document.write( "Question 672987: The braking distance of a sample of Ford F-150's are normally distributed. On a dry surface, the mean braking distance was 158 feet with a standard deviation of 7.23 ft. What is the longest braking distance on a dry surface one of these F-150 trucks could have an still be in the best 1%? \n" ); document.write( "
Algebra.Com's Answer #418428 by stanbon(75887)\"\" \"About 
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The braking distance of a sample of Ford F-150's are normally distributed. On a dry surface, the mean braking distance was 158 feet with a standard deviation of 7.23 ft. What is the longest braking distance on a dry surface one of these F-150 trucks could have and still be in the best 1%?
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\n" ); document.write( "I'm guessing the best 1% means shorter braking distances.
\n" ); document.write( "I may be wrong about what is \"best\".
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\n" ); document.write( "The z-value with a left-tail of 1% is -2.4363
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\n" ); document.write( "The corresponding distance value is x = z*s+u
\n" ); document.write( "x = -2.4363*7.23+158
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\n" ); document.write( "x = 141.18 feet
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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