SELECT
    up.university,
    difficult_level,
    COUNT(qpd.result) / COUNT(distinct qpd.device_id) avg_answer_cnt
FROM user_profile up
JOIN question_practice_detail qpd
ON up.device_id = qpd.device_id
JOIN question_detail qd
ON qpd.question_id = qd.question_id
WHERE university = "山东大学"
and qpd.result IS NOT NULL
GROUP BY difficult_level;