posted by monk .
a tire manufacturer believes that the tread life of its tires are normally distributed and will lst an average of 60,000 miles with a standard deviation of 3,000 miles. sixty-four randomly selected tires were tested and the average miles where the tire failed were recorded. what is the probability that the mean miles recorded when the tires failed will be more than 59,500 miles?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)(To make things simpler, you could use n.)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to that Z score.