select university,difficult_level,count(qpd.question_id) / count(distinct qpd.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 university = '山东大学'
group by university,difficult_level;