select 
c.difficult_level as difficult_level
,round(sum(if(a.result = 'right',1 ,0)) / count(a.device_id), 4) as correct_rate
from question_practice_detail a
left join user_profile b
on a.device_id = b.device_id
left join question_detail c
on a.question_id = c.question_id
where university = '浙江大学'
group by c.difficult_level
order by correct_rate