# select university, difficult_level, # round(count(q.question_id)/count(distinct q.device_id), 4) as avg_answer_cnt # from user_profile u, question_practice_detail q, question_detail qd # where u.device_id = q.device_id and q.question_id = qd.question_id # group by # u.university, qd.difficult_level; select university, difficult_level, round(count(q.question_id)/count(distinct q.device_id), 4) as avg_answer_cnt from user_profile u join question_practice_detail q on u.device_id = q.device_id join question_detail qd on qd.question_id = q.question_id group by u.university, qd.difficult_level;