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I am a little unclear if this question makes sense. Say I have a fair die with sides 1 to 6. Can I ask what is the variance of a single roll of the die? The calculation I was thinking was the following. $\mu = 3.5$

$$\frac{1}{6}\times\left[2.5^{2} + 1.5^{2} + .5^{2}\right]\times 2 = 2.91$$

So then the standard deviation is 1.70. Does this further mean that within 3.5 $\pm$ 1.7 is 68% of all the outcomes? (Not sure if this makes sense in this example where prob are same for each outcome)

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    $\begingroup$ Why are you posing this as though you have a normal distribution. $\endgroup$
    – Michael R. Chernick
    Commented Apr 15, 2017 at 5:31
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    $\begingroup$ @Michael Although indeed most references do focus on the normal distribution, the 68-95-99.7 rule applies to many non-normal distributions, especially symmetric ones. The potential misunderstanding here really concerns the way in which a small discrete random variable is likely not to conform to this rule. However, since $3.5\pm 1.7=[1.8, 5.2]$ comprehends four of the six possible results, or $67\%$, the rule works remarkably well. Furthermore $100\%$ of the results lie within two (or three) sds of the mean and both $95$ and $99.7$ are excellent approximations to $100$ in this context. $\endgroup$
    – whuber ♦
    Commented Apr 15, 2017 at 14:13
  • $\begingroup$ Here the OP is talking about a + or - 1 sigma of 68%. Nothing here about 2 or 3 sigma. The "empirical" works approximately for some symmetric unimodal distributions. As you say @whuber, this is a discrete distribution that is not necessarily symmetric. $\endgroup$
    – Michael R. Chernick
    Commented Apr 15, 2017 at 14:25
  • $\begingroup$ @Michael The OP is implicitly referring to the extremely well known "68-95-99.7" rule taught in many (if not most) stats textbooks. I doubt they learned a "68" only rule. BTW, because this is a uniform distribution, it's obviously symmetric. $\endgroup$
    – whuber ♦
    Commented Apr 15, 2017 at 14:35
  • $\begingroup$ So I get it that the 68-95-99.7 rule applies only if you have a sufficiently large sample. But in this case it is just a coincidence that the mean +/-expected standard deviation (is this the right term?) happens to fall in the correct interval? $\endgroup$
    – user3079275
    Commented Apr 15, 2017 at 21:53

2 Answers 2

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I am not sure your question makes sense. Wanting that, here are some ideas about variance:

Variance is a measure of how spread out the sample data are about the mean, or, alternately, how spread out the population values are about the population mean.

If by "a single value" you mean "a single (sample) observation," then the variance must be zero, since the sample mean is just the value of the one observation, and there is no spread of observations about it.

If by "a single value" you mean a single value from the distribution of a fair six-sided die, then, while there is no spread of data around (i.e. a distribution of values above and below) the population mean ($\mu=3.5$), there is a deviation of the single observation from that population mean: $x-\mu$.

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    $\begingroup$ I would even less inclined to attribute sense to the answer 3.5 ± 1.7 is 68%. Take the mean, perhaps there is a circumstance in which this makes sense, but it cannot be an outcome from a die, which only assume integer values. Without also inserting a context, e.g., two adjudicators and a dirty die, there is no way to get 3.5. Out of context, there is no meaning, at least no primary one. $\endgroup$
    – Carl
    Commented Apr 16, 2017 at 11:13
  • $\begingroup$ @Carl We agree... my answer hoped to kindle little sparks of light for the OP, nothing more. :) $\endgroup$
    – Alexis
    Commented Apr 17, 2017 at 18:27
  • $\begingroup$ +1 For making sense in your comment, I would suggest modifying your answer slightly to reflect that as well. $\endgroup$
    – Carl
    Commented Apr 17, 2017 at 18:48
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This is a classic example of 'experts' deliberately misinterpreting the intent of the OP's question and refusing to provide a clear answer on the basis that the original question lacked technical precision. I suspect this is the actual answer you are looking for:

Yes, your calculation is correct. However, your 68% proposal implicitly assumes a normal distribution. For a single die roll, this doesn't make much sense. However, were you to roll a die say 100 times, then a normal distribution would provide a very close approximation to the actual distribution of results. The more times you roll the die, the better the approximation becomes.

However, now you have to scale the Variance according to the number of die rolls. The variance of the total scales according to n (100), while the variance of the average scales according to 1/n. Therefore, if you roll a die 100 times: Total sum : Expected value 350, Variance roughly 17 (101.7) Average : Expected value 3.5, Variance roughly .17 (1/101.7)

And your 68% rule applies. Notice how the expected value and variance of the total sum is 100 * those of the average.

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