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Assume that accordning to a national statistic people in germany have a $mean$ of $10,000$ € on their bank account and we call that their "assets". Unfortunattely, only the $mean$ is given but not the standard deviation ($sd$). Assuming that the assets are normally distributed, is it possible to say what the $sd$ is? Note that in this example negative values are possible and would mean that someone has debts (maybe there is a more correct word than assets for what I mean but I don't know).

My thoughts so far:

A standard normal distribution has a mean of $0$ and a sd of $1$. Now we can think of the distribution of the assets as a standard normal distribution that has been shifted by $+10,000$. The resulting distribution has $mean$ of $10,000$ and a $sd$ of $1$ and is normally distributed. The problem with that idea is that every distribution can be seen as a shifted normal distribution, hence we would always assume a $sd$ of $1$. This neglects the scale of the variable. For example, for a $mean$ of $10,000$ € the $sd$ would $1$ and for a $mean$ of $1,000,000$ cents the $sd$ would be $1$, too. This doesn't seem correct to me.

Edit:

  1. Because of the answer I want to further explain why I was thinking it would be possible to tell the sd. We call only distributions with certain shapes a normal distribution, i.e. the distribution must have a kurtosis of 3 and a skewness of 0 (both approximately). The sd reflects the shape of a distribution, too, hence, I thought a distribution can't be normal with any sd (or does it?). This lead me to the idea that there must be some sd that the distribution must have if it is supposed to be a normal distribution. What is the mistake in my reasoning?

  2. I changed the example because the variable of the first example can not be normal, hence, the question wouldn't make that much sense (as pointed out by Glen_b).

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    $\begingroup$ In order to estimate the standard deviation, you would need the dataset behind the given mean. There is no way of estimating the standard deviation just given the mean. $\endgroup$ – C. Refsgaard May 6 '19 at 7:24
  • $\begingroup$ I realize you are assuming a normal distribution, but wealth is never distributed normally. It's usually modeled as a Pareto with even fatter tails. $\endgroup$ – John May 6 '19 at 12:03
  • $\begingroup$ Maybe I should have made the question general, i.e. without any example because the problem with the first example was that it coulnd't have negative values (meat consumption) and now the assumption a normality does not hold in reality. Anyway, I leave it as it is and stress an other time that it is rather a general question and I mentioned the example just to clarify what I mean. Obviously, that didn't work out well because the examples were bad. $\endgroup$ – user247002 May 6 '19 at 12:09
  • $\begingroup$ The existing answers correctly point out that the standard deviation is not linked to the mean, for normal distributions. But for e.g. Poisson distributions, the standard deviation is exactly equal to the mean. $\endgroup$ – MSalters May 6 '19 at 12:33
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    $\begingroup$ Perhaps worth pointing out that normal distributions are very often useful distributions to describe common, everyday phenomena, even though they are theoretically incompatible with the possible range of values. Note that even in the bank account example, there's a non-zero probability that someone has more assets/debts than the entire world GDP combined, which is clearly impossible. Many normal distributions "in the wild" have truncated tails at one or both ends - they are not truly normal, but the approximation is oftentimes good enough. $\endgroup$ – Nuclear Hoagie May 6 '19 at 13:33
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The mean and standard deviation of a normal distribution are not linked in any way. Normal distributions can have any real value as a mean and any positive value as a standard deviation.

It follows that if we only know the mean, we can say nothing whatsoever about the standard deviation.

For instance, any of the following four histograms of the simulated meat consumption of 100,000 Germans are consistent with a mean of 132. They just have different standard deviations, which I simply pulled out of my hat.

meat consumption

Now, we can be pretty confident that the SD is not 50, because that would imply an appreciable amount of people with negative meat consumption. Assuming a normal distribution, that is. But even a very small SD of 1 does not completely rule out a negative consumption, because every normal distribution's density extends to the negative numbers.

If I hadn't told R to use the same horizontal axis on all four plots (using the breaks parameter), then the axes would scale to each simulated dataset separately, and all four plots would look like the typical bell shape. See the density plots for the same data:

all normal

R code: 

sds <- c(1,10,20,50)
set.seed(1)
meat_consumption <- sapply(sds,function(sd)rnorm(1e5,132,sd))

par(mfrow=c(2,2))
for ( ii in seq_along(sds) ) {
    hist(meat_consumption[,ii],,xlab="",main=paste("SD:",sds[ii]),
        breaks=seq(floor(min(meat_consumption)),ceiling(max(meat_consumption))))
}

# Density plots
for ( ii in seq_along(sds) ) {
  plot(density(meat_consumption[,ii]),,xlab="",main=paste("SD:",sds[ii]))
}
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    $\begingroup$ Thank you so much! I think I see what my mistake was: From the plots that you show I would assume only the one with sd=50 being a normal distribution because this is how a normal distribution is usually depicted. On the other side, I would not assume the plot with sd= 1 to be normal. But to be sure I've got it right: All of the four distributions are normal, right? $\endgroup$ – user247002 May 6 '19 at 7:40
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    $\begingroup$ Yes, exactly. If I hadn't told R to use the same horizontal axis on all four plots (using the breaks parameter), then the axes would scale to each simulated dataset separately, and all four plots would look like the typical bell shape. $\endgroup$ – Stephan Kolassa May 6 '19 at 7:42
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    $\begingroup$ The very fact that distributions are give as $N(\mu,\sigma)$ indicates that they are not linked; if $\sigma$ could b derived from $\mu$, we would save time and just write $N(\mu)$ (see, for instance, Poisson distributions, which are given in terms of just one parameter since the variance can be found from the mean). $\endgroup$ – Acccumulation May 6 '19 at 15:40
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Meat consumption cannot be exactly normally distributed at any standard deviation; we know this for certain --

  1. it's simply impossible to consume negative amounts of meat, but every normal distribution assigns positive probability to negative numbers.

    If we move instead to say it's approximately normal, then there are limits to how large you can make the standard deviation before it can't be very close to normal, but we can't say how large that is without first saying what we regard as not close enough to be counted as approximately normal.

    To satisfy most people for typical purposes, you'd need the mean to be at least several standard deviations away from 0.

  2. However, it is possible to consume no meat, and any distribution of meat consumption among many countries' populations (German or otherwise) will include at least a small proportion of people who consume none. A recent survey found that 10% of Germans are vegetarian and 1.1% are vegan [1], so if those figures are accurate fully 1/9th of people don't consume meat -- a substantial proportion. Any non-zero fraction at exactly zero consumption means it can't be exactly normal, and 1/9 is not small.

    In particular, a spike of 1/9 at 0 consumption makes something of a mockery of even an assertion of approximate normality.

Leaving issues with the example (which imposed a positivity constraint) aside; if there are no other constraints or requirements other than the mean, the standard deviation may be anything at all; it is not restricted.

[1]: O'Riordan, Tim; Stoll-Kleemann, Susanne (31 August 2015). "The Challenges of Changing Dietary Behavior Toward More Sustainable Consumption". Environment: Science and Policy for Sustainable Development. 57 (5): 4–13.

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    $\begingroup$ "it's simply impossible to consume negative amounts of meat": You are right and this shows that my example is a bad one. As I mentioned, I still have this question in general. You say "here are limits to how large you can make the standard deviation before it can't be very close to normal": This is exact what I was thinking. How can one tell what the limits or what the "perfect" sd for a "perfect" normal distribution is (if this is possible)? $\endgroup$ – user247002 May 6 '19 at 7:44
  • $\begingroup$ As soon as there's any bound on the variable, it can't be perfectly normal. $\endgroup$ – Glen_b May 6 '19 at 7:45
  • $\begingroup$ Again, you can forget my example and see the question as "Assuming a normal distribution: what is the sd for a given mean?" In fact, I change the question so that this doesn't interfere with the idea of normal distribution. $\endgroup$ – user247002 May 6 '19 at 7:46
  • $\begingroup$ That comment was leaving aside your example. If you impose no constraint on the possible values then - even given the mean - the sd may be anything whatever. As soon as you impose any constraint on what values the random variable takes, the distribution cannot be perfectly normal. This applies to every possible case not just the one you mention. $\endgroup$ – Glen_b May 6 '19 at 7:49
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Without considering other data sources, there is no way of estimating the standard deviation of a sample without the actual sample.
The train of thought that you outline simply assumes the standard deviation to be 1, but there's no way of verifying or refuting that without data.

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I would like to pushback on a comment in another answer that

The mean and standard deviation of a normal distribution are not linked in any way.

They might be, in the sense that it might be reasonable that $p\left(\sigma\mid\mu\right) \neq p\left(\sigma\right)$. For the problem at hand, knowledge of the mean perhaps gives you some information about plausible magnitudes of the standard deviation.

Imagine you were are unfamiliar with the currency. For example, let's consider the same problem this time in South Korea. What's the standard deviation of the savings? The problem is that you've perhaps got no idea about the relevant scale. How much is a lot of South Korean Won? How much is little? Well, if I tell you the mean savings are 10 million Won, I think you now have some idea of plausible magnitudes of the deviation. I chose South Korea because I know the relevant scale was bigger than you might expect.

Obviously, though, information about $\mu$ is rarely enough to uniquely determine $\sigma$, unless you have very specific background information about a relationship between the two.

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