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) ![]() You can put this solution on YOUR website! 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( "------ \n" ); document.write( "1/50 is 2% \n" ); document.write( "-------- \n" ); document.write( "Find the z-value with a left-tail of 98% \n" ); document.write( "invNorm(0.98) = 2.0537 \n" ); document.write( "--- \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) \n" ); document.write( "======================================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "============ \n" ); document.write( " \n" ); document.write( " |