select
    difficult_level,
    sum(if(result='right',1,0))/count(result) as correct_rate
from
    user_profile u
    left join question_practice_detail qp on u.device_id = qp.device_id
    left join question_detail qd on qp.question_id = qd.question_id
where
    university = '浙江大学' and difficult_level is not null
group by difficult_level
order by correct_rate