Here's how Michigan voted for President George H.W. Bush in 1988

President George H.W. Bush won 53.6 percent of the vote in Michigan in 1988 en route to winning the race for U.S. President. 

That was worth 20 electoral votes at the time. Bush received nearly 300,000 more votes in Michigan than his Democratic competitor Michael Dukakis. 

Overall, Bush won 53.4 percent of the national vote that year, putting Michigan right there at the national average. 

This map by Wikimedia author 7partparadigm indicates how each Michigan county voted in the 1988 presidential election: 

Dukakis won between 60 and 70 percent of the vote in Wayne County, the darkest shade of blue on the map. Ottawa County on the far west side of the state was particularly red that year. 

Technically, the 1988 election was the 3rd time Michigan voters selected Bush for office. He was President Ronald Reagan's running mate in both 1980 and 1984. 

Bush ran again in 1992 but lost the state of Michigan's 18 electoral votes to Democrat Bill Clinton, who won his first presidential term that year. Clinton won Michigan again in 1996.

Bush's son, President George W. Bush, lost twice in Michigan in 2000 and 2004. 

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