select
   qd.difficult_level,
   sum(if(result = 'right',1,0))/count(result) as correct_rate
from
    question_practice_detail as upd
    left join user_profile as up on up.device_id = upd.device_id
    left join question_detail as qd on qd.question_id = upd.question_id
where
    university = '浙江大学'
group by
    qd.difficult_level
order by correct_rate asc