impossible distribution statistics? I'm currently reviewing an article where authors presented distribution statistics that look erroneous to me. But I'm not able to find a way to ascertain it. The article presented results with a mean of 95% and standard deviation of 25%. Maximum value can't be more than 100% and minimum value can't be less than 0%. I don’t have the sample number.
I tried to generate lognormal random numbers with these statistics without success: 
# R Code
require("Runuran")
d1 <- urlnorm(n = 1000, meanlog = log(95), sdlog = log(25), lb = 0, ub = 100)

How can I generalize this conclusion whatever the distribution? 
 A: If you are working in the context that there are $n$ observations for which $\hat{\mu} = 0.95$ and $\hat{\sigma} = 0.25$, then I think you are correct. Playing around with sample sizes, the sample standard deviation is strictly less than 0.25 (can get around .22). The largest variance will be when the observations are either $1$'s or $0$'s. In that case, we need $\frac{19}{20}$ of the $n$ observations to be $1$ with the rest $0$, and $\frac{19}{20}\cdot 1 + \frac{1}{20}\cdot 0 = 0.95$.
For the standard deviation to be $0.25$, we need a variance of $0.0625$, and thus we need:
$$
\begin{align}
&=\sum_n\left(x_i - 0.95\right)^2 = 0.0625\\
&=\frac{1}{n - 1}\left(\frac{19n}{20}.05^2 + \frac{n}{20}.95^2\right) = 0.0625\\
&=\left(\frac{19n}{20}.05^2 + \frac{n}{20}.95^2\right) = 0.0625n - 0.0625\\
&= n\left(.95\cdot{.05}^2 + .05\cdot{.95}^2 - 0.0625\right) = -0.0625\\
&\Rightarrow n = 4.1666666
\end{align}
$$
Which is an non-integral number and impossible.
A: Without the original context I'm speculating, but I suspect this is an odds vs probability issue?  
If event A has a 95% chance of happening then what is 25% more likely to happen than event A?  That's the same as calculating something 75% less likely than A' or (1-.95)*(1-.25)= 0.0375 which means that .9625 is 25% more likely than .95 and is 1 standard deviation above your distribution's mean.
Hope this helps.
A: If you are reviewing for a journal and the data look incorrect, this is something you can raise in your written response.  Even if the data are not in error, it is still a problem because it appears to be an error or an incorrect / unconventional way of reporting data. I think you have done due diligence by thinking carefully about the paper and the results, but the burden is ultimately on the authors to make the results clear to the readership.
