An intermediate exercise in Mathematical statistics is to construct a (100-$latex \alpha$)% confidence interval for the mean of a random variable distributed according to the log normal law ($latex X\tilde{\ }LN\left( \mu,\sigma _{{}}^{2}\right)$.)

The mean is

$latex E\left( X \right)=\exp \left( \mu +\frac{\sigma _{{}}^{2}}{2} \right)$.

Then

$latex \log \theta =\mu +\frac{\sigma _{{}}^{2}}{2}$.

The MVUE is

$latex T\left( Y \right)=\bar{Y}+\frac{s_{{}}^{2}}{2}$

, where $latex Y=\log X,{{s}^{2}}=\frac{1}{n-1}\sum{{{\left( Y-\bar{Y} \right)}^{2}}}$.

The variance of $latex T\left( Y \right)$ is

$latex Var\left\{ T\left( Y \right) \right\}=\frac{s_{{}}^{2}}{n}+\frac{s_{{}}^{4}}{2\left( n-1 \right)}$.

Then evidently

$latex \frac{\bar{Y}+\frac{s_{{}}^{2}}{2}-\log \theta }{\sqrt{\frac{s_{{}}^{2}}{n}+\frac{s_{{}}^{4}}{2\left( n-1 \right)}}}\xrightarrow{d}Z$

, where $latex Z\tilde{\ }N\left( 0,1 \right)$.

To sum thinks up the confidence interval for  mean of X fixing  $latex \alpha=.05$ is

$latex \left( \exp \left\{ \bar{Y}+\frac{s_{{}}^{2}}{2}-1.96\sqrt{\frac{s_{{}}^{2}}{n}+\frac{s_{{}}^{4}}{2\left( n-1 \right)}} \right\},\exp \left\{ \bar{Y}+\frac{s_{{}}^{2}}{2}+1.96\sqrt{\frac{s_{{}}^{2}}{n}+\frac{s_{{}}^{4}}{2\left( n-1 \right)}} \right\} \right)$.