** 统计每个用户的平均刷题数**

select university,difficult_level,round(count(question_detail.question_id)/count(distinct(question_practice_detail.device_id)),4) as avg_answer_cnt

from user_profile

join question_practice_detail

on user_profile.device_id = question_practice_detail.device_id

join question_detail

on question_practice_detail.question_id = question_detail.question_id

where university='山东大学'

group by university,difficult_level