CREATE TABLE table_name_79 (interview NUMBER, evening_gown VARCHAR2, average VARCHAR2, swimsuit VARCHAR2)

Which mean interview number had an average of more than 9.233, a swimsuit stat of more than 9.473, and an evening gown score of more than 9.671?
SELECT AVG(interview) FROM table_name_79 WHERE average > 9.233 AND swimsuit = 9.473 AND evening_gown > 9.671
SELECT AVG("interview") FROM "table_name_79" WHERE "average" > 9.233 AND "evening_gown" > 9.671 AND "swimsuit" = 9.473