Estimate population mean from "best of N" samples If I have a data set for which I know all measurements represent the largest of N observations, is there a good method for estimating the mean of all observations? So for example if N=10 and I have 3 data points
234
351
299

Then I know that there were 30 original data points in the source population and 9 of them were smaller than 234, 9 were smaller than 351, and 9 were smaller 299.
The goal is to estimate the original population mean/variance from this biased sample. It seems like it should be possible since the bias is known, but I'm having trouble convincing myself of what the relationship between the sample variance and the population variance is. I think the sample mean should be approximating the 90th percentile boundary for the population. Assuming that I have that right, if I knew the population variance I'd then be able to estimate the mean, but I'm not quite sure how to think about the relationship of the (biased) sample variance to the (unbiased) population variance.
I'm willing to assume that the source population is normally distributed etc, though a method for testing for normality of the source based on the sample would be good to have as well.

Here is a moment-matching estimate assuming normality:
For a standard normal distribution, the best-of-ten order statistic has mean $10\int_{-\infty}^\infty \Phi(x)^9\, \phi(x)\,x\, dx = 1.54$ and variance $10\int_{-\infty}^\infty \Phi(x)^9\, \phi(x)\,x^2\, dx-1.54^2=.344$
So for a normal distribution $N(\mu,\sigma)$, the best-of-ten order statistic has mean $\mu+1.54\sigma$ and variance $.344 \sigma^2$.
In this sample of best-of-tens, the mean is $\sim 295$ and the variance is $\sim 3436$.
So by solving two equations for $\mu$ and $\sigma$, we can estimate a parent distribution with $\mu\sim 141$, $\sigma\sim 100$.
