select  university, difficult_level , round(count(q.question_id) / count(distinct q.device_id) ,4) as avg_answer_cnt
from user_profile u 
inner join question_practice_detail q 
on u.device_id=q.device_id
inner join question_detail q1
on q.question_id=q1.question_id
where u.university='山东大学' 
group by university, q1.difficult_level;