select
u.university,
qd.difficult_level,
count(q.question_id) / count(distinct q.device_id) as avg_answer_cnt
from
user_profile as u,
question_practice_detail as q,
question_detail as qd
where
u.device_id = q.device_id
and q.question_id = qd.question_id
group by
university, difficult_level