select u.university,qq.difficult_level,
round(count(q.question_id)/count(distinct q.device_id),4) as avg_answer_cut

from user_profile u
join question_practice_detail q
on u.device_id = q.device_id
join question_detail qq
on q.question_id = qq.question_id
where u.university ='山东大学'
group by qq.difficult_level

和上一题没啥区别，只是加了山东大学的限定