The average life of a brand of automobile tires is 30,000 miles, with a standard deviation of 2,000 miles. If a tire is selected and tested, find the probability that it will have a lifetime between 25,000 and 31,000 miles. Assume the variable is normally distributed. Give your answer rounded to four decimal places (to the nearest ten-thousandth).

Find the z scores for 25000 and 28000.

z1 = (x - average) / standard deviation = (25000-30000) / 2000 = - 2.5

z2 = (x - average) / standard deviation = (31000-30000) / 2000 = 0.5

Find the corresponding P (z).

P (z1) = 0.00621

P (z2) = 0.15866

Subtract the probabilities to find the probability for the range,

(25000< = x <=28000)

P (x) = 0.15866-0.00621 = 0.1524