select university , round(count(question_id) / count(distinct(u.device_id)),4) as avg_answer_cnt from user_profile as u join question_practice_detail as q on q.device_id = u.device_id group by university order by university asc 先看结果的第一行要什么，然后select，这道题的重点是求平均数（注意审题），后面就是常规的join，记得order和group
