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select --TagName, --count(*) N, --avg(ViewCount) avg_views,
       -- http://stackoverflow.com/a/8846679/1438
       PERCENTILE_CONT(0.5) 
         WITHIN GROUP (ORDER BY ViewCount)
         OVER (PARTITION BY TagName) median_views     
from Posts p
     join PostTags on p.Id = PostId
     join Tags t on t.Id = TagId
where PostTypeId = 1
      and TagName in (##tag##)    
group by TagName, ViewCount

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