正确解法：

select university, 
difficult_level,
count(a.question_id) / count(distinct a.device_id) as avg_answer_cnt
from question_practice_detail as a
left join user_profile as b on a.device_id=b.device_id
left join question_detail as c on a.question_id=c.question_id
where university = '山东大学'
group by difficult_level

和上一道题的思路差不多，主要是学会了三个表的left join，这道题就一次做出来了，激动！