SELECT t1.university , t3.difficult_level,
       ROUND(COUNT(t2.question_id)/COUNT(DISTINCT t2.device_id),4) AS avg_answer_cnt
FROM user_profile AS t1
JOIN question_practice_detail AS t2
ON t1.device_id=t2.device_id
JOIN question_detail AS t3
ON t2.question_id=t3.question_id
GROUP BY t3.difficult_level,t1.university
HAVING t1.university="山东大学";