select  max(my_column) as [my_column], quartile
from    (select my_column, ntile(4) over (order by my_column) as [quartile]
         from   my_table) i
--where quartile = 2
group by quartile

This will give you the median and interquartile range in one fell swoop. If you really only want one row that is the median then uncomment the where clause.

When you stick that into an explain plan, 60% of the work is sorting the data which is unavoidable when calculating position dependent statistics like this.

My original quick answer was:

select  max(my_column) as [my_column], quartile
from    (select my_column, ntile(4) over (order by my_column) as [quartile]
         from   my_table) i
--where quartile = 2
group by quartile

This will give you the median and interquartile range in one fell swoop. If you really only want one row that is the median then uncomment the where clause.

When you stick that into an explain plan, 60% of the work is sorting the data which is unavoidable when calculating position dependent statistics like this.

I've amended the answer to follow the excellent suggestion from Robert Ševčík-Robajz in the comments below:

;with PartitionedData as
  (select my_column, ntile(10) over (order by my_column) as [percentile]
   from   my_table),
MinimaAndMaxima as
  (select  min(my_column) as [low], max(my_column) as [high], percentile
   from    PartitionedData
   group by percentile)
select
  case
    when b.percentile = 10 then cast(b.high as decimal(18,2))
    else cast((a.low + b.high)  as decimal(18,2)) / 2
  end as [value], --b.high, a.low,
  b.percentile
from    MinimaAndMaxima a
  join  MinimaAndMaxima b on (a.percentile -1 = b.percentile) or (a.percentile = 10 and b.percentile = 10)
--where b.percentile = 5

This should calculate the correct median and percentile values when you have an even number of data items. Again, uncomment the final where clause if you only want the median and not the entire percentile distribution.