Robust Design from ASQ Questions Bank
Hi Friends,

Please help me with this question:

Assuming the variation of the width of each disk in a stack of three disks is normally distributed with a standard deviation of 1 inch, and a specified tolerance for the width of each disk is set at three standard deviation, their statistical tolerance for the width of a stack of three disks would be most nearly which of the followings? The answer is 5.20 inches; How?
3 Replies
Lu Gan
5 Posts
Standard Error equals standard deviation/sqrt of sample size, in this case sample size is 3, square root of 3 is 1.732, so, the specified tolerance is 3, the final number is 3*1.732=5.196
Yes, Lu calculation is correct. There is a difference between the standard deviation of a sample and the standard deviation of a population of individuals.

TIP: If they are talking about tolerance or specification, you need the standard deviation of the distribution of individuals of the population. If you are talking about standard deviation of samples, like subgroups on control charts you use the standard deviation of samples.

This Standard Error of the Mean*, where the standard deviation of samples is equal to the standard deviation of the population divided by the square root of the sample size, is very useful on other ASQ questions.

*Ignoring the "Finite Correction Factor".

Good luck,

A Friend
Hi, Saloumeh. In the case of an Xbar & R control chart, this will be defined by Xbar=Xbarbar+/-A2*Rbar; as Rbar would be the standard deviation of the three disks, and A2 with a sample size = 3 (9 disks), which has an associated A2=1.732; then the variation for the stack of three disks will be 1.732*3=5.20. Tolerance limits and control limits are different things, I think this question is directed to know if you understand the concept of normal distribution and its additive properties.


Humphrey Martinez

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