select university,
        count(question_id)/count(distinct t1.device_id) as avg_answer_cnt
        from user_profile t1 join question_practice_detail t2
        on t1.device_id = t2.device_id
        group by university
        order by university;

count()聚合函数计数；

distinct 前缀去重；

from...join...on...表连接；

group by分组；

order by排序。