SELECT university,
        qd.difficult_level,
        COUNT(qpd.question_id)/COUNT(DISTINCT qpd.device_id) AS avg_answer_cnt 
FROM user_profile up
JOIN question_practice_detail qpd
    USING(device_id)
JOIN question_detail qd
    USING(question_id)
WHERE university = "山东大学"
GROUP BY qd.difficult_level