The Minnesota Timberwolves have scheduled a major press conference for Thursday at which, according to multiple media reports, Andrew Wiggins will be announced as the NBA's Rookie of the Year.
The Star Tribune reports that it received the news from a league source and the official announcement will be made on Thursday.
Wiggins will become the first Timberwolves player in franchise history to win the rookie of the year award. He led all NBA rookies this season in points, minutes, field goals and free throws. He also was the only Wolves player to appear in all 82 games this season.
Wiggins also won four Western Conference rookie of the month awards, scored in double figures in 69 of 82 games, scored 20 or more points on 31 different occasions and scored 30-plus four times.
He averaged 16.9 points, 2.1 assists and 4.6 rebounds per game for the Wolves. Additionally Wiggins shot 43.7 percent from the field and 31 percent from behind the 3-point line.
More importantly, he got to the free throw line 466 times – ranking sixth in the NBA. The only players to finish ahead of him on that statistic are James Harden, Russell Westbrook, DeMarcus Cousins, LeBron James and DeAndre Jordan.
The Wolves acquired Wiggins, Anthony Bennett and Thaddeus Young in the trade that sent Kevin Love to Cleveland. The Cavaliers had selected Wiggins with the No. 1 overall pick.