with tmp as (
    select TOTAL_VALUE as SA_MEDIAN, dense_rank() over (order by TOTAL_VALUE desc) as dense
    from CONVENIENCE
    where SURVEY_YEAR = 2019 and KIND_CODE = '100'
)
select case when max(dense) % 2 = 0 then max(dense) / 2 + 0.5 else max(dense) / 2 + 1 end as center
from tmp
;