striphe said:
Since you specifically referred to the normal distribution in your most recent question, Chebyshev's theorem is not needed to answer any question about the percentage within one standard deviation of the mean - it wouldn't even be appropriate since, if you know or are assuming normality, you an leverage that to get the 68% value. (Whether normality is a reasonable assumption is an entirely different question.) However, even this is poorly worded.
* If you are discussing only the population, then you need to work with the parameters, and \mu \pm \sigma contains roughly the central 68% of the distribution
* If you have a sample which you've deemed to be mound-shaped and symmetric in its distribution, then \overline x \pm sd contains roughly 68%
of the sample values
Now the population MD is \sigma \sqrt{\, \frac 2 {\pi}} for a normal distribution, so (population again)
<br />
\mu \pm MD \sqrt{\, \frac{\pi} 2} <br />
will contain roughly the central 68% of the population. A similar comment could be made for the sample versions
when the sample distribution is mound-shaped and symmetric
However, if the sample is skewed, there is not (that I know of) anything like Chebyshev's theorem for using \overline x and MD. (The idea above won't work, since the
simple relationship between SD and MD doesn't hold without the normality assumption).