select university, difficult_level, 
       round(count(t2.question_id) / count(DISTINCT t1.device_id),4)avg_answer_cnt
from user_profile t1
inner join question_practice_detail t2
on t1.device_id = t2.device_id
inner join question_detail t3
on t2.question_id = t3.question_id
group by university, difficult_level
having university = '山东大学'