select difficult_level,
       sum(if(result='right',1,0))/count(*) correct_rate
from question_practice_detail a
left join question_detail b
on a.question_id = b.question_id
where device_id in (select device_id 
                    from user_profile 
                    where university ='浙江大学')
group by difficult_level
order by correct_rate