expected lowest value of 10 normally distributed values Consider 10 values that follow a standard normal distribution. What would you expect to be the lowest value?
I tried to simulate this problem in R. I basically just simulated 100000 standard normal distribution with 10 values and took the mean of the each lowest value.
> mean(replicate(100000,min(rnorm(10))))
[1] -1.536875

This corresponds to a probability to get a lower value of 
> pnorm(-1.536875)
[1] 0.06216196

I tried to reach these values analytically but I really have no idea how to approach this.
I thought about it for quite some time now and also tried to look it up. But I can't find a solutions to this simple problem. Probably just overlooking something obvious.
Someone that can help me?
 A: As has been clarified by Steck (1962), results contained in Ruben (1954) lead to the following expression for the expected value of the $1$st (minimum) and $n$th (maximum) order statistic, of a sample of $n$ i.i.d standard normal random variables: 
$$E[X_{(n)}] = -E[X_{(1)}] = \frac {n(n-1)}{2\sqrt{\pi}}\mathcal Z_{n-2}(0,...,0;\rho=1/3) $$ 
where $\mathcal Z_{n-2}() $ represents the multivariate standard normal cumulative distribution function of $n-2$ identical and equicorrelated standard normal random variables with common correlation coefficient $\rho=1/3$, all r.v's evaluated at $0$.
This is not an analytical solution, but this integral can be (and has been) evaluated.  For $n=10$ the value of the integral, as given in Gupta (1963) is
$\mathcal Z_{8}(0,...,0;\rho=1/3) = 0.06061$ and so we obtain
$$E[X_{(10)}] = -E[X_{(1)}] = \frac {10\cdot 9}{2\sqrt{\pi}}\cdot 0.06061 = 1.538799$$ 
Chen & Tyler (1999)  provide the following approximation to these expected values:
$$E[X_{(n)}] = -E[X_{(1)}] \approx \Phi^{-1}(0.5264^{1/n})$$
Where $\Phi^{-1}(\cdot)$ is the inverse (not the reciprocal) of the standard normal cumulative distribution function. They report that for $5<n<10$ this has less than $2$% approximation error, while for $n\ge 10$ the approximation error falls below $1$%. They derived this formula by the following approach: They asked "what would the expected value be if the distribution of the extreme order statistics was symmetric around the mean?" If it was symmetric the expected value would equal the value at the median, which is $\Phi^{-1}(0.5^{1/n})$. But since the distribution is not symmetric around the mean, they searched for a correction term, which lead them to the expression above. For $n=10$, this approximation gives
$$E[X_{(10)}] = -E[X_{(1)}] \approx \Phi^{-1}(0.5264^{1/10}) = \Phi^{-1}(0.937846) =1.536941$$
which has percentage approximation error  $-1.2074\cdot10^{-3}$, much less than $1$% $=10^{-2}$.
