United States presidential election in Minnesota, 1996
|Elections in Minnesota|
The 1996 United States presidential election in Minnesota took place on November 5, 1996, as part of the 1996 United States presidential election. Voters chose ten representatives, or electors to the Electoral College, who voted for President and Vice President.
A Democratic-leaning state, Minnesota was comfortably won by incumbent Democratic President Bill Clinton. Clinton took 51.10% of the popular vote over Republican challenger Bob Dole, who took 34.96%, a victory margin of 16.14%. Reform Party candidate Ross Perot finished in third with 11.75% of the popular vote.
|United States presidential election in Minnesota, 1996|
|Democratic||Bill Clinton (incumbent)||1,120,438||51.10%||10|
|U.S. Taxpayer||Howard Phillips||3,416||0.16%||0|
|Natural Law||John Hagelin||1,808||0.08%||0|
|Ind. Grassroots||John Birrenback||787||0.04%||0|
|Socialist Workers||James Harris||684||0.03%||0|
|Socialist Equality||Jerome White||347||0.02%||0|
- "1996 Presidential General Election Results - Minnesota". U.S. Election Atlas. Retrieved 30 January 2013.