SELECT difficult_level,
sum(IF(result = 'right',1,0)) / COUNT(result) as correct_rate
from user_profile as up
right JOIN question_practice_detail as qpd
on up.device_id = qpd.device_id
RIGHT JOIN question_detail as qd
on qpd.question_id = qd.question_id
where up.university = '浙江大学'
group by difficult_level
order by correct_rate