select u.university, qd.difficult_level,
count(qpd.question_id) / count(distinct qpd.device_id) as avg_answer_cnt from question_practice_detail qpd join user_profile u on qpd.device_id = u.device_id join question_detail qd on qpd.question_id = qd.question_id group by university, difficult_level；

虽然运行通过，但是看别的大神还加了平均值精度：保留4位小数round(x, 4)。 所以加粗的地方可以修改成： round（count(qpd.question_id) / count(distinct qpd.device_id), 4) as avg_answer_cnt;

还是要记住不要少了distinct