select Ques.difficult_level as difficult_level ,sum(IF(Ques.result = 'right',1,0))/count(Ques.result) as correct_rate
from user_profile A
left join (select B.device_id,	B.question_id,	B.result, C.difficult_level from 
question_practice_detail B
left join  question_detail C
on B.question_id= C.question_id) as Ques
on A.device_id = Ques.device_id
where A.university = "浙江大学" and Ques.difficult_level IN('easy','medium','hard')
group by Ques.difficult_level
order by  correct_rate ASC