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 up 
on up.device_id=qpd.device_id

left join question_detail qd
on qd.question_id=qpd.question_id

where university="山东大学"
group by difficult_level

自己独立分析逻辑做出的sql题