select
    university,
    difficult_level,
    round(
        count(qpd.device_id) / count(distinct qpd.question_id),
        4
    ) as avg_answer_cnt
from
    user_profile up,
    question_practice_detail qpd,
    question_detail qd
where
    up.device_id = qpd.device_id
    and qpd.question_id = qd.question_id
    and university = '山东大学'
group by
    difficult_level;