SELECT t3.difficult_level,
COUNT(CASE WHEN t2.result="right" THEN t2.question_id ELSE NULL END)/COUNT(t2.question_id) AS correct_rate
FROM user_profile AS t1
JOIN question_practice_detail AS t2
ON t1.device_id=t2.device_id
JOIN question_detail AS t3
ON t2.question_id=t3.question_id
WHERE t1.university="浙江大学"
GROUP BY t3.difficult_level
ORDER BY correct_rate;

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