Percentile of the arithmetic mean of a lognormal distribution The arithmetic mean of a lognormal distribution is greater than the geometric mean (which equals the median). So its percentile is greater than 50%. But how much greater? It depends on the geometric standard deviation, which defines the asymmetry of the distribution.
What function defines the percentile of the arithmetic mean (of a lognormal distribution) from the geometric standard deviation?
 A: It's easiest to work on the log-scale, with a normal distribution.
The log of the mean of a lognormal in the usual parameterization is $\mu+\frac12\sigma^2$. Its z-value is therefore $\frac12\sigma$, so I believe this quantile should be at the $\Phi(\frac12\sigma)$ point of the cdf.
A little more formally, let $Y\sim \text{logN}(\mu,\sigma^{2})$, then:
$$\begin{align}P(Y\leq E[Y]) &= P(Y\leq e^{\mu+\frac12 \sigma^2})\\
 &= P(\log(Y)\leq\mu+\frac12 \sigma^2)\\
 &= P(\frac{\log(Y)-\mu}{\sigma}\leq\frac12 \sigma)\\
 &= P(Z\leq\frac12 \sigma)\\
 &= \Phi(\frac12 \sigma)\end{align}$$
e.g. at $\sigma=0.2$, the mean is at about the 54th percentile.
Here's a quick simulation in R as a check on my work -
 nsim = 1000000
 mu = 0
 sig = 0.2
 mnln = exp(mu+sig^2/2)
 y = rlnorm(nsim,mu,sig)
 c(sim = mean( y < mnln ), exact = pnorm( sig/2 ) )
       sim     exact 
 0.5392810 0.5398278 

Note, however, that $\sigma$ here is the standard deviation of the natural logs of the lognormal random variable. The geometric standard deviation equals $\exp(\sigma)$ (https://en.wikipedia.org/wiki/Geometric_standard_deviation#Relationship_to_log-normal_distribution).
So now if $\sigma_g = \exp(\sigma)$ then $\Phi(\frac12 \sigma)=\Phi(\frac12 \log(\sigma_g))$, where "log" is the natural logarithm. This equation computes the quantile; multiply by 100 to obtain the percentile, graphed below.

