A company advertises two car tire models. The number of thousands of miles that the standard model tires last has a mean \mu_S =
60μ S =60 and standard deviation \sigma_S = 5σ S =5. The number of miles that the extended life tires last has a mean \mu_E = 70μ E =70 and standard deviation \sigma_E = 7σ E =7. If mileages for both tires follow a normal distribution, what is the probability that a randomly selected standard model tire will get more mileage than a randomly selected extended life tire?
First, we see the closest multiple of 7 under 100 is 98 and that divided by 7 is 14. there are 2/7 as our remainder. We do 2/7 times 160 to get 320/7 and 320/7 = 45 and 5/7. We convert 5/7 into a decimal like 5/7 = .7. The answer is 45.7 milligrams of sodium.
