How do you determine the sample size of a normal distribution?

Question

I am presented with 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.

(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?

I thought a) was 1-normalcdf(9,11,10,1), but that answer appears to be wrong... I'm completely stumped on the others - I don't evene know where to start. Any chance someone could explain how to go about answering this question? As always, I'm trying to understand it - not just look for answers.

Answer 1

You have certain observations $X_1,\cdots,X_n$ which follow $N(10,1)$ distribution, where $X_i$ denote the thickness of the $i^{th}$ wafer. Then the the average thickness of the $n$ wafers $\bar{X}_n=\sum_{i=1}^nX_i/n$ follows $N(10,1/n)$ distribution. Then the probability that $$P(\bar{X}_n<11~or~\bar{X}_n>9)=P(-1<\bar{X}_n-10<1)=\Phi(\sqrt{n})-\Phi(-\sqrt{n})=2\Phi(\sqrt{n})-1$$ where $\Phi$ is the standard normal CDF. Put $n=10$ to get the value of $(a)$. $$P(\bar{X}_n>11)=P\left(\frac{\bar{X}_n-10}{1/\sqrt{n}}>\sqrt{n}\right)=1-\Phi(\sqrt{n})$$ So for $(b)$ find $n$ such that $1-\Phi(n)=0.01$ or $\sqrt{n}=\Phi^{-1}(0.99)$.

For $(c)$ suppose the required standard deviation is $s$. Then solve $s$ for the equation $$P(\bar{X}_n<11~or~\bar{X}_n>9)=P\left(-\frac{\sqrt{n}}{s}<\frac{\bar{X}_n-10}{s/\sqrt{n}}<\frac{\sqrt{n}}{s}\right)=2\Phi(\sqrt{n}/s)-1=0.001$$ when $n=10$. Or $$s=\frac{\sqrt{10}}{\Phi^{-1}(1.001/2)}$$

Answer 2

Suppose $t_k$ are the wafer thicknesses, each has distribution ${\cal N} (\mu, \sigma^2)$, and the $t_k$ are independent.

Then $t_1+\cdots +t_n$ will then have distribution ${\cal N} (n\mu, n\sigma^2)$, and so the quantity $\frac{t_1+\cdots + t_n-n \mu}{\sqrt{n} \sigma}$ will have distribution ${\cal N}(0,1)$, and has CDF $\Phi$. That is, $P(\frac{t_1+\cdots t_n-n \mu}{\sqrt{n}} \le x ) = \Phi(x)$. We note that $\Phi(x) = 1-\Phi(-x)$.

(a) We wish to compute the probability that $t_1+\cdots +t_n \notin [n(\mu-\sigma), n(\mu+\sigma)]$, which is the same as the probability that $\frac{t_1+\cdots + t_n-n \mu}{\sqrt{n} \sigma} \notin [-\sqrt{n}, \sqrt{n}]$. Hence the probability is $\Phi(-\sqrt{10}) + (1-\Phi(\sqrt{10})) = 2 \Phi(-\sqrt{10}) \approx 0.0015$.

(b) First we compute the probability that $t_1+\cdots + t_n \ge n(\mu+\sigma) $, and this is the same as the probability that $\frac{t_1+\cdots + t_n-n \mu}{\sqrt{n} \sigma} \ge \sqrt{n}$, which is $1-\Phi(\sqrt{n})$. Then we compute the minimum $n$ such that $1-\Phi(\sqrt{n}) \le 0.01$, and this gives $n=6$.

(c) First we compute the probability that $t_1+\cdots + t_n \notin [9n,11n]$ which is the same as the probability that $\frac{t_1+\cdots + t_n-n \mu}{\sqrt{n} \sigma} \notin [-\frac{\sqrt{n}}{\sigma}, \frac{\sqrt{n}}{\sigma}]$, which is $2 \Phi( -\frac{\sqrt{n}}{\sigma})$. Then we attempt to solve $2 \Phi( -\frac{\sqrt{10}}{\sigma}) = 0.001$, which gives $\sigma \approx 0.96102$.