Context: CREATE TABLE table_name_25 (points_for_higher INTEGER, points_for_foundation VARCHAR, points_for_ordinary VARCHAR, grade VARCHAR)
Question: What average points for highers has 0 has points for ordinary, and Ng as the grade, and less than 0 as points for foundation?
Answer: SELECT AVG(points_for_higher) FROM table_name_25 WHERE points_for_ordinary = 0 AND grade = "ng" AND points_for_foundation < 0