就三表联结，没啥好说的。。。

select c.difficult_level
      ,sum(if(a.result= 'right',1,0))/count(c.question_id) 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 b.university = '浙江大学'
group by c.difficult_level
order by correct_rate