document.write( "Question 438102: A tire manufacturer believes the tread life of snow tires has a normal distribution with mean of 32000 and standard deviation of 7000 miles.If the dealer is willing to give refunds to no more than 1 in 50 tires sold ,what mileage can he guarantee each tire to last? \n" ); document.write( "
Algebra.Com's Answer #303024 by stanbon(75887)\"\" \"About 
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A tire manufacturer believes the tread life of snow tires has a normal distribution with mean of 32000 and standard deviation of 7000 miles.If the dealer is willing to give refunds to no more than 1 in 50 tires sold ,what mileage can he guarantee each tire to last?
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\n" ); document.write( "1/50 is 2%
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\n" ); document.write( "Find the z-value with a left-tail of 98%
\n" ); document.write( "invNorm(0.98) = 2.0537
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\n" ); document.write( "solve for the corresponding x-value:
\n" ); document.write( "x = zs + u
\n" ); document.write( "x = 2.04537*7000+32000
\n" ); document.write( "x = 46376.24 miles (the mileage he can guarantee)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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