# What does standard deviation mean in this case?

Suppose that someone was collecting samples and he was trying to estimate the average amount of money people spend on purchasing fast food meals. The calculated mean was USD 50 and the standard deviation was USD 7. I want to know the right meaning of standard deviation from the below answers because I understand the definition but I don't know how this can be applied or what is the benefit of SD in real life.

• Does SD show that, on average, people spend between USD 43 and USD 57 on meals?
• Or, on average, they spend USD 50 per purchase but their purchase amount has a relatively large standard deviation of USD 7?
• Or, on average, they spend USD 50 per purchase but my estimate has a relatively large standard deviation of USD 7?

#### Interpretation of the Mean

When we say that the average value spent on meals was 50 USD - it means that if we take the total amount spent on fast foods and equally divide the sum among all the people who made the purchase - each person would get 50 USD.

However, this number hides a lot of information. We can get an average of 50 USD in a lot of different situations. One extreme is when everyone spends exactly 50 USD. Another extreme is when half the people spend 0 USD and another half spend 100 USD. And there are infinite number of situations in between that would give us a mean value of 50.

#### Average Deviation

Hence, we are interested in the variability of those amounts. One intuitive way to quantify how much variability there is is to calculate the average deviation from that mean value. So when we know the mean value, for each person we can calculate the difference between their spent amount and the mean value, and get the average of that:

$$MAD = \frac{\sum_i | x_i - \bar{x} |}{n}$$

This is "Mean Absolute Deviation" (MAD). It answers the question: among the customers - what is the average difference between their purchase and the average?

We can check what this score would be in the two extreme scenarios. If every purchase was equal to 50 USD then the average would be 50, and the MAD would be 0. And if half of the purchases were 0 and another half 100 then the mean would be 50 and the MAD would be 50.

#### Standard Deviation

Standard deviation is a variant of the MAD, but it's harder to interpret. Note that when we look for averages differences from the mean in MAD calculation - we take the absolute value. We want to get rid of the sign, because otherwise roughly half the deviations will be negative and half - positive, and so they would cancel out. Standard deviation, instead of taking the absolute value, uses the square, which, just like absolute value, transforms negative numbers into positive. And then transforms-back by taking the square root:

$$SD = \sqrt{\frac{\sum_i ( x_i - \bar{x} )^2}{n}}$$

The idea is the same. It is harder to interpret, but it has some nice properties. The reasons why standard deviation is used more often are discussed here: Why square the difference instead of taking the absolute value in standard deviation?

#### Questions

Does SD show that, on average, people spend between USD 43 and USD 57 on meals?

No. What would "on average people spend between X and Y" mean, exactly? Average is a point estimate, not a range. If the amount spent followed a normal distribution that we could derive that around 68% of customers spent between 43 and 57 USD. However, dollar amounts certainly do not follow normal distributions (i.e. they don't have negative values).

Or, on average, they spend USD 50 per purchase but their purchase amount has a relatively large standard deviation of USD 7?

This is correct. But does this answer your question? It restates that the average was 50, and SD was 7. And only adds some external interpretation that 7 is relatively large.

Or, on average, they spend USD 50 per purchase but my estimate has a relatively large standard deviation of USD 7?

No, there is a separate measure for that, called Standard Error.

• It's been a while since I took a stats class... \$7 out of \$50 feels like a fairly large amount; is it fair to say that a standard deviation of \$7 with a mean of \$50 would suggest that the variability in per-person spending is fairly high, where a standard deviation of, say \\$1 would indicate that the actual prices were more tightly grouped? Nov 3, 2021 at 3:20
• @minnmass high versus low depends more on the context of the problem, and maybe prior knowledge we have about the situation. There are ways to try and quantify if a standard deviation is high or low compared with the mean. One such measure is coefficient of variation. But I think we have to accept that at one point we will have to add our interpretation to the numbers. Without that - they will remain just numbers. Nov 3, 2021 at 8:30
• Depending on how 'roughly' is interpreted, "roughly half the errors are positive and half negative" might be deceptive. I would read it casually as saying that, out of 100 samples, maybe not exactly 50 are positive, but maybe 52, or 47. But, of course, all we know for sure is that the positive errors have the same total as the magnitude of the negative errors; we don't know how many of either there are, and there could be, for example, 99 tiny positive errors and 1 large negative error out of 100. Nov 3, 2021 at 17:04
• For my USD, this portion of your answer seems to address op question the best: "If the amount spent followed a normal distribution that we could derive that around 68% of customers spent between 43 and 57 USD" Nov 4, 2021 at 1:03

I'm surprised no one answered with the rule-of-thumb for regarding the standard deviation in data with normal distributions: The 68-95-99.7 rule. (See Wikipedia article.

If you have normally-distributed data, about 68% of observations fall within the range of the mean ± the standard deviation. And about 95% of observations fall within the mean ± two standard deviations.

As far as I can tell, for a uniform distribution, about 58% of observations fall within μ ± σ, and 100% fall within μ ± 2σ.

It appears that for skewed distributions, more of the data would fall within μ ± σ, so that for a common log-normal distribution, perhaps 75% of observations fall within μ ± σ.

The data could also have a bimodal distribution. For a pretty obviously bimodal distribution, perhaps more that 50% of observations fall within μ ± σ.

EDIT: To make these observations more explicitly tied to the question: It depends on the distribution of the data, and on what you mean by "on average", but probably more than 50% of observations fall within μ ± σ , so "on average, people spend between USD 43 and USD 57 on meals" isn't a bad interpretation of the statistics presented.

EDIT 2: I bolded the mentioned distributions to make the logic of my answer more apparent.

• "If you have normally-distributed data..." For the amount of money spent on meals, we probably do not have normally-distributed data or something close. So that's why no one answered with the 68-95-99.7 rule. Nov 2, 2021 at 13:00
• @SextusEmpiricus , I literally addressed four different shapes of distributions in my answer. For each of them, at least for distributions that aren't very extreme, something like 50% to 75% of observations fall within μ ± σ . Given the limited information in the question, I thought this was a very fair way to assess the statement "on average, people spend between USD 43 and USD 57 on meals". ... I'll highlight those distributions in my answer to make the logic more apparent for the reader. Nov 2, 2021 at 14:58
• @SalMangiafico I am referring to your surprise and I gave a reason why it is not so surprising. Nov 2, 2021 at 15:06
• These meals are very likely not normal or uniform distributed distributed. The expressions that you give for the other two distributions, skewed/bimodal, are not exact at all. I would not connect the standard deviation to mean something like "probably more than 50% of observations fall within μ ± σ" (with Chebyshev's inequality we can see that possibly almost 0% falls within μ ± σ). This is why statistics like interquartile range are used. Nov 2, 2021 at 15:10
• The amount of probability mass within one SD of the mean is tied more to kurtosis than skew. Nov 2, 2021 at 23:21

Standard deviation refers to the distribution of the data (or the distribution from which the data were drawn).

Standard error refers to estimating a parameter.

It is not quite right to say that people tend to spend between 43 and 57, but that is closer to the correct interpretation. Some of the confusion comes from the fact that the terms have similar names and that beginning students only see the standard error of the mean, the calculation of which involves the standard deviation. However, you can have a standard error of any parameter you estimate, and the calculation may not involve the standard deviation of the distribution (such as a correlation coefficient).

• So, the best answer from the above 3 options is that people tend to spend between 43 and 57? Or It's not possible to estimate how much the purchase amount varies? Nov 2, 2021 at 10:33
• Under certain assumptions, particularly a normal distribution, it is reasonable to say that the average purchase amount is 50 and typical purchases range from 43 to 57. This looks like your second of the three descriptions.
– Dave
Nov 2, 2021 at 10:37

Suppose that someone was collecting samples and he was trying to estimate the average amount of money people spend on purchasing fast food meals. The calculated mean was USD 50 and the standard deviation was USD 7.

This doesn't tell us enough information. But that is ok, we can tease out the likely answer.

So, our problem is decoding the calculated mean of what was USD 50? Was it some sampling of people spending money on fast food meals? Or, was it every single fast food meal purchased?

In the first case, 50 USD is the sample mean. They checked 10 people, and the average spend was 50 USD.

In the second case, 50 USD is the population mean. They checked every person, and the average spend was 50 USD.

Now, the SD probably means the standard deviation of the points that are averaged. But in the first case, statisticians sometimes get fancy, and attempt to calculate the distribution of the actual population mean given the sampling data gathered. Those statisticians, trying to produce useful information.

Here, 50 USD is the mean of the estimate of the population mean, and 7 USD is the standard deviation on the population mean given the sample data.

• Does SD show that, on average, people spend between USD 43 and USD 57 on meals?

If 50 is the population mean, and the SD is 7, then we are pretty certain that a good percentage of the population spends between USD 43 and 57 on meals; that is 1 SD away from the mean. As it is almost certain that the distribution of meal purchases is going to be somewhat normal, this is a safe bet.

• Or, on average, they spend USD 50 per purchase but their purchase amount has a relatively large standard deviation of USD 7?

This again is a population mean of 50 USD. It is another way to say the previous point, with more precision and less assumptions.

• Or, on average, they spend USD 50 per purchase but my estimate has a relatively large standard deviation of USD 7?

This is a case where we read the USD 50 as an estimate of the population mean.

Which of these is in play is not actually described well by the question.

But based on the test taking strategy (first two answers are variations on each other) and the fact that the initial paragraph talked about estimating the average, the difficulty of actually getting data for the entire population, and the fact that statisticians do this operation a lot, I'd go with this one.

Now, statisticians attempt to estimate the actual average based on the sample data; but in doing so, they make assumptions about the underlying distribution (quite reasonable ones). As an example of how you might do this, you might use Student's T distribution, which assumes a-priori that the purchases made are normal, but with unknown mean and standard deviation.

Then from a set of samples, you can generate an estimate of the population mean and a standard deviation of the error in your estimate.

I'm guessing this is what the situation is being described.