# How can I calculate cut-off points from a normal distribution?

I'm trying to calculate the upper percentage points for the 0.99 percentile for samples drawn from a normal distribution, with a sample size of 500. How can I calculate the expected values for skewness, kurtosis, studentized range, which would be expected for a drawing of 500 values?

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When you say "calculate" do you want mathematical formulae, or a way of getting a numerical approximation? (And is this homework?) –  guest Apr 7 '12 at 20:46
en.wikipedia.org/wiki/Student_t –  Gschneider Apr 7 '12 at 20:48
no this is no homework :P. yes a formula would be nice, I'm somehow unsure how to integrate the sample size here. The objective is to know the approximative numerical value which can be expected, when 500 values are drawn.. –  PeriodicProgrammer Apr 7 '12 at 21:05
This question appears to contradict itself: are you drawing samples from a normal distribution or a Student t distribution? Which one? (PS: You're right: this site is a better choice for the question than the Mathematica site. Once you get your question straightened out here, if you need specific help with MMA in coding a solution you will know precisely what to ask over there as a follow-up.) –  whuber Apr 7 '12 at 21:08
I could do an Confidence interval but i'm somehow missing the standard deviation which would be needed to derive theorethically? I assume it to be close to zero for 500 values for the skewness, respectively to 3 for the kurtosis..? –  PeriodicProgrammer Apr 7 '12 at 21:27
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You can always simulate this kind of thing first, e.g. in R

set.seed(1)
samat <- matrix(rnorm(1000000), ncol=500)
means <- rowMeans(samat)
varis <- rowMeans((samat-means)^2)
skews <- rowMeans((samat-means)^3) / varis^(3/2)
kurts <- rowMeans((samat-means)^4) / varis^2  -  3

and find the quantiles

> quantile(means, probs=c(0.01, 0.5, 0.99) )
1%          50%          99%
-0.096977914 -0.001302285  0.109797211
>
> quantile(varis, probs=c(0.01, 0.5, 0.99) )
1%       50%       99%
0.8494985 0.9970259 1.1495887
>
> quantile(skews, probs=c(0.01, 0.5, 0.99) )
1%          50%          99%
-0.257665499 -0.002052328  0.260948808
>
> quantile(kurts, probs=c(0.01, 0.5, 0.99) )
1%        50%        99%
-0.4286435 -0.0320487  0.5399296

and while it is clear that you can probably only treat the first couple of decimal places as useful, it also shows that skewness of 0.60 would be extremely high. It also suggests that the "skewness 0.129" you quote is rather low for the 99th percentile; I make it nearer 0.35 or 0.36 for a sample of 250 and about 0.26 for a sample of 500.

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