# Parametrising LogNormal distribution knowing only sample mean, and 90th percentile

#### thatdannyt

##### New Member
I'm a little stuck, and hope someone will be able to explain this to me.

I have sample mean, and value of the 90th percentile of the sample. If distribution is assumed to be LogNormal, is there a way of estimating parameters of mu and sigma of that LogNormal distribution?

I'm trying to de-construct work that was previously done by someone else, and their solution is this:

(1) Prob=NORMDIST((LN(percentile)-mu)/sigma,0,1,1)
(2) sigma=SQRT(2*(LN(mean)-mu))
(3) mu=[found iteratively using solver]

Solver changes value of (3) mu, trying to equate (1) to 90% (after all, the 90th percentile). And this does seem to work (as in, if you plug the sigma and mu parameters, statistics look reasonable), however I'm unsure how formula for (2) sigma came to be.

For example, if we assume that sample mean =30, and 90th percentile is 93, the parameters are
(2) sigma=1.30,
(3) mu=2.85,

while (1) Probability=90.3%

What am I missing?

Cheers
Dan

#### BGM

##### TS Contributor
Re: Parametrising LogNormal distribution knowing only sample mean, and 90th percentil

I think you are just equating the sample quantile to the population quantile, and equating the sample mean to the population mean, similar to method of moment.

Let $$z_{\alpha}$$ be the number satisfying $$\Pr\{Z > z_{\alpha}\} = \alpha$$ where $$Z \sim \mathcal{N}(0, 1)$$

Then you equate $$X_{((1-\alpha) n)} = e^{\mu + \sigma z_{\alpha}}$$

and $$\bar{X} = e^{\mu + \frac {\sigma^2} {2}}$$