select qd.difficult_level,
(sum(case when qpd.result="right" then 1 else 0 end) / count(up.answer_cnt))as "correnct_rate"
from user_profile up 
left join question_practice_detail qpd on up.device_id = qpd.device_id
left join question_detail qd on qpd.question_id = qd.question_id
where up.university = "浙江大学"
and qd.difficult_level is not null
group by qd.difficult_level
order by correnct_rate asc