SELECT uu university,dl difficult_level,avg(cc) avg_answer_cnt FROM(SELECT u.university uu,u.device_id,qpd_qd.d_lev dl,count(*) cc FROM user_profile u JOIN ( SELECT qpd.device_id d_id,qpd.question_id q_id,qd.difficult_level d_lev FROM question_practice_detail qpd JOIN...