OCP CBT Question 533

Please solve the following question.

You need to categorize employees based on their salary into three groups: 'High Salary' (salary >= 15000), 'Medium Salary' (salary >= 8000 and < 15000), and 'Low Salary' (salary < 8000). For employees with `NULL` salary, they should be categorized as 'No Salary Data'. Which SQL statement correctly implements this logic?

EMPLOYEES Table Sample Data (relevant columns)
EMPLOYEE_ID SALARY
----------  ------
100         24000
101         17000
102         17000
103         9000
104         6000
105         4800
106         4200
107         NULL
108         10000
109         12000
110         7000

SELECT first_name, last_name, salary, CASE WHEN salary >= 15000 THEN 'High Salary' WHEN salary >= 8000 THEN 'Medium Salary' WHEN salary < 8000 THEN 'Low Salary' ELSE 'No Salary Data' END AS salary_category FROM employees;

SELECT first_name, last_name, salary, CASE WHEN salary >= 15000 THEN 'High Salary' WHEN salary >= 8000 AND salary < 15000 THEN 'Medium Salary' WHEN salary < 8000 THEN 'Low Salary' WHEN salary IS NULL THEN 'No Salary Data' END AS salary_category FROM employees;

SELECT first_name, last_name, salary, IF(salary >= 15000, 'High Salary', IF(salary >= 8000, 'Medium Salary', IF(salary < 8000, 'Low Salary', 'No Salary Data'))) AS salary_category FROM employees;

SELECT first_name, last_name, salary, DECODE(SIGN(salary - 15000), 1, 'High Salary', DECODE(SIGN(salary - 8000), 1, 'Medium Salary', 'Low Salary')) AS salary_category FROM employees;