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