SQL Server STDEV() Function

The SQL Server (Transact-SQL) STDEV() function returns the statistical standard deviation of all values in the specified expression.

Syntax

The syntax for using STDEV() function is given below:

SELECT STDEV(column_name) 
FROM table_name
WHERE condition(s);

The SQL Server (Transact-SQL) GROUP BY clause is used to arrange result table into identical groups when one or more columns are used. Please note that it is must to include those column names in a GROUP BY clause which are not encapsulated within the STDEV() function. See the syntax below:

SELECT column1, column2, ...
       STDEV(column_name) 
FROM table_name
WHERE condition(s)
GROUP BY column1, column2, ...;

Parameters

Return Value

Returns the statistical standard deviation of all values in the specified expression.

Example - With Single Column

Consider a database table called Employee with the following records:

To get the standard deviation of salary of all employees whose age is greater than 25, the following query can be used:

SELECT STDEV(Salary) AS STDEV_Salary 
FROM Employee
WHERE Age > 25;

This will produce the result as shown below:

Example - Using DISTINCT

The DISTICT clause can be used with STDEV() function. For example - To get the standard deviation of distinct (unique) salaries, the following code can be used:

SELECT STDEV(DISTINCT Salary) AS STDEV_UniqueSalary 
FROM Employee;

This will produce the result as shown below:

Example - Using Formula

The expression contained within the STDEV() function does not need to be a single field. A formula can also be used with this function. For example - Consider a bonus (15% of Salary) is given to each employee, and to calculate the standard deviation of bonus amount, the following query can be used:

SELECT STDEV(Salary * 0.15) AS STDEV_BonusAmount
FROM Employee;

This will produce the result as shown below:

Example - Using GROUP BY

To get the standard deviation of salary of employees group by their age, the following query can be used:

SELECT Age, STDEV(Salary) AS STDEV_SalaryByAge 
FROM Employee
GROUP BY Age;

This result of the above code will be: