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