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Q&A for electronics and electrical engineering professionals, students, and enthusiasts

SELECT * FROM 
  (SELECT Reputation,
    100.0 * SUM(COUNT(*)) OVER (ORDER BY Reputation) /
    SUM(COUNT(*)) OVER() AS CumPct
    FROM Users
    GROUP BY Reputation) AS x


-- Originally Forked from: smithmartin
-- https://meta.stackoverflow.com/questions/349805/where-do-i-stand-in-the-reputation-distribution
-- https://data.stackexchange.com/stackoverflow/query/676644/rep-distribution#resultSets

fork query

edited jun 16 Adam.Gerard

created may 27 17 smithmartin