Is it possible to find the 99th percentile using other percentiles? 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!

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.
