The thickness of a part is to have an upper specification of 0.925 and a lower specification of 0.870 mm. the average of the process is currently 0.917 with a standard deviation of 0.005. what is the probability that the product specification will be above 0.90 mm?

To solve this problem, we make use of the z statistic. The formula for the z score is:z score = (x – u) / swhere x is the sample value = 0.90, u is the sample mean = 0.917, and s is the standard deviation = 0.005
Therefore:z score = (0.90 – 0.917) / 0.005z score = -3.4
From the standard probability tables, the p-value for a right tailed test of z = -3.4 is:P = 0.9997
Therefore there is a 99.97% chance that it will be above 0.90 mm

I'm not sure exactly but my best guess would be 1/4

The answer is 225 miles per 4.5 hours. hope this ; )