The photoresist thickness in semiconductor manufacturing has

The photoresist thickness in semiconductor manufacturing has a mean of 10 micrometers and a standard deviation of 1 micrometer. Assume that the thickness is normally distributed and that the thicknesses of different wafers are independent.

(a) Determine the probability that the average thickness of 10 wafers is either greater than 11 or less than 9 micrometers.

(b) Determine the number of wafers that needs to be measured such that the probability that the average thickness exceeds 11 micrometers is 0.01

c) If the mean thickness is 10 micrometers, what should the standard deviation of thickness equal so that the probability that the average of 10 wafers is either greater than 11 or less than 9 micrometers is 0.001?

Solution

A. find the z scores of both 11 and 9 micromoeters since they are the same distance from the mean the z score of one will be the negative of the other. Z = (xbar-mu)/(s/sqrt(n)) = (11-10)/(1/sqrt(10)) = 3.16227766 The z score for 9 is -3.16227766 The porbabilitliy that it is outside of these points is .0016 B. the probability that it exceeds 11 is .001 so we know that the z score must be approximatley 2.326 this will equal (xbar-mu)/(s/sqrt(n)) = (11-10)/(1/sqrt(n)) so we will have to solve for n algebraicly. 1/(1/sqrt(n)) = 2.326 1= 2.326(1/sqrt(n)) 1=2.326/sqrt(n) sqrt(n) = 2.326 n = 2.326^2 = 5.41026 whenever you calculate n you should always round up, so n=6 C. Using a similar method as the previous problem except in this case Z = 3.891 3.891 = (11-10)/(s/sqrt(10)) 3.891 = 1/(s/sqrt(10)) 3.891(s/sqrt(10)) = 1 3.891s = sqrt(10) s = sqrt(10)/3.891 s= .812715924 the z of 3.891 is associated with the tail area of .00005 we use this because 9 and 11 are equal distance from the mean so we can use the p-value above one side instead of trying to figure out the algebra for testing both of them at the same time.