Sports

Mookie Betts wins 2018 American League MVP

BOSTON, MA - JULY 12: Mookie Betts #50 of the Boston Red Sox celebrates after hitting a grand slam against the Toronto Blue Jays during the fourth inning at Fenway Park on July 12, 2018 in Boston, Massachusetts. (Photo by Maddie Meyer/Getty Images)

Capping off an incredible season for the World Series champion Red Sox, outfielder Mookie Betts has added another honor to his resume as he was named 2018 American League MVP.

Betts took home the award Thursday night, winning over Los Angeles outfielder Mike Trout and Cleveland Indians infielder Jose Ramirez.

Betts finished the 2018 season with a .346 batting average, .640 slugging percentage and 84 extra-base hits on the year, all good enough to lead the American League.

>>RELATED: Red Sox take home three Gold Glove Awards

He piled on 80 RBI on 180 hits, adding on 32 home runs to round out an incredible regular season, and also earned Gold Glove honors in the outfield.

Betts and teammate J.D. Martinez also took home Silver Slugger awards on the year, awarded to the best hitters at each position.

>>RELATED: J.D. Martinez, Mookie Betts win Silver Slugger Awards

Betts finished second in MVP voting in 2016, and finally earned the hardware after five years in the league.

In the National League, Christian Yelich of the Milwaukee Brewers took home MVP honors in his first year with the club after spending his first few seasons with the Miami Marlins.