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Given a normal distribution, how would I determine what the expected highest sample would be out of N samples?

Presently I'm doing some strange calculations that I think are incorrect; I'm solving (for x):

P(X > x) = 1/N

And stating that x is the expected highest (in reality, it's likely the expected minimum).

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Are you asking: What is the expected value of the sample maximum, in a sample of size $n$? –  wolfies Oct 15 '13 at 13:10

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For $N$ i.i.d. random variables, $\left(Pr(X \le x) \right)^N$ is the probability all $N$ values are less than or equal to $x$, so the solution to $$Pr(X \le x) = \sqrt[N]{\frac12}$$ will give you the median of the distribution of the maximum, while $$\int_{0}^{\infty} \left(1-\left(Pr(X \le x) \right)^N \right) \;dx - \int_{-\infty}^{0} \left(Pr(X \le x) \right)^N \;dx$$ will give you the mean of the distribution of the maximum.

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Somebody made a comment and then deleted it - this made me think about the mean again with a consequential edit –  Henry Oct 15 '13 at 15:16

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