select t3.difficult_level,sum(case when t2.result="right" then 1 else 0 end)/count(t2.question_id) as correct_rate from user_profile t1 join question_practice_detail t2 on t1.device_id=t2.device_id join question_detail t3 on t2.question_id=t3.question_id where university="浙江大学" g...