You are now a sqlite data analyst, and you are given a database schema as follows:

    【Schema】
    【DB_ID】 food_inspection_2
【Schema】
# Table: violation
[
(inspection_id:INTEGER, Primary Key, Examples: [44247, 44248, 44249]),
(point_id:INTEGER, Primary Key, Examples: [30, 32, 33]),
(fine:INTEGER, Examples: [100, 0, 500]),
(inspector_comment:TEXT)
]
# Table: employee
[
(employee_id:INTEGER, Primary Key, Examples: [103705, 104633, 106581]),
(first_name:TEXT, Examples: [Anastasia, Joshua, Zach]),
(last_name:TEXT, Examples: [Hansen, Rosa, Barber]),
(address:TEXT, Examples: [6023 S Elizabeth St, 5000 N Wolcott Ave, 7522 N Oleander Ave]),
(city:TEXT, Examples: [Chicago, Park Forest, Hoffman Estates]),
(state:TEXT, Examples: [IL]),
(zip:INTEGER, Examples: [60636, 60640, 60631]),
(phone:TEXT, Examples: [(773) 424-8729, (773) 293-6409, (219) 473-0757]),
(title:TEXT, Examples: [Sanitarian, Supervisor, Division Manager]),
(salary:INTEGER, Examples: [79300, 82000, 79900]),
(supervisor:INTEGER, Examples: [177316, 186742, 179582])
]
# Table: inspection
[
(inspection_id:INTEGER, Primary Key, Examples: [44247, 44248, 44249]),
(inspection_date:DATE, Examples: [2010-01-05]),
(inspection_type:TEXT, Examples: [Complaint]),
(results:TEXT, Examples: [Pass, Pass w/ Conditions, Fail]),
(employee_id:INTEGER, Examples: [141319, 143902, 143487]),
(license_no:INTEGER, Examples: [1222441, 1336561, 1334073]),
(followup_to:INTEGER, Examples: [67871, 248210, 233215])
]
# Table: establishment
[
(license_no:INTEGER, Primary Key, Examples: [1, 2, 9]),
(dba_name:TEXT, Examples: [HARVEST CRUSADES MINISTRIES]),
(aka_name:TEXT, Examples: [COSI]),
(facility_type:TEXT, Examples: [Special Event, Restaurant, TAVERN]),
(risk_level:INTEGER, Examples: [2, 1, 3]),
(address:TEXT, Examples: [118 N CENTRAL AVE , 230 W MONROE ST , 116 S MICHIGAN AVE ]),
(city:TEXT, Examples: [CHICAGO, Chicago, chicago]),
(state:TEXT, Examples: [IL]),
(zip:INTEGER, Examples: [60644, 60606, 60603]),
(latitude:REAL, Examples: [41.8828450747188, 41.8807571586472, 41.8803958382596]),
(longitude:REAL, Examples: [-87.7650954520439, -87.6347092983425, -87.6245017215946]),
(ward:INTEGER, Examples: [29, 42, 18])
]
# Table: inspection_point
[
(point_id:INTEGER, Primary Key, Examples: [1, 2, 3]),
(Description:TEXT),
(category:TEXT, Examples: [Food Protection]),
(code:TEXT, Examples: [7-38-005 (B) (B-2), 7-38-005 (A), 7-38-010 (A) (B)]),
(fine:INTEGER, Examples: [500, 250, 100]),
(point_level:TEXT, Examples: [Critical, Serious , Minor   ])
]
【Foreign keys】
violation.inspection_id=inspection.inspection_id
violation.point_id=inspection_point.point_id
employee.supervisor=employee.employee_id
inspection.employee_id=employee.employee_id
inspection.license_no=establishment.license_no
inspection.followup_to=inspection.inspection_id

    【Question】
    What is the average number of inspections carried out in the year 2010 by a sanitarian whose salary is over 70000? in the year 2010 refers to inspection_date like '2010%'; salary is over 70000 refers to salary > 70000; average number = divide(sum(inspection where inspection_date like '2010%'), sum(employee_id where salary > 70000))

    【Evidence】
    in the year 2010 refers to inspection_date like '2010%'; salary is over 70000 refers to salary > 70000; average number = divide(sum(inspection where inspection_date like '2010%'), sum(employee_id where salary > 70000))

    Please read and understand the database schema carefully, and generate an executable SQL based on the user's question and evidence. 

SQL: SELECT CAST(SUM(CASE WHEN T2.inspection_date LIKE '2010%' THEN 1 ELSE 0 END) AS REAL) / SUM(CASE WHEN T1.salary > 70000 THEN 1 ELSE 0 END) FROM employee AS T1 INNER JOIN inspection AS T2 ON T1.employee_id = T2.employee_id