select university ,difficult_level,count(qpd.question_id)/count(DISTINCT qpd.device_id) as avg_answer_cnt
from user_profile as u
inner join question_practice_detail as qpd
on u.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 difficult_level