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 need 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 more than 11 or less

than 9 micrometers is 0.001?

Step 1 of 3:

Given that the thickness in semiconductor (X) manufacturing has mean is 10 micrometers, and standard deviation is 1 micrometer.

Our goal is:

a). We need to find the probability of thickness of 10 wafer is either greater than 11 or less than 9.

b). We need to find the number of wafers that need to be measured.

c). We need to find the standard deviation of thickness.

a). We assume that the thickness is normally distributed and thickness of 10 wafer is either greater than 11 or less than 9 micrometer.

We know that standard deviation is 1 and the sample size n is 10.

Now we need to find the sample standard deviation.

Then the sample standard deviation formula is

The probability of thickness of 10 wafer is either greater than 11 or less than 9 is

P(

Then,

From the table value,.

Hence the required probability is

P(

P(

P(

Therefore, the probability of thickness of 10 wafer is either greater than 11 or less than 9 is 0.0016.

Step 2 of 3:

b). Given the average thickness exceeds 11 micrometers is 0.01.

Then the probability of the average thickness exceeds 11 micrometers is

P() = P=0.001

P() = P=0.001

Here, P

(using area under the normal curve table 0.99 value fall into 2.33)

= 2.33

We know that value is 1.

= 2.33

n = 2.332.33

n = 5.4289

Hence the nearest value is 6.

Therefore, the number of wafers that need to be measured is 6.