1
$\begingroup$

I'm pretty sure I'm seeking standard deviation and finding the mass of my 'widgets' that exist outside my SD. Here's the context.

I just sorted a bunch of widgets into two piles. One pile is known bad widgets and are being deconstructed. I want to quality control/quality assurance the other pile of widgets.

These widgets are made up of four distinct components, components a, b, c, and d. Three of said components are determinant and their mass will vary only slightly. For simplicity their mass is negligible thus of course the characteristics of every completed widget can vary only slightly piece to piece.

The differences between the masses of components a, b and c, should be negligible to the difference in component d; ie: a known good mass of component d should be 22.2 grams. Component a will almost always equal 55 grams +/- a few hundredths of a gram, the same goes for components b and c; and thus should remain relatively constant. The greatest difference in weight from widget to widget should be seen in component d.

If I can calculate the population mean of the mass then I can collect the standard deviation of the population.

For any widget that has a mass outside 1 or 2 standard deviations those widgets are candidates for deconstruction.

I think my logic is sound. I just need to do the math now. This means taking records of every widgets weight, finding the SD, and then using that to pull any anomalies out of the population.

Am I on the right track here? Or should I be considering some sort of distribution to help me judge what of my widgets I should be deconstructing?

Moderators, I'm not sure if this sort of question belongs here, or in Math. Please advise. Thank you.

$\endgroup$
7
  • 1
    $\begingroup$ As I understand your problem, you have a bunch of widgets indexed by $i=1, \ldots, n$. For each widget, you have some measurement $x_i$. It seems to me that it's imprecise to reject a widget simply if $x_i \notin (-2\sigma, 2\sigma)$ where $\sigma = \sqrt{\frac{1}{n} \sum_i (x_i - \bar{x})}$? For example, if $x$ follows the normal distribution, you'd always reject about 5 percent of your sample, regardless of how well the widgets work since approx. 5 percent of observations are outside 2 two standard deviations. That's simply a property of the normal distribution. $\endgroup$ Sep 11 '17 at 21:00
  • $\begingroup$ It seems to me you want to find a region $(x_a, x_b)$ such that if $x \in(x_a, x_b)$ then your widgets fail with sufficiently low probability? What really matters isn't how far $x_i$ is from the mean in units of standard deviation but how far $x_i$ can be from some target value $x_{\mathrm{reference}}$ and still work? $\endgroup$ Sep 11 '17 at 21:00
  • $\begingroup$ I would expect that some of my widgets fall within an acceptable range. But that some of the widgets will have outlandish masses being candidates for deconstruction, those candidates being 'outliers'. $\endgroup$
    – gh0st
    Sep 11 '17 at 21:04
  • 1
    $\begingroup$ If you know the distribution of $x$, you can use your approach to identify the most extreme $p$ percent of observations. Perhaps that's useful? But should you be pulling 5% of widgets? 0.5% of widgets? 50% of widgets? Isn't what really matters whether the widgets work, not their normalized z-score? BTW I know nothing about manufacturing etc... :) so my comments should be understood in that context. $\endgroup$ Sep 11 '17 at 21:07
  • 1
    $\begingroup$ My intuition is that you'd want to know the distribution of $x$ (eg. plot a simple histogram for starters) just because that's interesting. And then you'd want to know how the failure rate is a function of distance from some $x_\mathrm{reference}$? Maybe that data is hard to get? If you can classify what widgets work vs. what widgets don't work, you could run logistic regression to model how failure rate is a function of distance $d_i = |x_i - x_\mathrm{reference}|$. Maybe it turns out that failure rate gets high $2\sigma$ out but maybe not? $\endgroup$ Sep 11 '17 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.