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

1、根据难度进行分组

2、选出山东大学

3、计算上面两个条件下的平均用户答题数