United States presidential election in Maryland, 1988
Main article: United States presidential election, 1988
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County Results
Dukakis—70-80%
Dukakis—60-70%
Dukakis—50-60%
Bush—50-60%
Bush—60-70%
Bush—70-80% | ||||||||||||||||||||||||||||||||
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Elections in Maryland | ||||||||||
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Elections by year |
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The 1988 United States presidential election in Maryland took place on November 8, 1988, as part of the 1988 United States presidential election. Voters chose ten representatives, or electors to the Electoral College, who voted for President and Vice President.
Maryland was won by Vice President George H.W. Bush (R-Texas), with 51.11% of the popular vote, over Massachusetts Governor Mike Dukakis (D-Massachusetts) with 48.20% of the popular vote.[1] Bush ultimately won the national vote, defeating Governor Dukakis.[2]
Results
United States presidential election in Maryland, 1988[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Republican | George H.W. Bush | 876,167 | 51.11% | 10 | |
Democratic | Mike Dukakis | 826,304 | 48.20% | 0 | |
Libertarian | Ron Paul | 6,748 | 0.39% | 0 | |
New Alliance | Lenora Fulani | 5,115 | 0.29% | 0 | |
N/A | Write-ins | 24 | <0.01% | 0 | |
Totals | 1,714,358 | 100.0% | 10 |
References
- 1 2 "1988 Presidential General Election Results - Maryland". U.S. Election Atlas. Retrieved 8 June 2012.
- ↑ "1988 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.
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