select

    university, difficult_level, count(qpd.question_id) / count(distinct qpd.device_id)

    from user_profile as up

    inner join question_practice_detail as qpd on up.device_id = qpd.device_id

    inner join question_detail as qd on qpd.question_id = qd.question_id

    group by university, qd.difficult_level having university = "山东大学";