select university,difficult_level,round(count(question_id)/count(distinct device_id),4) as avg_answer_cnt
from
(select a.device_id,a.university,b.question_id,c.difficult_level
from user_profile a
left join question_practice_detail b on a.device_id = b.device_id 
left join question_detail c on b.question_id = c.question_id
where university = '山东大学' ) as t
group by university,difficult_level