SELECT qd.difficult_level,(
SUM(CASE WHEN qpd.result='right' THEN 1
    ELSE 0
    END
)/COUNT(qpd.result)
) correct_rated
FROM user_profile up,
     question_practice_detail qpd,
     question_detail qd
WHERE     up.university='浙江大学'
      AND up.device_id=qpd.device_id
      AND qpd.question_id=qd.question_id
GROUP BY  qd.difficult_level
order by correct_rated asc