Finding average / median time duration in a dataset - time

Question

I am using BigQuery and doing a capstone project for a course which needs us to analyze the data for a fictional cycling company.
Of the data that we're given, we're given the start and end time of the trips per month including date, hour, minute and second. I have the data in SQL with TIMESTAMP type for started_at and ended_at and TIME type for trip_duration
I would like to find the average and median trip per month for the data through SQL.
I was able to find the max and min trip, however I could not use simply AVG function to find the average trip duration.
What would be the best way to find the average and median times for trips?
I tried converting the duration into minutes by :
SELECT
ended_at, started_at, (ended_at-started_at)*1440,
FROM
`case-study-367714.case_study.yearly_data`
This gave the following result :
but this does not make sense as the first row is supposed to be 1 hr 26 minutes or 86 minutes, but it is showing 2064 minutes.

Answer 1

Consider the approach below:
with sample_data as (
select timestamp("2022-07-14 21:31:00") as ended_at, timestamp("2022-07-14 20:05:00") as started_at
union all select timestamp("2022-07-12 22:14:00") as ended_at, timestamp("2022-07-12 21:25:00") as started_at
union all select timestamp("2022-05-28 23:31:00") as ended_at, timestamp("2022-05-28 22:38:00") as started_at
union all select timestamp("2022-05-11 15:59:00") as ended_at, timestamp("2022-05-11 14:26:00") as started_at
union all select timestamp("2022-08-19 17:31:00") as ended_at, timestamp("2022-08-19 16:43:00") as started_at
union all select timestamp("2022-05-03 16:45:00") as ended_at, timestamp("2022-05-03 15:59:00") as started_at
union all select timestamp("2022-08-04 21:59:00") as ended_at, timestamp("2022-08-04 21:22:00") as started_at
union all select timestamp("2021-10-18 15:52:00") as ended_at, timestamp("2021-10-18 14:45:00") as started_at
union all select timestamp("2022-08-20 17:06:00") as ended_at, timestamp("2022-08-20 16:28:00") as started_at
),
cte as (
select
*,
concat(extract(year from ended_at),"-" ,extract(month from ended_at)) as month_date,
timestamp_diff(ended_at,started_at,minute) as duration_minutes,
from sample_data
)
select
month_date,
duration_minutes,
avg(duration_minutes) over (partition by month_date) as average_duration_per_month,
percentile_cont(duration_minutes, 0.5) over () as median
from cte
Output:

Answer 2

Thank you Ricco. I used a bit different code than what you had posted and was able to get the answer that I wanted for average trip duration, however the median is still giving me issues.
I used :
WITH dataset AS
(
SELECT
started_at,
ended_at,
member_casual,
timestamp_diff(ended_at, started_at, MINUTE) as Minute_Trip_Duration,
EXTRACT(MONTH FROM started_at) AS month,
FROM
case-study-367714.case_study.yearly_data
)
select
month,
member_casual,
avg(Minute_Trip_Duration) AS average_trip_duration,
from dataset
GROUP BY month,member_casual
Using this code I was able to get the following data. I was able to get average data for each month by whether the rider is a member or a casual rider:
enter image description here
The only issue is that if I enter the median code to it like below, I get an error message saying "SELECT list expression references column Minute_Trip_Duration which is neither grouped nor aggregated"
WITH dataset AS
(
SELECT
started_at,
ended_at,
member_casual,
timestamp_diff(ended_at, started_at, MINUTE) as Minute_Trip_Duration,
EXTRACT(MONTH FROM started_at) AS month,
FROM
case-study-367714.case_study.yearly_data
)
select
month,
member_casual,
avg(Minute_Trip_Duration) AS average_trip_duration,
percentile_cont(Minute_Trip_Duration,0.5) OVER () AS Median_Trip_Duration
from dataset
GROUP BY month,member_casual