1. The problem statement, all variables and given/known data The top-selling Red and Voss tire is rated 60000 miles, which means nothing. In fact, the distance the tires can run until wear-out is a normally distributed random variable with a mean of 75000 miles and a standard deviation of 5000 miles. A. What is the probability that the tire wears out before 60000 miles? B. What is the probability that a tire lasts more than 85000 miles? 2. Relevant equations z = [X - (mean of X)]/Standard deviation 3. The attempt at a solution This is another simple question that I can't seem to be getting correct. For part A, I did z = (60000 - 75000)/5000 = -3 Then P(z < -3) = 0.0013 according to the chart.. well 0.0013 is P(z= -3.0).. I tried entering the answer for P( z= -2.99) = 0.0014 since it's the last value on the chart lower than -3 also but that was also incorrect. For part B, z = [85000 - 75000]/5000 = 2 Then P(z > 2) = 1- P(z = 2) = 1 - 0.9772 = 0.0288 Also incorrect..