United States presidential election in Oregon, 1992

The 1992 United States presidential election in Oregon took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for president and vice president.

Oregon was won by Governor Bill Clinton ( D- Arkansas) with 42.48% of the popular vote over incumbent President George H. W. Bush ( R- Texas) with 32.53%. Businessman Ross Perot (Independent politician, I-Texas) finished in third, with 24.21% of the popular vote. Clinton ultimately won the national vote, defeating incumbent President Bush.

, this is the last occasion when Curry County, Oregon, Curry and Jefferson County, Oregon, Jefferson counties have given a plurality to the Democratic Party nominee.Sullivan, Robert David
‘How the Red and Blue Map Evolved Over the Past Century’
''America Magazine'' in ''The National Catholic Review''; June 29, 2016 It was also the first election since 1888 United States presidential election in Oregon, 1888 when Crook County, Oregon, Crook County had not backed the winning candidate, although Bush’s margin was less than 200 votes and was only a plurality. Nevertheless, this would mark the county becoming a Republican bastion.Lewis-Beck; Michael S.; ‘Election Forecasts in 1984: How Accurate Were They’; ''Political Science and Politics'' 18(1) (January 1985), pp. 53-63 Deschutes County, Oregon, Deschutes County, which was narrowly won by Clinton via plurality, would not vote for a Democrat again until 2020.

Results

Results by county

See also

* United States presidential elections in Oregon
* Presidency of Bill Clinton

Notes

References

{{1992 United States elections

1992 United States presidential election by state, Oregon
United States presidential elections in Oregon, 1992
1992 Oregon elections, President