
  • $\begingroup$ The variances of the terms differ, which is why a straight sum of squares is incorrect. It can be used as an approximation, but it's an inadmissible procedure. BTW, "inadmissible" may sound strong and "normative," but it's not: it is a technical term of art, and in this sense "better," although normative, is well-defined and justifiable. $\endgroup$ – whuber Apr 25 '20 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.