Return 30% of results?

[Nick <nboutelier@xxxxxxxxx>]

Is there a way to return a percentage of the rows found? I tried
window functions but get an error...

ERROR:  syntax error at or near "OVER"

SELECT id, cume_dist FROM (
  SELECT id, cume_dist() OVER (ORDER BY id) FROM employees
) s
WHERE cume_dist < 0.3

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