SELECT up.university as university, qd.difficult_level as 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 INNER JOIN question_detail as qd ON qd.question_id=qpd.question_id GROUP BY up.university, qd.difficult_level;