-- Database schema
| businesses : business_id [ INTEGER ] primary_key , name [ TEXT ] , address [ TEXT ] , city [ TEXT ] , postal_code [ TEXT ] , latitude [ REAL ] , longitude [ REAL ] , phone_number [ INTEGER ] , tax_code [ TEXT ] , business_certificate [ INTEGER ] , application_date [ DATE ] , owner_name [ TEXT ] , owner_address [ TEXT ] , owner_city [ TEXT ] , owner_state [ TEXT ] , owner_zip [ TEXT ] | inspections : business_id [ INTEGER ] inspections.business_id = businesses.business_id , score [ INTEGER ] , date [ DATE ] , type [ TEXT ] | violations : business_id [ INTEGER ] violations.business_id = businesses.business_id , date [ DATE ] , violation_type_id [ TEXT ] , risk_category [ TEXT ] , description [ TEXT ] |
-- -- Which establishment has the highest number of inspections done? Give the name of the establishment and calculate for its average score per inspection.

SELECT T2.name, AVG(T1.score) FROM inspections AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id GROUP BY T2.name ORDER BY COUNT(T2.business_id) DESC LIMIT 1;