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I am working with real estate asset values and prices are in millions of \$.

Typical commercial Asset values are \$10m, \$12m, \$15m, \$9m. Would my standard deviation in this case also be in millions and would it be relative to the mean?

I typically hear 1/2/3 std deviations, so wanted to confirm how std is expressed for large values such as asset prices.

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  • $\begingroup$ Standard deviations aren't relative to anything! Although it's common to compute them with formulas involving mean centering, they can be computed without reference to the mean at all: and that demonstrates the mean is not an essential feature of the SD. $\endgroup$
    – whuber ♦
    Jul 14, 2021 at 15:23

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Standard deviation is just the number outputted by that $s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n(x_i-\bar x)^2}$ formula. If it’s in the millions, so be it.

(There are other ways to calculate standard deviation, but this is the most common.)

What you hear is when people use standard deviation as a unit. If the standard deviation is a million, then two standard deviations above the mean is two million above the mean.

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  • $\begingroup$ Yeah, when I calculate it using that formula, it's is in millions. However, i was wondering if that's how it's expressed as well. Do I say std of 3 million? $\endgroup$
    – kms
    Jul 14, 2021 at 0:44
  • $\begingroup$ No, it’s just 3 standard deviations, the value of which happens to be a million. It’s a measure of how far from the mainstream you are. Being a million above average might sound like a lot, but if you have a standard deviation of a million and a normal distribution, you are not that far above average relative to the variability of the distribution. $\endgroup$
    – Dave
    Jul 14, 2021 at 0:48
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    $\begingroup$ @kms If the variable is in millions of dollars (say), then yes, the standard deviation you calculate will also be in millions of dollars. It's a measure of spread - of 'typical distance' values are apart (but not the actual average distance between values). It's not added to the mean unless you choose to add it to the mean. Why would you be doing that? $\endgroup$
    – Glen_b
    Jul 14, 2021 at 1:26

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