SELECT
    difficult_level,
    sum(if(result="right",1,0))/count(*) as correct_rate
from user_profile a
left join question_practice_detail b
on a.device_id=b.device_id
inner join question_detail c
on b.question_id=c.question_id
where university="浙江大学"
group by difficult_level
order by correct_rate asc