select u.university, count(question_id) / count(distinct u.device_id) as avg_answer_cnt from user_profile u right join question_practice_detail q on u.device_id = q.device_id group by u.university

刚开始做的时候少些了distinct，后来发现要和question_p_detail表里面的device_id相除，一直除以了user_profile表里的device_id，所以忘了写distinct

做题的时候要保持清醒头脑..