select up.university,qd.difficult_level,COUNT(qp.question_id) / count(DISTINCT(up.device_id)) as avg_answer_cnt from user_profile as up, question_practice_detail as qp, question_detail as qd where up.device_id = qp.device_id AND qp.question_id =qd.question_id group by university,difficult_level having university = "山东大学"