I'm also trying to create a 75 percentile function and here is what I come with so far I wrote this query and I wasn't able to use the alias for the 0.75, all_count and so on. I'm just learning QuestDB and SQL in general so all this is new to me.

WITH ordered_fid AS(SELECT metric_name, values FROM vitals WHERE metric_name = 'FID' ORDER BY values),
calc_data AS (SELECT 
  metric_name,
  last(values) last_item,
  count() - 1 all_count,
  (0.75 * count() - 1) x,
  (0.75 * count() - 1) % 1 r,
  round(0.75 * count() - 1) i FROM ordered_fid)
SELECT 
*,
CASE
    WHEN i = all_count THEN last_item
    ELSE ordered_fid[i] + r * (ordered_fid[i + 1] - ordered_fid[i]) -> the issue is here
END as percentile
FROM calc_data;

I'm using the formula from below

# uses C=1 variant, like percentile_cont
# <https://en.wikipedia.org/wiki/Percentile#The_linear_interpolation_between_closest_ranks_method>        
# [1, 2, 3].percentile(0.75)

        sorted = map(&block).sort
        x = percentile * (sorted.size - 1)
        r = x % 1
        i = x.floor
        if i == sorted.size - 1
          sorted[-1]
        else
          sorted[i] + r * (sorted[i + 1] - sorted[i])
        end

SQL is not a procedural language . You can do this type of computation with e.g. window function

WITH ordered_fid AS ( SELECT values, row_number() over(order by values) rn FROM vitals WHERE metric_name = 'FID' ),
     cnt as ( SELECT count(*) val FROM vitals WHERE metric_name = 'FID' )
select * 
from 
cnt join
ordered_fid on  cnt.val*0.75 = rn

but it won't be as efficient as a dedicated function.

I understand and makes sense! Thank you very much for your precious help!