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by jjmarr 743 days ago
> Statistically you're correct,

The Central Limit Theorem (which says if we add a bunch of random numbers together they'll converge on a bell curve) only guarantees that you'll get a normal distribution. It doesn't say where the mean of the distribution will be.

Correct me if I'm wrong, but if your resistor factory has a constant skew making all the resistances higher than their nominal value, a bunch of 6.8K + 6.8K resistors will not on average approximate a 13.6K resistor. It will start converging on something much higher than that.

Tolerances don't guarantee any properties of the statistical distribution of parts. As others have said, oftentimes it can even be a bimodal distribution because of product binning; one production line can be made to make different tolerances of resistors. An exactly 6.8K resistor gets sold as 1% tolerance while a 7K gets sold as 5%.
2 comments
nomel 743 days ago
> Tolerances don't guarantee any properties of the statistical distribution of parts.

That's incorrect. They, by definition, guarantee the maximum deviation from nominal. That is a property of the distribution. Zero "good" parts will be outside of the tolerance.

> It will start converging on something much higher than that.

Yes' and that's why tolerance is used, and manufacturer distributions are ignored. Nobody designs circuits around a distribution, which requires luck. You guarantee functionality by a tolerance, worst case, not a part distribution.

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Dylan16807 743 days ago
> The Central Limit Theorem (which says if we add a bunch of random numbers together they'll converge on a bell curve) only guarantees that you'll get a normal distribution. It doesn't say where the mean of the distribution will be.

That's kind of overstating and understating the issue at the same time. If you have a skewed distribution you might not be able to use the central limit theorem at all.

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_dain_ 742 days ago
>If you have a skewed distribution you might not be able to use the central limit theorem at all.

The CLT only requires finite variance. Skew can be infinite and you still get convergence to normality ... eventually. Finite skew gives you 1/sqrt(N) convergence.

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