No. P(max(x_i)>z)=1-P(max(x_i)<z)=1-(G1(X<z))^n


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送交者: 湘女 于 2007-09-27, 15:55:44:

回答: correct, so we are worse than we thought! :) 由 chouqilozi 于 2007-09-27, 15:26:47:

=> larger n produces lager P(max(x_i)>z).




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