### Database Schema:
CREATE TABLE unemployed (
	name	TEXT	NOT NULL	DEFAULT ''	PRIMARY KEY,
	foreign key (name) references person (name)
);

CREATE TABLE disabled (
	name	TEXT	NOT NULL	DEFAULT ''	PRIMARY KEY,
	foreign key (name) references person (name)
);

CREATE TABLE male (
	name	TEXT	NOT NULL	DEFAULT ''	PRIMARY KEY,
	foreign key (name) references person (name)
);

CREATE TABLE no_payment_due (
	name	TEXT	NOT NULL	DEFAULT ''	PRIMARY KEY,
	bool	TEXT	NULL,
	foreign key (bool) references bool (name),
	foreign key (name) references person (name)
);

CREATE TABLE enrolled (
	name	TEXT	NOT NULL	PRIMARY KEY,
	school	TEXT	NOT NULL	PRIMARY KEY,
	month	INTEGER	NOT NULL	DEFAULT 0,
	foreign key (name) references person (name)
);

CREATE TABLE longest_absense_from_school (
	name	TEXT	NOT NULL	DEFAULT ''	PRIMARY KEY,
	month	INTEGER	NULL	DEFAULT 0,
	foreign key (name) references person (name)
);

CREATE TABLE bool (
	name	TEXT	NOT NULL	DEFAULT ''	PRIMARY KEY
);

CREATE TABLE enlist (
	name	TEXT	NOT NULL,
	organ	TEXT	NOT NULL,
	foreign key (name) references person (name)
);

CREATE TABLE filed_for_bankrupcy (
	name	TEXT	NOT NULL	DEFAULT ''	PRIMARY KEY,
	foreign key (name) references person (name)
);

CREATE TABLE person (
	name	TEXT	NOT NULL	DEFAULT ''	PRIMARY KEY
);

### User's Query:
Calculate the average number of disabled female students enrolled in UCI.

### Extra Knowledge:
average = DIVIDE(SUM(disabled.name who are not in male.name WHERE school = 'uci'), COUNT(enrolled.name)); female students refers to disabled.name who are NOT in male.name; UCI refers to school = 'uci';
SELECT CAST(SUM(IIF(T1.school = 'uci' AND T4.name IS NULL, 1, 0)) AS REAL) / COUNT(T1.name) FROM enrolled AS T1 INNER JOIN disabled AS T2 ON T1.name = T2.name INNER JOIN person AS T3 ON T1.name = T3.name LEFT JOIN male AS T4 ON T3.name = T4.name