Q:

How many details should be measured

How many details do you need to measure, to gain a probability more than 0.9, that the chosen details mean length differs from the whole batch expected length mean not more than 0.001 centimetres? You also know, that the standard deviation of the detail length does not exceed 0.04.
What I did:
Found $DX = 0.0016$
The mean - expected mean cannot exceed 0.001
Divided $DX / a^2$, where a = 0.001.
Got the answer: >= 16000
(this most likely is wrong).

A:

Sorry for the confusion on the post. This is how I solved it in the end.
$p >= 1-DX/(n*a^2)$
p is 0.9, DX = 0.0016, a = 0.001, n is missing.
$0.9 >= 1 - 0.0016/(n*0.001^2)$
$-0.1n >= -0.016/0.001^2$
$n >= 16000$
I stated that answer before as well, but I got that answer with a typo in the calculation, this should be the somewhat more of a correct way to do it.

