和上道题一样，多了个 判断 学校

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