SELECT u.university,e.difficult_level,
(COUNT(q.question_id) / COUNT(DISTINCT(q.device_id))) AS avg_answer_cnt
FROM user_profile AS u
JOIN question_practice_detail AS q
ON u.device_id = q.device_id
JOIN question_detail AS e
ON e.question_id = q.question_id
WHERE u.university = '山东大学'
GROUP BY e.difficult_level