<p>download instant at www.easysemester.com Assignment #2 String and Aggregate Functions</p><p>1. Create a list of all students with middle initials (specifically, with a first name AND then a middle initial) AND remove the middle initial from the first name. Take care: some first names are a single initial only. [HINT: use the INSTR function embedded within other functions). (7 rows).</p><p>SELECT first_name Name, SUBSTR(first_name,1,INSTR(first_name,' ') - 1) "Instr Name" FROM student WHERE INSTR(first_name,' ') < INSTR(first_name,'.') AND INSTR(first_name,' ') <> 0 AND INSTR(first_name,'.') <> 0</p><p>2. Create a list of courses, sections and their capacity. Produce the result in the following format: (78 rows)</p><p>Course and Section CAPACITY ------350 Section 3 ------25 10 Section 2 ------15 20 Section 2 ------15 …</p><p>SELECT LPAD(course_no||' Section '||section_no, 20) "Course and Section", RPAD('-',10,'-') " ",LTRIM(capacity) Capacity FROM section</p><p>3. Count the number of students with a “v” in their last name. Use INSTR to produce the answer. (16)</p><p>SELECT COUNT(*) "Count of V’s" FROM student WHERE INSTR(LOWER(last_name),'v') > 0</p><p>4. Show the average cost of a course with no prerequisites (1195)</p><p>SELECT AVG(cost) "Average Cost" FROM course WHERE prerequisite IS NULL</p><p>5. Display the course number, number of sections and total capacity for courses having more than 3 sections.( 9 rows).</p><p>SELECT course_no Course, COUNT(section_no) "Section Count", SUM(capacity) Capacity FROM section GROUP BY course_no HAVING COUNT(section_no) > 3</p><p>6. List all instructors and how many sections they teach. (8 rows).</p><p>SELECT instructor_id ID, COUNT(*) FROM section GROUP BY Instructor_id</p><p> download instant at www.easysemester.com download instant at www.easysemester.com 7. Do #6 but limit it to instructors teaching more than 9 sections. (6 rows).</p><p>SELECT instructor_id ID, COUNT(*) FROM section GROUP BY Instructor_id HAVING COUNT(*) > 9</p><p> download instant at www.easysemester.com</p>