Do people care about "Six Sigma" for probability distributions that are not Normal?

Question

Background

At work there is a "Six Sigma" team that focuses on reducing quality defects. They're named that way because six sigma covers 99.7% around the mean (for a normal distribution).

So if a manufacturing process covers six sigma of deviations when the randomness is normally distributed then the process is perhaps good.

Question 1

Is it a thing to talk about how many sigmas are needed to "cover" other distributions? For example, the Gamma or the Poisson?

Question 2

Is the following correct?

• $X$ is a continuous Uniform RV from $(0,1)$ which means the variance is $1/12$ and the SD is $\frac{1}{\sqrt{12}}$

• Therefore "six sigma" covers $6\frac{1}{\sqrt{12}} \approx 1.73$ or all observations

Question 3

I'm trying to calculate how many sigmas to cover the Exponential for any $\lambda$.

Is the following correct?

• This is tough for me to figure out because unlike the binomial and normal, the exponential is not symmetrical around the mean

• I'm going to cop out and calculate how many sigmas cover the pdf starting from zero instead of around the mean...

• The variance of an exponential is $\frac{1}{\lambda^2}$ so sigma is $\frac{1}{\lambda}$ and I want to know how many multiples $m$ of sigma cover 99.7%

$$P(X \le \frac{m}{\lambda}) = 0.997$$

$$1 - e^{-\lambda \frac{m}{\lambda}} = 0.997$$

$$1- e^{-m} = 0.997$$

$$ e^{-m} = 0.003$$

$$ m = -ln(0.003)$$

Therefore, regardless of lambda, to cover 99.7% of the pdf you need 5.8 sigmas (which is close to six sigmas).

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Thanks for your help!!

Answer 1

The "Empirical Rule" states that if $X$ has a 'mound-shaped` distribution, then $P(\mu - \sigma < X \le \mu + \sigma) \approx .68,\,$ $P(\mu - 2\sigma < X \le \mu + 2\sigma) \approx .95,\,$ and $P(\mu - 3\sigma < X \le \mu + 3\sigma) \approx 1.$ These probabilities are nearly correct for a normal distribution.

[Of course Chebyshev's Inequality puts bounds on the second and third probabilities, but the bounds are not always useful in applications. Because Chebyshev's Inequality is a Theorem that applies to very many different distributions, it does not give 'tight' bounds for most distributions.]

Moreover, similar statements for proportions of observations in a normal sample, in terms of sample means and standard deviations, are often close enough to correct to be useful.

If the PDF or a distribution or histogram of a sample is 'mound-shaped' in the sense that it is very nearly normal, it is realistic to expect the approximations to be good approximations. As you have noticed, there some distributions that are not 'mound shaped' for which some of the probabilities happen to match reality. The Empirical Rule is not meant as a theorem, but as the word empirical suggests, it is often useful in practice.

(1) How much attention is paid to the Empirical Rule (ER) in general, and to the third part of it in particular, depends on circumstances. Some elementary texts mention it in passing (sometimes with a contrived sample), and others return to it in a thematic way whenever nearly normal distributions are mentioned. It is important to recognize that a normal sample (especially a small one) may not behave in the same way as the normal distribution from which it taken.

[As a possible example of excessive emphasis: In my opinion, the so-called 'six-sigma' quality management movement places unwarranted attention on the last of the the three statements of the ER, emphasizing cases in which the "effective support" of a distribution spans an interval that is about six standard deviations long, while placing too little attention on situations in which the actual support is an 'interval' of infinite length.]

(2) A uniform distribution is in no sense 'mound shaped', so it is not surprising that the third part of the ER does not apply--even approximately. As you say, if $U \sim \mathsf{Unif}(0,1),$ then $P(1/4 < X \le 3/4) = 1/2,$ which is far from 1.

(3) If $X \sim \mathsf{Exp}(rate = 1/4),$ then $\mu = \sigma = 4,$ and $$P(\mu - 3\sigma < X \le \mu + 3\sigma) = P(-8 < X \le 16) = P(X \le 16) = 0.9817.$$ Also, $P(X \le 6\sigma =24) = 0.9975.$ So there happen to be two senses in which a 'six-sigma' rule works for this exponential distribution. [It may be cheating to invoke the ER here, because an exponential is certainly not 'mound shaped'.]

pexp(16, 1/4)
## 0.9816844
pexp(24, 1/4)
## 0.9975212

Finally, let's explore briefly how well the ER works when applied to sample of size $n = 50$ from $\mathsf{Norm}(\mu=100, \sigma=15).$ The following simulation looks at 100,000 such samples. For each sample we find the proportion ER1 of the 50 observations falling within the interval $(\bar X - S, \bar X + S).$ And similarly for the second and third parts of the ER. It turns out that about 68% of the samples satisfy the first part of the ER, about 95% satisfy the second part, and over 99% satisfy the third part. Many other interpretations and simulations might have been chosen, but this one simulation shows that the ER works about as advertised for samples from a normal population.

set.seed(308)
m = 10^5;  n = 50
ER1 = ER2 = ER3 = numeric(m)
for (i in 1:m) {
   x = rnorm(n, 100, 15);  a=mean(x);  s=sd(x)
   ER1[i] = mean(x > a-s & x < a+s)
   ER2[i] = mean(x > a-2*s & x < a+2*s)
   ER3[i] = mean(x > a-3*s & x < a+3*s) }
mean(ER1); mean(ER2); mean(ER3)
## 0.6825468
## 0.9578796
## 0.9983232