select up.university,qd.difficult_level,round(count(qpd.device_id)/count(distinct(qpd.device_id)),4) as avg_answer_cnt
from user_profile up
inner join question_practice_detail qpd
on up.device_id = qpd.device_id
inner join question_detail qd
on qd.question_id = qpd.question_id
where up.university = '山东大学'
group by qd.difficult_level;