select university,difficult_level,round(count(qpd.question_id)/count(distinct qpd.device_id),4) avg_answer_cnt
from user_profile up
join question_practice_detail qpd on up.device_id=qpd.device_id
join question_detail qd on qpd.question_id=qd.question_id
group by university,difficult_level
having university='山东大学';