From last several days I am working on various Denali Analytical functions and it is indeed really fun to refresh the concept which I studied in the school.
Earlier I wrote article where I explained how we can use PERCENTILE_CONT() to find median over here SQL SERVER – Introduction to PERCENTILE_CONT() – Analytic Functions Introduced in SQL Server 2012. Today I am going to ask question based on the same blog post. Again just like last time the intention of this puzzle is as following:
Learn new concept of SQL Server 2012
Learn new concept of SQL Server 2012 even if you are on earlier version of SQL Server.
On another note, SQL Server 2012 RC0 has been announced and available to download SQL SERVER – 2012 RC0 Various Resources and Downloads.
Now let’s have fun following query:
USE AdventureWorks
GO
SELECT SalesOrderID, OrderQty, ProductID,
PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY ProductID)
OVER (PARTITION BY SalesOrderID) AS MedianCont
FROM Sales.SalesOrderDetail
WHERE SalesOrderID IN (43670, 43669, 43667, 43663)
ORDER BY SalesOrderID DESC
GO
The above query will give us the following result:
The reason we get median is because we are passing value .05 to PERCENTILE_COUNT() function. Now run read the puzzle.
Puzzle:
Run following T-SQL code:
USE AdventureWorks
GO
SELECT SalesOrderID, OrderQty, ProductID,
PERCENTILE_CONT(0.9) WITHIN GROUP (ORDER BY ProductID)
OVER (PARTITION BY SalesOrderID) AS MedianCont
FROM Sales.SalesOrderDetail
WHERE SalesOrderID IN (43670, 43669, 43667, 43663)
ORDER BY SalesOrderID DESC
GO
Observe the result and you will notice that MidianCont has different value than before, the reason is PERCENTILE_CONT function has 0.9 value passed. For first four value the value is 775.1.
Now run following T-SQL code:
USE AdventureWorks
GO
SELECT SalesOrderID, OrderQty, ProductID,
PERCENTILE_CONT(0.1) WITHIN GROUP (ORDER BY ProductID)
OVER (PARTITION BY SalesOrderID) AS MedianCont
FROM Sales.SalesOrderDetail
WHERE SalesOrderID IN (43670, 43669, 43667, 43663)
ORDER BY SalesOrderID DESC
GO
Observe the result and you will notice that MidianCont has different value than before, the reason is PERCENTILE_CONT function has 0.1 value passed. For first four value the value is 709.3.
Now in my example I have explained how the median is found using this function. You have to explain using mathematics and explain (in easy words) why the value in last columns are 709.3 and 775.1
Hint:
SQL SERVER – Introduction to PERCENTILE_CONT() – Analytic Functions Introduced in SQL Server 2012
Rules
Leave a comment with your detailed answer by Nov 25's blog post.
Open world-wide (where Amazon ships books)
If you blog about puzzle’s solution and if you win, you win additional surprise gift as well.
Prizes
Print copy of my new book SQL Server Interview Questions Amazon|Flipkart
If you already have this book, you can opt for any of my other books SQL Wait Stats [Amazon|Flipkart|Kindle] and SQL Programming [Amazon|Flipkart|Kindle].
Reference: Pinal Dave (http://blog.SQLAuthority.com)
Filed under: Pinal Dave, PostADay, SQL, SQL Authority, SQL Function, SQL Puzzle, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, T SQL, Technology
