<html><head><style type="text/css"><!-- DIV {margin:0px;} --></style></head><body><div style="font-family:times new roman, new york, times, serif;font-size:12pt"><DIV>Just to clarify this, for the nth "permeantile", I think you'd weight each point on the n side (100-n)^2 and on the (100-n) side it would be n^2 .<BR></DIV>
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<B><SPAN style="FONT-WEIGHT: bold">From:</SPAN></B> Toby Pereira <tdp201b@yahoo.co.uk><BR><B><SPAN style="FONT-WEIGHT: bold">To:</SPAN></B> electorama list <election-methods@lists.electorama.com><BR><B><SPAN style="FONT-WEIGHT: bold">Sent:</SPAN></B> Sat, 9 July, 2011 0:19:30<BR><B><SPAN style="FONT-WEIGHT: bold">Subject:</SPAN></B> [EM] Median-based Proportional Representation<BR></FONT><BR>
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<DIV>*I was thinking about how you would calculate permeantiles. In a uniform distribution between 0 and 1, the 25th permeantile would be 0.25. If you weight the averages of each side 3 to 1 in favour of the smaller side of the permeantile (0 to 0.25), and average these, then you get 0.25. (3*0.125 + 1*0.625) / 4 = 0.25. So for the 10th permeantile, you have (9*0.05 + 1*0.55) / 10 = 0.1 and so on. I imagine this would work for non-uniform distributions too. (Sorry for going off topic.)</DIV></DIV></DIV></DIV></div></body></html>