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| The expected value of a Normal Order Statistic may be computed as the absolute value of | The expected value of the i-th Normal Order Statistic from a sample of size, n, may be computed as the absolute value of |
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| $$\Phi^{-1}(U(i))$$ | $$\Phi^{-1}(U(i,n))$$ |
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| where $$\Phi^{-1}$$ is the inverse Normal cumulative density function which converts a probability to a z-score. | where $$\Phi^{-1}$$ is the inverse Normal cumulative density function which converts a probability, such as U(i,n), to a z-score. |
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| where U(i) is given [ http://en.wikipedia.org/wiki/Normal_probability_plot here.] | where the U(i,n) are given as U(i) [http://en.wikipedia.org/wiki/Normal_probability_plot here.] U(i,n) = \left\{ \begin{tabular}{ll} 1- 0.5^{1/n} & i=1 \\ \frac{i-0.3175}{n+0.365} & i=2, \ldots, n-1 \\ 0.5^{1/n} & i=n \end{tabular} \right. |
Quick formula for the expected value of a Normal order statistic
The expected value of the i-th Normal Order Statistic from a sample of size, n, may be computed as the absolute value of
$$\Phi^{-1}(U(i,n))$$
where $$\Phi^{-1}$$ is the inverse Normal cumulative density function which converts a probability, such as U(i,n), to a z-score.
where the U(i,n) are given as U(i) [http://en.wikipedia.org/wiki/Normal_probability_plot here.]
U(i,n) = \left\{ \begin{tabular}{ll} 1- 0.5^{1/n} & i=1 \\ \frac{i-0.3175}{n+0.365} & i=2, \ldots, n-1 \\ 0.5^{1/n} & i=n \end{tabular} \right.