document.write( "Question 302971: Truck tire life is normally distributed with a mean of 60,000 miles and a standard deviation of 4,000 miles.
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document.write( "- You bought 4 tires. What is the probability that the average mileage of the 4 tires exceeds 66,000 miles?
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document.write( "a.) 0.0013
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document.write( "b.) 0.9987
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document.write( "c.) 0.4987
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document.write( "d.) 0.9544\r
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document.write( "- What is the probability that 1 tire will last 72,000 miles or more?
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document.write( "a.) 0.0013
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document.write( "b.) 0.9987
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document.write( "c.) 0.4987
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document.write( "d.) 0.9544\r
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document.write( "Thank you for your help! \n" );
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Algebra.Com's Answer #217172 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Truck tire life is normally distributed with a mean of 60,000 miles and a standard deviation of 4,000 miles.\r \n" ); document.write( "\n" ); document.write( "- You bought 4 tires. What is the probability that the average mileage of the 4 tires exceeds 66,000 miles? \n" ); document.write( "--- \n" ); document.write( "z(66,000) = (66000-60000)/[4000/sqrt(4)] = 6000/2000 = 3 \n" ); document.write( "P(x > 66000) = P(z> 3) = 0.0013\r \n" ); document.write( "\n" ); document.write( "a.) 0.0013 \n" ); document.write( "b.) 0.9987 \n" ); document.write( "c.) 0.4987 \n" ); document.write( "d.) 0.9544 \n" ); document.write( "---------------------------------- \n" ); document.write( "- What is the probability that 1 tire will last 72,000 miles or more? \n" ); document.write( "z(72000) = (72000-60000)/4000 = 12000/4000 = 3 \n" ); document.write( "P(z> 72000)= P(z > 3) = 0.0013 \n" ); document.write( "--- \n" ); document.write( "a.) 0.0013 \n" ); document.write( "b.) 0.9987 \n" ); document.write( "c.) 0.4987 \n" ); document.write( "d.) 0.9544 \n" ); document.write( "Thank you for your help! \n" ); document.write( "================================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |