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I have the 10th, 20th, 25th, 30th, 40th, 50th (I assume this is the same as the mean), 60th, 70th, 75th, 80th and 90th percentile values of a data set. This is for the distribution of salaries in the UK, which I assume is a Bell curve, but I might be wrong.

I would like to know the 99th. Is there a formula or an online tool for this?

Thank you and all the very best!

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    $\begingroup$ It is very unlikely that salaries across the UK are normally distributed - Table 5 of this ONS dataset shows quite visible right-skewness in full-time employee pay. In this case the mean salary will be higher than the median. $\endgroup$
    – B.Liu
    Apr 5, 2022 at 9:52
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    $\begingroup$ Without making any other distributional assumptions, it will be difficult to obtain the 99th percentile simply from other percentile values. One might want to consider other data sources. $\endgroup$
    – B.Liu
    Apr 5, 2022 at 9:56
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    $\begingroup$ The median (50th percentile) is substantially less than the mean. In terms of equivalenced household disposable income (not salaries), the mean is about $1.23$ times the median: see ons.gov.uk/resource?uri=/peoplepopulationandcommunity/… $\endgroup$
    – Henry
    Apr 5, 2022 at 10:26
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    $\begingroup$ In terms of the survey of personal incomes, the pre-tax income (not just salaries) of those at the 99th point seems to be of the order of 3 times that of those at the 90th point gov.uk/government/statistics/… You will not be able to deduce that from the data you have $\endgroup$
    – Henry
    Apr 5, 2022 at 10:30
  • $\begingroup$ I interpret the question as: "Given these eleven percentiles, what is a reasonable estimate for the 99th percentile?" -- and in that form, I think it is reasonable and interesting. $\endgroup$
    – Matt F.
    Apr 14, 2022 at 19:48

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If you assume your data come from a Gaussian distribution (bell curve) then yes, you can get the 99th percentile from others.

  • Set $\mu = q_{50\%}$. We are here using the fact that the median $q_{50\%}$ of a Gaussian is equal to it's mean $\mu$.

  • Set $\sigma = \frac{q_{75\%} - q_{25\%}}{1.34896}$. We are here using the fact that the interquartile range $q_{75\%} - q_{25\%}$ of a Gaussian is 1.349 times its standard deviation $\sigma$.

Then you can get any quantile using the formula

$$ q_{\alpha} = \mu + \sigma \times z_{\alpha}$$ where $z_\alpha$ is the corresponding quantile of a standard normal distribution, which you can find on tables online. For the 99-th quantile, you have $z_{0.99} = 2.326$.

However, if I may make a remark, I think the Gaussian distribution may not be the best fit for a salary distribution, which is usually skewed and heavy tailed. A Pareto distribution, for example, may be a better fit.

To get the Pareto quantiles you then must use the formula $$ q_\alpha = x_0 (1 - \alpha) ^{- \frac{1}{k}}\,\, .$$

  • First find the $x_0$ and $k$ parameters of the Pareto distribution. For instance: $$k = \frac{\log(0.75) - \log(0.25)}{q_{75\%} - q_{25\%}}\,\, \textrm{ and } \,\,x_0 = {q_{50\%}}\times {0.5 ^{\frac{1}{k}}}\,\, .$$

  • Then use the same formula again to get $q_{99\%} = x_0 0.01 ^ {-\frac{1}{k}}$ where you plug in the $x_0$ and $k$ values you found.

A log-normal distribution could also be a good candidate for the salaries distribution.

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    $\begingroup$ This is technically correct, but misleading without major caveats. Salaries would have to fit one of these distributions very closely indeed for the suggested extrapolations to be at all meaningful. If the OP wanted e.g. an interpolation to the 45th percentile, then a calculation like this would be a good approach. But the 99th percentile is well outside the range of the data points OP has — well into the upper tail of the distribution, which is very likely (in general, but especially in this case) to behave differently from other parts of the distribution. $\endgroup$
    – PLL
    Apr 5, 2022 at 20:12
  • $\begingroup$ I guess you are right. And even if data do come from a gaussian or Pareto distribution, the confidence interval for the 99% quantile could be huge as a tiny change in $\sigma$ or $k$ would yield a very large change in the quantile. $\endgroup$
    – Pohoua
    Apr 5, 2022 at 20:54
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    $\begingroup$ This interpretation of the question was posed and answered at stats.stackexchange.com/questions/48973. Comments and methods using all percentiles are posted at stats.stackexchange.com/questions/6022. Another solution is described at stats.stackexchange.com/questions/130156. $\endgroup$
    – whuber ♦
    Apr 5, 2022 at 23:24

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