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Consider the following relations:
student(ID, name, dept_name, credits)
course(ID, course_ID, sec_ID, semester, year, grade)
Which of the following is the correct SQL for "For each course section offered in 2009, find the average total credits of all students enrolled in the section, if the section had at least 2 students".
  • a)
    select course_id, semester, year, sec_id, avg (credits)
    from course natural join student where year=2009 
    having count (ID) >=2; 
  • b)
    select course_id, semester, year, sec_id, avg (credits)
    from course natural join student where year=2009
    group by course_id, semester, year, sec_id
    having count (ID) >=2; 
  • c)
    select course_id, semester, year, sec_id, avg (credits)
    from course
    group by course_id, semester, year, sec_id 
    having count (ID) >=2; 
  • d)
    select course_id, semester, year, sec_id, avg (credits)
    from course natural join student where year=2009 
    group by course_id, semester, year, sec_id ;
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Consider the following relations:student(ID, name, dept_name, credits)...
The correct answer is option 2.
The sequence of operations is as follows:
  • The from clause is first evaluated to get a relation.
  • The predicate in the where clause is applied to the result relation of the from clause. 
  • Tuples satisfying the where predicate are then placed into groups by the group by clause.
  • The having clause is applied to each group; the groups that do not satisfy the having clause predicate are removed. 
  • The select clause uses the remaining groups to generate tuples of the result of the query, applying the aggregate functions to get a single result tuple for each group.
Satisfying the given sequence of operations, only option 2 performs the desired operation.
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Most Upvoted Answer
Consider the following relations:student(ID, name, dept_name, credits)...
Explanation of Option B:

Select Statement:
- The SELECT statement is used to retrieve the average total credits of all students enrolled in each course section offered in 2009.

From Clause:
- The FROM clause specifies the tables involved in the query, in this case, the course and student tables are being used.

Join:
- The query uses a natural join between the course and student tables, connecting them based on their common attribute, which is the course ID.

Where Clause:
- The WHERE clause filters the results to only include course sections offered in the year 2009.

Group By Clause:
- The GROUP BY clause groups the results by course ID, semester, year, and section ID.

Having Clause:
- The HAVING clause is used to filter out the results based on the condition that the count of student IDs (number of students enrolled) in each course section must be greater than or equal to 2.

Aggregate Function:
- The AVG(credits) function calculates the average total credits of all students enrolled in each course section meeting the specified conditions.

Conclusion:
- Option B is the correct SQL query for finding the average total credits of all students enrolled in each course section offered in 2009, where the section had at least 2 students.
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Consider the following relations:student(ID, name, dept_name, credits)course(ID, course_ID, sec_ID, semester, year, grade)Which of the following is the correct SQL for "For each course section offered in 2009, find the average total creditsof all students enrolled in the section, if the section had at least 2 students".a)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009having count (ID) >=2;b)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009group by course_id, semester, year, sec_idhaving count (ID) >=2;c)select course_id, semester, year, sec_id, avg (credits)from coursegroup by course_id, semester, year, sec_idhaving count (ID) >=2;d)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009group by course_id, semester, year, sec_id ;Correct answer is option 'B'. Can you explain this answer?
Question Description
Consider the following relations:student(ID, name, dept_name, credits)course(ID, course_ID, sec_ID, semester, year, grade)Which of the following is the correct SQL for "For each course section offered in 2009, find the average total creditsof all students enrolled in the section, if the section had at least 2 students".a)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009having count (ID) >=2;b)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009group by course_id, semester, year, sec_idhaving count (ID) >=2;c)select course_id, semester, year, sec_id, avg (credits)from coursegroup by course_id, semester, year, sec_idhaving count (ID) >=2;d)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009group by course_id, semester, year, sec_id ;Correct answer is option 'B'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider the following relations:student(ID, name, dept_name, credits)course(ID, course_ID, sec_ID, semester, year, grade)Which of the following is the correct SQL for "For each course section offered in 2009, find the average total creditsof all students enrolled in the section, if the section had at least 2 students".a)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009having count (ID) >=2;b)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009group by course_id, semester, year, sec_idhaving count (ID) >=2;c)select course_id, semester, year, sec_id, avg (credits)from coursegroup by course_id, semester, year, sec_idhaving count (ID) >=2;d)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009group by course_id, semester, year, sec_id ;Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the following relations:student(ID, name, dept_name, credits)course(ID, course_ID, sec_ID, semester, year, grade)Which of the following is the correct SQL for "For each course section offered in 2009, find the average total creditsof all students enrolled in the section, if the section had at least 2 students".a)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009having count (ID) >=2;b)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009group by course_id, semester, year, sec_idhaving count (ID) >=2;c)select course_id, semester, year, sec_id, avg (credits)from coursegroup by course_id, semester, year, sec_idhaving count (ID) >=2;d)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009group by course_id, semester, year, sec_id ;Correct answer is option 'B'. Can you explain this answer?.
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Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Consider the following relations:student(ID, name, dept_name, credits)course(ID, course_ID, sec_ID, semester, year, grade)Which of the following is the correct SQL for "For each course section offered in 2009, find the average total creditsof all students enrolled in the section, if the section had at least 2 students".a)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009having count (ID) >=2;b)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009group by course_id, semester, year, sec_idhaving count (ID) >=2;c)select course_id, semester, year, sec_id, avg (credits)from coursegroup by course_id, semester, year, sec_idhaving count (ID) >=2;d)select course_id, semester, year, sec_id, avg (credits)from course natural join student where year=2009group by course_id, semester, year, sec_id ;Correct answer is option 'B'. 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