select 
	up.university,
	qd.difficult_level,
	round(count(qpd.question_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 qpd.question_id=qd.question_id
where up.university='山东大学'
group by qd.difficult_level
order by qd.difficult_level