Question: List the income and number of inhabitants of customers with a reference ID greater than the 50% of average of number of false response?
SQL: SELECT T2.INCOME_K, T2.INHABITANTS_K FROM Customers AS T1 INNER JOIN Demog AS T2 ON T1.GEOID = T2.GEOID INNER JOIN Mailings1_2 AS T3 ON T1.ID = T3.REFID WHERE T3.REFID > ( SELECT 0.5 * COUNT(CASE WHEN RESPONSE = 'false' THEN 1 ELSE NULL END) / COUNT(RESPONSE) FROM Mailings1_2 )