select u.university, ( count(q.question_id) / count(distinct u.device_id) ) as avg_answer_cnt from user_profile as u, question_practice_detail as q where u.device_id = q.device_id group by u.university order by u.university
select u.university, ( count(q.question_id) / count(distinct u.device_id) ) as avg_answer_cnt from user_profile as u, question_practice_detail as q where u.device_id = q.device_id group by u.university order by u.university