select 
	qd.difficult_level as difficult_level, 
	sum(case when qpd.result = "right" then 1 else 0 end)/ count(qpd.device_id)
	as correct_rate
from user_profile as u

inner join question_practice_detail as qpd
on u.device_id = qpd.device_id

inner join question_detail as qd
on qpd.question_id = qd.question_id

where u.university = "浙江大学"
group by difficult_level
order by correct_rate

需要链接三个表。重点在于计算正确率，一道题答对的次数比总答题次数