select
university,
difficult_level,
round(count(qpd.id) / count(distinct u.id), 4) as avg_answer_cnt
from
user_profile u
join
question_practice_detail qpd on u.device_id = qpd.device_id
join
question_detail q on q.question_id = qpd.question_id
group by
u.university,
difficult_level
order by
u.university,
difficult_level

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