Database schema:

CREATE TABLE beers (
	id INTEGER NOT NULL -- example: ['2688', '2687'],
	brewery_id INTEGER NOT NULL -- example: ['0', '1'],
	abv REAL -- example: ['0.087', '0.065'],
	ibu REAL -- example: ['85.0', '65.0'],
	name TEXT NOT NULL -- example: ['Gordon Ale (2009)', "Dale's Pale Ale"],
	style TEXT -- example: ['American Pale Ale (APA)', 'American Double / Imperial IPA'],
	ounces REAL NOT NULL -- example: ['12.0', '16.0'],
	PRIMARY KEY (id), 
	CONSTRAINT fk_beers_breweries FOREIGN KEY(brewery_id) REFERENCES breweries (id)
)


CREATE TABLE breweries (
	id INTEGER NOT NULL -- example: ['0', '1'],
	name TEXT -- example: ['Against the Grain Brewery', 'NorthGate Brewing '],
	city TEXT -- example: ['Louisville', 'Minneapolis'],
	state TEXT -- example: ['KY', 'MN'],
	PRIMARY KEY (id)
)

Question: When compared to the total number of breweries in the US producing American Blonde Ale, how many in the state of Wisconsin produces American Blonde Ale? Indicate your answer in percentage (%).
Evidence: Percentage of the state of Wisconsin produces American Blonde Ale could be computed by count(breweries in Wisconsin producing American Blonde Ale) / count(all breweries)

SQL:
SELECT CAST(SUM(IIF(T2.state = 'WI', 1, 0)) AS REAL) * 100 / COUNT(T1.id) FROM beers AS T1 INNER JOIN breweries AS T2 ON T1.brewery_id = T2.id WHERE T1.style = 'American Blonde Ale'