document.write( "Question 204952: The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000 miles and a standard deviation of 1600 miles. What is the probability a randomly selected tire of this brand will last between 56,640 miles and 57,120 miles?
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\n" ); document.write( "i cant seem to figure this out or even know where to begin!... i would be so,so,so greatful if someone would point me in the right direction and at least give me some tools!please! \n" ); document.write( "
Algebra.Com's Answer #154695 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000 miles and a standard deviation of 1600 miles. What is the probability a randomly selected tire of this brand will last between 56,640 miles and 57,120 miles?
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\n" ); document.write( "Same procedure as on your other problem.
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\n" ); document.write( "Find the z-scores for 56640 and 57120
\n" ); document.write( "z(56640)= (56640-60000)/1600 = -2.1
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\n" ); document.write( "z(57120) = (57120-60000)/1600 = -1.8
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\n" ); document.write( "P(56640 < x < 57120)= P(-2.1 < z < -1.8) = normalcdf(-2.1,-1.8) = 0.018066..
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \n" ); document.write( "
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