# Consider a database that has the relation schema EMP (EmpId, EmpName, and DeptName). An instance of the schema EMP and a SQL query on it are given bel

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Consider a database that has the relation schema EMP (EmpId, EmpName, and DeptName). An instance of the schema EMP and a SQL query on it are given below.

 EMP EmpId EmpName DeptName 1 XYA AA 2 XYB AA 3 XYC AA 4 XYD AA 5 XYE AB 6 XYF AB 7 XYG AB 8 XYH AC 9 XYI AC 10 SYJ AC 11 XYK AD 12 XYL AD 13 XYM AE

 SELECT AVG(EC.Num FROM EC WHERE(DeptName, Num) IN           (SELECTED DeptName, COUNT(EmpId)AS                                                     EC(DeptName, Num)            FROM EMP            GROUP BY DeptName)

The output of executing the SQL query is _________.

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Concept:

Count the number of EmpIDs corresponding to a particular department and then write as attribute value of Num in new relational schema EC.

Explanation:

SQL query results

 EC DeptName Num AA 4 AB 3 AC 3 AD 2 AE 1

AVG(EC. Num) will find out the average of Num values in the returned query.,

$Average = \;\frac{{4 + 3 + 3 + 2 + 1}}{5} = 2.6$