I REALLY NEED HELP IN THIS PART OF THE QUESTION!

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.

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? Please round your answer to 4 decimal places.

I got this help but the answer is not correct:
Since the normal distribution is symmetrical, and the total probability at both ends is to be 0.001, we need the probability at each end to be 0.0005

i.e. we want the standard score corresponding to a cumulative probability of 0.9995

The table at
http://www.math.unb.ca/~knight/utility/N…
only goes up to 0.9990, but just below that is another table headed
"Far right tail probabilities"
It gives the probability of Z exceeding 3.3 as being 0.0004834, which is near enough to 0.0005
So we require that 1 micrometer margin to be equal to 3.3 standard deviations.
Therefore std dev = 1/3.3
........................ = 0.3 correct to one figure. and 0.3030 is not correct too

Random variable X (thickness of wafer) has mean = 10, and standard deviation σ

Your link got cut off, but using an online normal distribution calculator, I get
z = 3.291

3.291 s.d. = 1
s.d. = 1/3.291

Now your solution would be correct if we were dealing with only 1 wafer, but we are dealing with the average of 10 wafers.

Let Y = (X₁ + X₂ + . . . + X₁₀)/10

Standard deviation found above (1/3.291) is standard deviation for Y, not X

Expected value of Y is also 10. We can show this using following identities:
E[X₁+X₂] = E[X₁] + E[X₂]
E[aX] = a E[X]

E[X₁] = E[X₂] = . . . = E[X₁₀] = 10
E[Y] = E[(X₁ + X₂ + . . . + X₁₀)/10]
E[Y] = 1/10 E[X₁] + 1/10 E[X₂] + . . . + 1/10 E[X₁₀] = 10

So mean doesn't change

Var[X₁] = Var[X₂] = . . . = Var[X₁₀] = σ²

We want to find σ, so that standard deviation of Y = 1/3.291
Use following identities:
Var(aX) = a² Var(X)
Var(X₁+X₂) = Var(X₁) + Var(X₂) + 2 Cov(X₁,X₂)

Since thicknesses of different wafers are independent, Cov(X₁,X₂) = 0. Therefore:
Var(X₁+X₂+...+X₁₀) = Var(X₁) + Var(X₂) + . . . + Var(X₁₀)
Var(X₁+X₂+...+X₁₀) = σ² + σ² + . . . + σ² = 10 σ²

Var(Y) = Var(1/10 (X₁ + X₂ + . . . + X₁₀))
Var(Y) = 1/10² * 10 σ² = σ²/10

Now recall that s.d. of Y need to be 1/3.291, not s.d. of X

Var(Y) = σ²/10
S.D.(Y) = σ/√10
1/3.291 = σ/√10
σ = √10/3.291 = 0.960886557

So try 0.96

Mαthmφm

i dont know