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Is it possible for all the sample values to lie within 1 standard deviation of the sample mean? (Given that the distribution is normal)

Would like to know the answer both in therms of Theory and Practically.

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    $\begingroup$ The sample values will be whatever they are, and the calculation of their SD has absolutely nothing to do with your assumptions about the underlying distribution. With that clarification, a strong clue to the answer lies in Chebyshev's Inequality. A careful examination of its proof will lead you directly to the Bernoulli$(1/2)$ distribution, which will produce samples that have approximately equal numbers of zeros and ones: you will find it instructive to work out what the the mean and SD is. $\endgroup$ – whuber Jun 19 '17 at 14:16
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You've tagged this "normal-distribution", so I'll confine myself to answers which assume you're drawing your samples from an underlying normal distribution.

The trivial answer is yes: if all your samples are the same value, then they all fall within one standard deviation from the sample mean. This is easy to achieve if you have only one sample value. If you have more than one sample value, the probability of getting that sample value set is infinitesimal with the standard (real-valued) normal distribution - infinitesimal, but not theoretically impossible.

If we restrict ourselves to non-zero probability multi-valued samples, the answer hinges on which standard deviation you're talking about: the standard deviation of the sample, or the standard deviation of the underlying normal distribution.

If it's the standard deviation of the underlying normal distribution, the answer is still yes. In fact, for each sample value, there's a ca. 68.2% chance that a given value falls within one standard deviation (of the underlying distribution) from the underlying distribution's mean. So that's $0.682^N$ chance that all $N$ samples fall within one standard deviation of the underlying mean. This falls off rapidly, but is still a non-zero probability even for large N. Relaxing the requirement that it's within one (underlying) standard deviation of the underlying mean to needing to be within one (underlying) standard deviation of the sample mean increases the probability a bit further.

If it's the standard deviation of the sample, then the answer turns to no, for non-trivial multi-valued sample. Recall that the standard deviation is

$$ \sigma = \sqrt{ \frac{1}{N} \sum { (x - \bar x )^2 }}$$

If you have more than one sample value, and they don't all have the same absolute deviation from the mean, the average under the radical will be less than $(x_{max} - \bar x)^2$ and thus $x_{max}$ will fall outside the standard deviation around $\bar x$. (And you will almost never draw a multi-valued sample from an actual normal distribution with all entries having exactly the same absolute deviation from the sample mean.)

Things change slightly if instead of the "standard deviation of the sample" you use the "sample standard deviation". (That is, the unbiased estimator of the population standard deviation, calculated from the sample values.) This calculation is then:

$$ \sigma = \sqrt{ \frac{1}{N-1} \sum { (x - \bar x )^2 }}$$

For this to work, you need to have the absolute deviations of the $x$s close enough to each other that the $N-1$ correction extends the sample standard deviation enough to include all of them.

For example, -10, 10, -9, and 9 have a sample mean of 0, and a sample standard deviation of 10.98, meaning that they all lie within one sample standard deviation of the mean.

Note that this sort of arrangement is somewhat precarious. If the sample was instead -10, 10, -9, and 7 - or even -10, 10, -9, 9, and 0, the -1 correction to $N$ would no longer be enough to spread the sample standard deviation to cover the most extreme values. However, having a set of values which do fall in the appropriate is not a zero probability event, as there is a variety of numbers which would also work. (e.g. -10, 10, -9, and 9.5 or -10, 10, -9, and 8)

So while it is possible (non-zero probability) to get multiple values from a normal distribution which have a sample standard deviation large enough around the sample mean to cover all of them, the probability of doing so is rather low.

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  • $\begingroup$ +1 That's one way to deal with an ambiguous question--cover all the possible interpretations! $\endgroup$ – whuber Jun 19 '17 at 16:36

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