SELECT difficult_level,SUM(IF(result='right',1,0))/COUNT(*) AS correct_rate
FROM user_profile AS up
INNER JOIN question_practice_detail AS qpd
INNER JOIN question_detail AS qd
ON up.device_id=qpd.device_id AND qpd.question_id=qd.question_id
WHERE university='浙江大学'
GROUP BY difficult_level
ORDER BY correct_rate