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