Pregunta 1:Para el número total de viajes (incluyendo cada vez que se subió un pico)
SELECT t1.sex, AVG(t1.peak_count) AS average
FROM
(SELECT sex, COUNT(trip_id) AS peak_count
FROM climber c LEFT JOIN participated p ON c.name = p.name GROUP BY c.name, c.sex) t1
Por cada vez que se subió un pico ÚNICO:
SELECT t1.sex, AVG(t1.peak_count) AS average
FROM
(SELECT sex, COUNT(trip_id) AS peak_count
FROM climber c LEFT JOIN participated p ON c.name = p.name GROUP BY c.name, c.sex) t1
Pregunta 2:
SELECT P1.Name, P2.Name, COUNT(DISTINCT p1.trip_id) AS trips
FROM participated p1 INNER JOIN participated p2 ON p1.trip_id = p2.trip_id
WHERE p1.name > p2.name -- > instead of <> gets only one of the pairs
GROUP BY P1.Name, P2.Name
HAVING COUNT(DISTINCT p1.trip_id) > 0
ORDER BY trips DESC
Pregunta 3:
SELECT p.name, cl.when AS span_begin_date, DATEADD(day, 60, cl.when) AS span_end_date, count(c2.trip_id) AS peaks
FROM climbed cl LEFT JOIN
climbed c2 ON c2.when BETWEEN cl.when AND DATEADD(day, 60, cl.when)
GROUP BY p.name, cl.when, DATEADD(day, 60, cl.when)
HAVING COUNT(c2.trip_id) > 20
ORDER BY peaks
