Question

The thickness (in millimeters) of the coating applied to disk drives is one characteristic that determines the usefulness of the product. When no unusual circumstances are present, the thickness (x) has a normal distribution with a mean of 3 mm and a standard deviation of 0.05 mm. Suppose that the process will be monitored by selecting a random sample of 16 drives from each shift's production and determining x, the mean coating thickness for the sample. (a) Describe the sampling distribution of x for a random sample of size 16. The distribution of x is ---Select--- with mean mm and standard deviation mm. (b) When no unusual circumstances are present, we expect x to be within 20 of 3 mm, the desired value. An x value farther from 3 than 20 is interpreted as an indication of a problem that needs attention. Compute 3 20, 3 – 205 = mm 3 + 203 = mm (c) Referring to part (b), what is the probability that a sample mean will be outside 3 = 20 - just by chance (that is, when there are no unusual circumstances)? (Round your answer to four decimal places.) (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 x > 3 + 20 = or x < 3 – 205 when u = 3.05. Round your answer to four decimal places.)

Answer 1

Soll Given Meant 3mm Standard deviation=6 = 0.05mm a> distribution of I is approximatelly normal with men ly got 17= 3 6x=3 = 0.05 = 0.0125 J16 1.74x=3] JFx=0.0125] 63-255= 3-240.0125= 2.975 3+205= 3+2x0.0125= 3.025 O P(2.975<23.025)= 43.002523) -2.915-3) (0.0125) = $(2) - d62) = 0.9772 -0.0228 = 0.95 hence prob that m lies outside 3-(2X (q) 43 +255 10 Calculated as prob=1-P(2.975 <x<31025) = - 0.95 = 0.05 & Prob that lies outside 3+208 is 10.05

Pl 3+257) = P( X7 3.025) - P(273.025-3.05. 0.0125 - P(27-2) - 0.9772 P(X<3 -267) = P(Z <2.975-3.05 ) 0.0125 = P(Z <-6) = 0 Required probability = P(7 737208) + P (=<3-268) = 0.9772