select university, difficult_level, count(qpd.question_id) / count(distinct qpd.device_id) as avg_answer_cnt from question_practice_detail as qpd inner join user_profile as up on up.device_id=qpd.device_id and up.university="山东大学" inner join question_detail as qd on qd.question_id=qpd.question_id group by difficult_level;