题解 | #得分不小于平均分的最低分#

select date_format(start_time,'%Y%m') month , round(COUNT(DISTINCT uid,DATE_FORMAT(start_time,'%Y%m%d')) /count(DISTINCT uid) ,2) avg_active_days, COUNT(DISTINCT(uid))

from exam_record where year(submit_time)=2021 group by month