### Database Schema:
CREATE TABLE inspection (
	inspection_id	INTEGER	NULL	PRIMARY KEY,
	inspection_date	DATE	NULL,
	inspection_type	TEXT	NULL,
	results	TEXT	NULL,
	employee_id	INTEGER	NULL,
	license_no	INTEGER	NULL,
	followup_to	INTEGER	NULL,
	foreign key (followup_to) references inspection (inspection_id),
	foreign key (employee_id) references employee (employee_id),
	foreign key (license_no) references establishment (license_no)
);

CREATE TABLE inspection_point (
	point_id	INTEGER	NULL	PRIMARY KEY,
	Description	TEXT	NULL,
	category	TEXT	NULL,
	code	TEXT	NULL,
	fine	INTEGER	NULL,
	point_level	TEXT	NULL
);

CREATE TABLE establishment (
	license_no	INTEGER	NULL	PRIMARY KEY,
	dba_name	TEXT	NULL,
	aka_name	TEXT	NULL,
	facility_type	TEXT	NULL,
	risk_level	INTEGER	NULL,
	address	TEXT	NULL,
	city	TEXT	NULL,
	state	TEXT	NULL,
	zip	INTEGER	NULL,
	latitude	REAL	NULL,
	longitude	REAL	NULL,
	ward	INTEGER	NULL
);

CREATE TABLE violation (
	inspection_id	INTEGER	NULL	PRIMARY KEY,
	point_id	INTEGER	NULL	PRIMARY KEY,
	fine	INTEGER	NULL,
	inspector_comment	TEXT	NULL,
	foreign key (point_id) references inspection_point (point_id),
	foreign key (inspection_id) references inspection (inspection_id)
);

CREATE TABLE employee (
	employee_id	INTEGER	NULL	PRIMARY KEY,
	first_name	TEXT	NULL,
	last_name	TEXT	NULL,
	address	TEXT	NULL,
	city	TEXT	NULL,
	state	TEXT	NULL,
	zip	INTEGER	NULL,
	phone	TEXT	NULL,
	title	TEXT	NULL,
	salary	INTEGER	NULL,
	supervisor	INTEGER	NULL,
	foreign key (supervisor) references employee (employee_id)
);

### User's Query:
Calculate the percentage of inspections with the fine for a minor food safety problem.

### Extra Knowledge:
fine for a minor food safety problem refers to fine = 100; percentage = divide(count(inspection_id where fine = 100), sum(inspection_id)) * 100%
SELECT CAST(COUNT(CASE WHEN fine = 100 THEN inspection_id END) AS REAL) * 100 / COUNT(inspection_id) FROM violation