Distribution of the lengths of a component question I've been working through my notes on the normal distribution and I'm currently struggling with the whole of  question i). Would appreciate any suggestions or advice on how to tackle it.

There are the answers I have so far:
a) -3.5s
b) 3.2s
c) 0.000233
d) 0.000687
e) 0.000920
f) 0.3
g) LSL = 18.95, USL = 20.96

 A: Some fine details depend on the method you are using to compute the relevant probabilities. One can either use a program or tables. One small problem with using tables is that most of them do not go quite far enough. The following table does. From the precision of your answers, I assume you are using software.
For a) your answer is correct. A very minor quibble: it asked how many standard deviation units below the mean the LSL is. I would say $3.5$ standard deviation units. If someone's ideal weight is $150$ pounds, and the person weighs $132$ pounds, I would say that she is $18$ pounds below her ideal weight, not that she is $-18$ pounds below her ideal weight.
For b), c), d), e), and f) your answers are correct.  For g) your answer is correct, if one assumes that the difference given in f) still holds in question g).
So far very good.  For the last question, which probably should be called h), proceed as follows. 
For i) the proposal is to put the mean so that it is halfway between the LSL and the USL.  The distance between these is $6.7$ standard deviation units, so we want the LSL to be $3.35$ standard deviation units below the mean, and the USL to be $3.35$ standard deviation units above the mean. That involves shifting the mean downwards by $0.15$ standard deviation units, that is, by $0.045$ to $19.955$.
Then the USL would be $3.35$ standard deviation units above the mean, and the LSL $3.35$ standard deviation units below. My table gives that the probability of being $3.35$ standard deviation units above the mean is roughly $0.0004$. By symmetry the same applies to below the mean, for a total probability of $0.0008$.  The current rejection rate, as computed in e), is $0.000920$, so there is an improvement in rejection rate.  But not a big improvement.
Warning: The above answer may not mean much in real life, since the normal distribution is an approximation to the reality, it is not the reality.  The fit between the genuine "tail" probabilities and the ones for the normal may be particularly poor. Yes, they are both small, so the difference between the real tail probability and the one given by the normal is small. However, the ratio between the real tail probabilities and the ones given by the normal approximation may not be close to $1$.  So in real life, the calculation made above would probably not persuade me to reset the mean.
For (ii), the answer is yes, higher precision tools would certainly improve the rejection rate (assuming that the normal distribution model is exact or near to exact).  Higher precision tools decrease the standard deviation.  If we decrease the standard deviation to, for example, half of its current value, that is, from $0.3$ mm to $0.15$ mm, then the USL would be $6.4$ standard deviation units above the mean of $20$, and the LSL would be $7.0$ standard deviation units below the mean. With a normal, this kind of extreme deviation from the mean essentially never happens.
Even a far more modest change in the standard deviation, from $0.3$ mm to $0.25$ mm, would make a dramatic difference to the rejection rate. For then the USL is $3.84$ standard deviation units above the mean, and the LSL is $4.2$ standard deviation units below.  Tail probabilities in the normal after a while decrease very rapidly.  Rejections would be overwhelmingly because of the USL, and the probability of rejection would be about $0.00006$, which is one-fifteenth of the current one!
