Assume that $X$ has a lognormal distribution and that the median of $X$ is $.5$ and its 98th percentile is $2.30$. Is it possible to find the mean and variance of $X$?
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percentile of lognormal and its first moments
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probability
statistics
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0If you know quantiles .5 and .98 for a _normal_ distribution, you can solve two equations in two unknowns to find its mean and variance. Taking logs is a monotone transformation, so you can find those two quantiles of normal. Then use well-known connections btw normal and lognormal moments to finish. – 2017-02-06
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Let $X\sim\log {\mathcal {N}}(\mu ,\,\sigma ^{2})$. The median is $e^{\mu}=0.5$, so $\mu=\log(0.5)$.
As $X$ is lognormal, then $Y=\log X$ has the normal distribution $\mathcal {N}(\mu ,\,\sigma ^{2})$. Then we calculate $$0.98=P(X<2.3)=P(\log X < \log2.3)=P(Y<\log2.3)=P\left(\frac{Y-\mu}{\sigma}<\frac{\log2.3-\mu}{\sigma}\right).$$ Thus, because $\frac{Y-\mu}{\sigma}\sim \mathcal{N}(0,1)$, we have $0.98=\Phi\left(\frac{\log2.3-\log0.5}{\sigma}\right)$, where $\Phi$ is the cdf of the standard normal distribution.
From that we conclude: $$\Phi^{-1}(0.98)=\frac{\log4.6}\sigma \implies \sigma = \frac{\log4.6}{\Phi^{-1}(0.98)}=\frac{\log4.6}{2.054}.$$
So, we have calculated the parameters of $X$, so we can easily calculate the mean and variance using $$EX=e^{\mu+\frac{\sigma^2}2},\quad VarX=(e^{\sigma ^{2}}\!\!-1)e^{2\mu +\sigma ^{2}}.$$