Ponderate average MYSQL - mysql

Question

We have a little simulator of a tour-operator DB (MYSQL) and we are asked to get a Query that gives us the weighted avg of duration of the tours that we have.
https://en.wikipedia.org/wiki/Weighted_arithmetic_mean
Using subquery I got to this point where I have the days that each tour lasts and the weight of each tour from the total of tours, but I am stuck and don't know how to get the weighted avg from here. I know I have to use another select from the result I already got but I would appreciate some help.
SQLfiddle down here:
http://sqlfiddle.com/#!9/53d80/2
Tables and data
CREATE TABLE STAGE
(
ID INT AUTO_INCREMENT NOT NULL,
TOUR INT NOT NULL,
TYPE INT NOT NULL,
CITY INT NOT NULL,
DAYS INT NOT NULL,
PRIMARY KEY (ID)
);
CREATE TABLE TOUR
(
ID INT AUTO_INCREMENT NOT NULL,
DESCRIPTION VARCHAR(255) CHARACTER SET UTF8 COLLATE UTF8_UNICODE_CI
NOT NULL,
STARTED_ON DATE NOT NULL,
TYPE INT NOT NULL,
PRIMARY KEY (ID)
);
INSERT INTO TOUR (DESCRIPTION, STARTED_ON, TYPE) VALUES
('Mediterranian Cruise','2018-01-01',3),
('Trip to Nepal','2017-12-01',1),
('Tour in Nova York','2015-04-24',5),
('A week at the Amazones','2014-09-11',2),
('Visiting the Machu Picchu','2013-02-19',4);
INSERT INTO STAGE (TOUR, TYPE, CITY, DAYS) VALUES
(1, 1, 38254, 1),
(1, 2, 22460, 3),
(1, 2, 47940, 3),
(1, 2, 42600, 4),
(1, 3, 38254, 1),
(2, 1, 13097, 1),
(2, 2, 29785, 5),
(2, 3, 13097, 1),
(3, 1, 788, 2); ,
(3, 2, 48019, 6),
(3, 3, 788, 1),
(4, 1, 38254, 2),
(4, 2, 8703, 3);,
(4, 3, 38254, 4),
(5, 1, 10453, 1),
(5, 2, 32045, 5),
(5, 3, 10453, 2);
Query:
SELECT
AVG(TD.TOUR_DAYS) AS AVERAGE_DAYS,
COUNT(TD.TOUR_ID) AS WEIGHT
FROM
(
SELECT
TOUR.ID AS TOUR_ID,
SUM(DAYS) AS TOUR_DAYS,
COUNT(STAGE.ID) AS STAGE_DAYS
FROM
TOUR
INNER JOIN
STAGE
ON
TOUR.ID = STAGE.TOUR
GROUP BY
TOUR.ID
) AS TD
GROUP BY
TD.TOUR_DAYS
weigthed avg would be:
(1×7+1×8+2×9+1×12) / (1+1+2+1) = 9

Answer 1

Wheighted AVG can be calculated with SUM(value * wheight) / SUM(wheight). In your case:
SELECT SUM(AVERAGE_DAYS * WEIGHT) / SUM(WEIGHT)
FROM (
SELECT
AVG(TD.TOUR_DAYS) AS AVERAGE_DAYS,
COUNT(TD.TOUR_ID) AS WEIGHT
FROM
(
SELECT
TOUR.ID AS TOUR_ID,
SUM(DAYS) AS TOUR_DAYS,
COUNT(STAGE.ID) AS STAGE_DAYS
FROM
TOUR
INNER JOIN
STAGE
ON
TOUR.ID = STAGE.TOUR
GROUP BY
TOUR.ID
) AS TD
GROUP BY
TD.TOUR_DAYS
) sub
http://sqlfiddle.com/#!9/53d80/4
I'm not 100% sure, but it looks like the following query is doing exactly the same:
SELECT AVG(TOUR_DAYS)
FROM (
SELECT TOUR, SUM(DAYS) AS TOUR_DAYS
FROM STAGE
GROUP BY TOUR
) sub;
Or even without any subqueries:
SELECT SUM(DAYS) / COUNT(DISTINCT TOUR)
FROM STAGE;
That would mean, the requirement should be simplified to "Get average number of days per tour".