1888 United States presidential election in Minnesota

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1888 United States presidential election in Minnesota

← 1884 November 6, 1888 1892 →
 
Nominee Benjamin Harrison Grover Cleveland
Clinton Fisk
Party Republican Democratic Prohibition
Home state Indiana New York New Jersey
Running mate Levi P. Morton Allen G. Thurman
John A. Brooks
Electoral vote 7 0 0
Popular vote 142,192 104,385 15,511
Percentage 54.12% 39.65% 5.82%

County Results

President before election

Grover Cleveland
Democratic

Elected President

Benjamin Harrison
Republican

The 1888 United States presidential election in Minnesota took place on November 6, 1888, as part of the 1888 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for president and vice president.

Minnesota voted for the Republican nominee, Benjamin Harrison, over the Democratic nominee, incumbent President Grover Cleveland. Harrison won the state by a margin of 14.47%.

With 5.82% of the popular vote, Minnesota would be the Prohibition Party candidate Clinton Fisk’s strongest victory in terms of percentage in the popular vote.[1]

Results

1888 United States presidential election in Minnesota[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Benjamin Harrison of Indiana
Levi Parsons Morton of New York
142,492 54.12% 7 100.00%
Democratic Grover Cleveland of New York
Allen Granberry Thurman of Ohio
104,385 39.65% 0 0.00%
Prohibition
Clinton Bowen Fisk of New Jersey
John Anderson Brooks of Missouri
15,311 5.82% 0 0.00%
Labor Alson Streeter of Illinois Charles E. Cunningham of Arkansas 1,097 0.42% 0 0.00%
Total 263,285 100.00% 7 100.00%

See also

Notes

References

  1. ^ "1888 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. ^ "1888 Presidential General Election Results - Minnesota". U.S. Election Atlas. Retrieved December 23, 2013.