SELECT t3.difficult_level, SUM( result = 'right' ) / COUNT( t1.device_id ) AS correct_rate FROM question_practice_detail AS t1 LEFT JOIN user_profile AS t2 ON t1.device_id = t2.device_id LEFT JOIN question_detail AS t3 ON t1.question_id = t3.question_id WHERE t2.university = '浙江大学' GROUP BY...
