Project Employees I
<p>Table: <code>Project</code></p>

<pre>
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
(project_id, employee_id) is the primary key of this table.
employee_id is a foreign key to <code>Employee</code> table.
Each row of this table indicates that the employee with employee_id is working on the project with project_id.
</pre>

<p>&nbsp;</p>

<p>Table: <code>Employee</code></p>

<pre>
+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+
employee_id is the primary key of this table. It&#39;s guaranteed that experience_years is not NULL.
Each row of this table contains information about one employee.
</pre>

<p>&nbsp;</p>

<p>Write an SQL query that reports the <strong>average</strong> experience years of all the employees for each project, <strong>rounded to 2 digits</strong>.</p>

<p>Return the result table in <strong>any order</strong>.</p>

<p>The query result format is in the following example.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> 
Project table:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+
Employee table:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   | 1                |
| 4           | Doe    | 2                |
+-------------+--------+------------------+
<strong>Output:</strong> 
+-------------+---------------+
| project_id  | average_years |
+-------------+---------------+
| 1           | 2.00          |
| 2           | 2.50          |
+-------------+---------------+
<strong>Explanation:</strong> The average experience years for the first project is (3 + 2 + 1) / 3 = 2.00 and for the second project is (3 + 2) / 2 = 2.50
</pre>
Database
# Write your MySQL query statement
SELECT project_id, ROUND(AVG(experience_years), 2) AS average_years
FROM
    Project
    JOIN Employee USING (employee_id)
GROUP BY 1;
