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Applies To: SQL Server 2014, SQL Server 2016 Preview

Calculates a percentile based on a continuous distribution of the column value in SQL Server. The result is interpolated and might not be equal to any of the specific values in the column.

Applies to: SQL Server (SQL Server 2012 through current version) , Azure SQL Data Warehouse Public Preview.

Topic link icon Transact-SQL Syntax Conventions (Transact-SQL)

PERCENTILE_CONT ( numeric_literal ) 
    WITHIN GROUP ( ORDER BY order_by_expression [ ASC | DESC ] )
    OVER ( [ <partition_by_clause> ] )


The percentile to compute. The value must range between 0.0 and 1.0.

WITHIN GROUP ( ORDER BY order_by_expression [ ASC | DESC ])

Specifies a list of numeric values to sort and compute the percentile over. Only one order_by_expression is allowed. The expression must evaluate to an exact numeric type (int, bigint, smallint, tinyint, numeric, bit, decimal, smallmoney, money) or an approximate numeric type (float, real). Other data types are not allowed. The default sort order is ascending.

OVER ( <partition_by_clause> )

Divides the result set produced by the FROM clause into partitions to which the percentile function is applied. For more information, see OVER Clause (Transact-SQL). The <ORDER BY clause> and <rows or range clause> of the OVER syntax cannot be specified in a PERCENTILE_CONT function.


Under compatibility level 110 and higher, WITHIN GROUP is a reserved keyword. For more information, see ALTER DATABASE Compatibility Level (Transact-SQL).

Any nulls in the data set are ignored.

PERCENTILE_CONT is nondeterministic. For more information, see Deterministic and Nondeterministic Functions.

The following example uses PERCENTILE_CONT and PERCENTILE_DISC to find the median employee salary in each department. Note that these functions may not return the same value. This is because PERCENTILE_CONT interpolates the appropriate value, whether or not it exists in the data set, while PERCENTILE_DISC always returns an actual value from the set.

USE AdventureWorks2012;

SELECT DISTINCT Name AS DepartmentName
                            OVER (PARTITION BY Name) AS MedianCont
                            OVER (PARTITION BY Name) AS MedianDisc
FROM HumanResources.Department AS d
INNER JOIN HumanResources.EmployeeDepartmentHistory AS dh 
    ON dh.DepartmentID = d.DepartmentID
INNER JOIN HumanResources.EmployeePayHistory AS ph
    ON ph.BusinessEntityID = dh.BusinessEntityID

Here is a partial result set.

DepartmentName        MedianCont    MedianDisc

--------------------   ----------   ----------

Document Control       16.8269      16.8269

Engineering            34.375       32.6923

Executive              54.32695     48.5577

Human Resources        17.427850    16.5865

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