If my SD makes one standard deviation below zero and negative, where the data does not go, then does the 2/3 rule stand with 68% between zero and 1 SD and is the missing negative data -1 SD and – 2 SD weight in percent transferred to + 1 and + 2 SD holding 14 percent not 7 and so on.

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#### Best Answer

Note that 68% is the *population* proportion within one population standard deviation of the population mean *if you have a normal distribution*.

That proportion doesn't apply in general. It's sometimes presented as applying to samples from a normal distribution. In sufficiently large samples, it approximately applies (where 'sufficiently' depends on how 'approximately' you need).

You can in practice have any positive proportion strictly between 0 and 100% within one population standard deviation of the population mean. Similar* limits apply to calculations in terms of sample statistics.

Consider two cases:

`Case1: 10000 observations: a -1, a 1 and 9998 0's Case2: 10000 observations: two 0's, 4999 -1's and 4999 +1's `

In Case 1, 99.98% of the data lies within 1 sample sd of the sample mean.

In Case 2, 0.02% of the data lies within 1 sample sd of the sample mean.

As we see, if we don't have normality, we don't necessarily have anywhere near 68% within 1 sd of the mean. Since the distribution is skewed, it's plainly not normal, so there's no good reason to expect anything remotely close to 68% of the data to be within 1 sd.

Some skewed distributions may get reasonably close to that 68% figure, but 'skewed' doesn't tell us if we're near one of those cases.

The exponential distribution has 86% of its probability within one population 1 sd of the population mean, for example. That's actually a bit nearer to 1 than it is to 68%.

Other skewed distributions show other proportions within 1 sd of the mean. For example, the chi(1) (or half-normal) density gets very close to the same proportion within 1 sd as the normal distribution has:

Still other distributions may show a fair bit more or less.

—

* the $n-1$ denominator in the usual Bessel-corrected sample standard deviation makes a difference.

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