Final answer:
The process becomes incapable when the number of holes drilled exceeds (T - s0) / 0.001, where T is the desired tolerance and s0 is the initial standard deviation.
Step-by-step explanation:
In this problem, we are given that the standard deviation of the machine tooling increases at a rate of 0.001 mm per hole. We need to find the number of holes that can be drilled before the process becomes incapable.
Let's assume the initial standard deviation is s0 and the number of holes drilled is n. The standard deviation after drilling n holes can be represented as:
s = s0 + 0.001n
When the process becomes incapable, the standard deviation is too large for the desired tolerance. Let's assume the desired tolerance is T. The process becomes incapable when the standard deviation exceeds the tolerance, which means:
s > T
Substituting the expression for s, we get:
s0 + 0.001n > T
Now, we can solve the inequality for n:
n > (T - s0) / 0.001
Therefore, the process becomes incapable when the number of holes drilled exceeds (T - s0) / 0.001.