United States presidential election in Rhode Island, 1992

United States presidential election in Rhode Island, 1992

November 3, 1992

Nominee Bill Clinton George H.W. Bush Ross Perot Party Democratic Republican Independent Home state Arkansas Texas Texas Running mate Al Gore Dan Quayle James Stockdale Electoral vote 4 0 0 Popular vote 213,299 131,601 105,045 Percentage 47.0% 29.0% 23.2%

County Results   Clinton—40-50%

President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

The 1992 United States presidential election in Rhode Island took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

Rhode Island was won by Governor Bill Clinton (D-Arkansas) with 47.04% of the popular vote over incumbent President George H.W. Bush (R-Texas) with 29.02%. Businessman Ross Perot (I-Texas) finished in third with 23.16% of the popular vote.^[1] Clinton ultimately won the national vote, defeating incumbent President Bush.^[2]

Results

United States presidential election in Rhode Island, 1992[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic	Bill Clinton	213,299	47.04%	4
	Republican	George H.W. Bush (incumbent)	131,601	29.02%	0
	Independent	Ross Perot	105,045	23.16%	0
	New Alliance	Lenora Fulani	1,878	0.41%	0
	Libertarian	Andre Marrou	571	0.13%	0
	Ind. For LaRouche	Lyndon LaRouche	494	0.11%	0
	Natural Law	John Hagelin	262	0.06%	0
	U.S. Taxpayers'	Howard Phillips	215	0.05%	0
	Write-in	James Bo Gritz	3	0.00%	0
	Write-in	J. Quinn Brisben	2	0.00%	0
	Write-in	Ronald Daniels	1	0.00%	0
	N/A	Write-ins	106	0.02%	0
Totals			453,477	100.0%	4

References

1. "1992 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 9 June 2012.
2. "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 9 June 2012.