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 meanto 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???