select university, difficult_level, count(qpd.question_id) / count(DISTINCT qpd.device_id) as avg_answer_cnt
from question_practice_detail qpd 
LEFT JOIN user_profile u ON qpd.device_id = u.device_id
LEFT JOIN question_detail qd ON qpd.question_id = qd.question_id
WHERE university = '山东大学'
GROUP BY difficult_level

FROM - ON - JOIN - WHERE - GROUP BY - WITH - HAVING - SELECT - DISTINCT - ORDER BY - LIMIT