*The thickness (mm) of the coating applied to disk drives is a characteristic that determines the usefulness of the product. When no unusual circumstances are present, the thickness *

*X has a normal distribution with mean of 3 mm and standard deviation of 0.05 mm. Suppose that the coating process will be monitored by selecting a random sample of 16 drives from each shift's production and determining the sample mean coating.*

*(a) Describe the sampling distribution of the sample mean.*

*(b) When no unusual circumstances are present, we expect the sample mean**to be within 3SE (standard error) of 3 mm, the desired coating thickness value. A value of the sample meanfarther from 3 than 3SEis interpreted as an indication of a problem that needs attention. Compute 3+/-3SE.*

*(c) Referring to part (b), what is the probability that the sample meanwill be outside 3+/-3SEjust by chance (that is, when there are no unusual circumstances)?*

*(d) Suppose that a machine used to apply the coating is out of adjustment, resulting in a mean coating thickness of 3.05 mm. What is the probability that a problem will be detected when the next sample is taken?*

*(Hint: This will occur if sample mean > 3+3SEor sample mean < 3-3SEwhen mu = 3.05).*

So for (a) I wrote the sample mean is 3mm, and the SE of the sample mean is 0.0125.

For (b) 3+/-3(SE) = 2.9625, 3.0375

For (c) P(sample mean > 3+3SEor sample mean < 3-3SE) = 0.0026

For (d) I'm thinking that it's the same calculation as in (c) except I should use 3.05 when finding the z-score. For example,

z = (3.0375-3.05)/0.0125 = -1

z = (2.9625 - 3.05)/0.0125 = -7

Then it becomes P(z < -7) + [1 - P(z < -1)]

Is this right? If so, what do I write for P(z < -7), since it would be a very minuscule number???