select up.university, qd.difficult_level, 
round(count(upd.id)/count(distinct(up.id)),4) avg_answer_cnt
from user_profile up
    join question_practice_detail upd on up.device_id=upd.device_id and up.university='山东大学'
    join question_detail qd on upd.question_id=qd.question_id
group by up.university, qd.difficult_level;

第一步内连接,第二步过滤非山东大学数据,第三步分组合并,第四步计算平均值,第五步规范小数点