如题，使用三表连接，按照university和difficult_level分类，理解平均答题数量的含义。

select 
university,
difficult_level,
count(*)/count(DISTINCT device_id) AS avg_answer_cnt
from question_practice_detail
JOIN question_detail
USING(question_id)
JOIN user_profile
USING(device_id)
where university='山东大学'
group by university,difficult_level