Question 302971
 Truck tire life is normally distributed with a mean of 60,000 miles and a standard deviation of 4,000 miles.

- You bought 4 tires. What is the probability that the average mileage of the 4 tires exceeds 66,000 miles?
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z(66,000) = (66000-60000)/[4000/sqrt(4)] = 6000/2000 = 3
P(x > 66000) = P(z> 3) = 0.0013

a.) 0.0013
b.) 0.9987
c.) 0.4987
d.) 0.9544
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- What is the probability that 1 tire will last 72,000 miles or more?
z(72000) = (72000-60000)/4000 = 12000/4000 = 3
P(z> 72000)= P(z > 3) = 0.0013
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a.) 0.0013
b.) 0.9987
c.) 0.4987
d.) 0.9544 
Thank you for your help!
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Cheers,
Stan H.