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