select university, count(b.question_id)/count(distinct(b.device_id)) as avg_answer_cnt from user_profile a join question_practice_detail b where a.device_id = b.device_id group by university
select university, count(b.question_id)/count(distinct(b.device_id)) as avg_answer_cnt from user_profile a join question_practice_detail b where a.device_id = b.device_id group by university