SELECT 
    u.university,
    qd.difficult_level AS difficult_level,
    COUNT(qpd.question_id)/COUNT(DISTINCT qpd.device_id) AS avg_answer_cnt
FROM 
    user_profile u
JOIN 
    question_practice_detail qpd ON u.device_id = qpd.device_id
JOIN 
    question_detail qd ON qpd.question_id = qd.question_id
GROUP BY 
    u.university, qd.difficult_level;