Q&A for people living abroad on a long-term basis
SELECT TagName AS Tags, AVG(Age * 1.0) AS [Average Answerer's Age], COUNT(*) AS [Number of Answerers] FROM ( SELECT DISTINCT pa.OwnerUserId, pt.TagId FROM Users u JOIN Posts pa ON pa.OwnerUserId = U.Id JOIN Posts pq ON pq.Id = pa.ParentId JOIN PostTags pt ON pt.PostId = pq.Id ) ut JOIN Tags t ON t.Id = ut.TagId JOIN Users u ON u.Id = ut.OwnerUserId WHERE u.Age BETWEEN 10 AND 90 GROUP BY t.Id, t.TagName HAVING COUNT(*) > 1000 ORDER BY 2 DESC