# Understanding “variance” intuitively

What is the cleanest, easiest way to explain someone the concept of variance? What does it intuitively mean? If one is to explain this to their child how would one go about it?

It's a concept that I have difficulty in articulating - especially when relating variance to risk. I understand it mathematically and can explain it that way too. But when explaining real world phenomena how do you make one understand variance and it's applicability in the 'real world', so to speak.

Let's say we are simulating an investment in a stock using random numbers (rolling a die or using an excel sheet, doesn't matter). We get some 'return on investment' by associating each instance of the random variable to 'some change' in the return. Eg.:

Rolling a 1 implies a change of 0.8 per \$1 in investment, a 5 a change of 1.1 per \$1 and so on.

Now if this simulation is run for about 50 times (or 20 or 100) we will get some values and the final value of the investment. So what does 'variance' actually tell us if we were to calculate it from the above data set? What does one "see" - If the variance turns out to be 1.7654 or 0.88765 or 5.2342 what does this even mean? What did/can I observe about this investment?? What conclusions can I draw - in lay man terms.

Please feel free to augment the question with that for standard deviation too! Although I feel it's 'easier' to understand, but something that would contribute to making it also 'intuitively' clear would be greatly appreciated!

• Shouldn't we merge this question with the same one asked last year? – whuber Oct 26 '11 at 15:10
• @whuber I think these should be merged. Having several time the same question (even if here the context is different) reduces the average quality of answers. – robin girard Oct 26 '11 at 15:19
• I'm okay with it being merged but I know how to calculate variance and it's use in statistics too. I want to be able to articulate this concept to people who wouldn't know anything about it and it takes a long while to do so and hence the question. The intent is rather different from the question on SD, IMHO – PhD Oct 26 '11 at 19:10
• I don't think any of you are doing a very good job of answering this in a way that a Layman would understand. I see a lot of assumptions being made and almost every answer ends with something that needs to be interpreted. I'm not complaining, just trying to point that out. I too can't answer the question simply. Maybe it's too difficult? – user98013 Dec 12 '15 at 18:16
• I don't think any of the answers below answered the question here. The question, as I interpret it, is more about variance as a number, when is it considered large or small. The top answer below for example, addresses the question what large variance vs small variance means. If I give you a dataset that you cannot reasonably visualize, so that you have to rely on the numbers, how can you tell if the variance is large/small? – Pig Nov 6 '16 at 6:16

I would probably use a similar analogy to the one I've learned to give 'laypeople' when introducing the concept of bias and variance: the dartboard analogy. See below:

The particular image above is from Encyclopedia of Machine Learning, and the reference within the image is Moore and McCabe's "Introduction to the Practice of Statistics".

EDIT:

Here's an exercise that I believe is pretty intuitive: Take a deck of cards (out of the box), and drop the deck from a height of about 1 foot. Ask your child to pick up the cards and return them to you. Then, instead of dropping the deck, toss it as high as you can and let the cards fall to the ground. Ask your child to pick up the cards and return them to you.

The relative fun they have during the two trials should give them an intuitive feel for variance :)

• So what does it 'mean'? If someone would see the the statistical variance of the darts on the board, what would they conclude? What does it mean to have low/high variance intuitively speaking... – PhD Oct 26 '11 at 3:36
• I'd say something like: Let's say we threw 4 darts. The number of hands required to remove the darts from the board all at once increases as variance of the dart positions increases (Note: very informal argument here as there a number of counterexamples, such as when 3 darts are grouped together and the last dart is on the wall 3 feet from the darboard). – user5594 Oct 26 '11 at 21:05
• Your diagram also seems to resonate the classical way of distinguishing precision and accuracy too! It just hit me! – PhD Oct 26 '11 at 21:43
• AAAAAAAAAAAH! Nice exercise! Good way to show someone what it means to have low/high variance! The average distance from the average value (mean) of the data points :) – PhD Oct 26 '11 at 21:56
• (+1) The dartboard-analog to demonstrate the difference between bias and variance is simply brilliant – steffen Dec 1 '11 at 15:16

I used to teach statistics to a layman by jokes, and I found they learn a lot.

Suppose for variance or standard deviation the following joke is quite useful:

Joke

Once two statistician of height 4 feet and 5 feet have to cross a river of AVERAGE depth 3 feet. Meanwhile, a third statistician comes and said, "what are you waiting for? You can easily cross the river"

I am assuming that layman know about 'average' term. You can also ask them the same question that would they cross the river in this situation?

What are they missing that is 'variance' to decide "what to do in the situation?"

It's all about your presentation skills. However, jokes help a lot to the layman who wants to understand statistics. I hope it helps!

• Maybe I'm not good with statistical jokes (I'm am quite good with the others though :). But I don't think I understand what is meant by "what to do in the situation"? What 'exactly' should one do if they have an idea of the variance? How should one interpret it? – PhD Oct 26 '11 at 3:38
• @Nupul: Actually, "what to do in the situation" means either they cross a river or not? If you know the variance (or SD) then you could decide it easily. Suppose variance is 0.25 (SD=0.5) then they can cross the river safely because range of interval (dont confuse this with confidence Interval (CI)) is 3+0.5 or 3-0.5, and their heights are 4 and 5. If variance is 4 then better to not cross the river. By the way, just enjoy jokes here stats.stackexchange.com/questions/1337/statistics-jokes – Biostat Oct 26 '11 at 7:43
• Perfect! I got it! :) That makes a lot of sense. In fact combining the answers from various people helps me frame the understanding better... – PhD Oct 26 '11 at 21:44
• Or, if sharks don't 'on average' eat people, that's little comfort if they are very moody (highly variant behavior). In the river analogy it's about whether you will take a step that will put you over your head. – Dean Radcliffe Nov 26 '12 at 6:28

I would focus on the standard deviation rather than the variance; the variance is on the wrong scale.

Just as the average is a typical value, the SD is a typical (absolute) difference from the average. It's not unlike folding the distribution over at the average and taking the average of that.

• Agreed. Let's say we focus on SD. My question still stands as to how to make someone understand SD intuitively other than 'high SD doesn't seem good'...how would I explain SD to a lay person since it's the square root of variance!!! – PhD Oct 26 '11 at 3:39
• @Nupul - Read my second paragraph: I would explain the SD as the typical difference from the average. – Karl Oct 26 '11 at 15:39
• "It's not unlike folding the distribution over at the average and taking the average of that." That comment, like the rest of your post, seems to describe mean absolute deviation, not standard deviation. – Macro May 15 '12 at 18:59
• @Macro - yes; in trying to explain the SD, I would approximate it by the MAD. I think it's best not to quibble over root-mean-square vs mean absolute value. – Karl May 19 '12 at 6:53

I disagree with a lot of the answers advocating people to purely think of variance as spread. As smart people (Nassim Taleb) have pointed out, when people think of variance as spread they just assume it is MAD.

Variance is a description of how far members are from the mean, AND it judges each observation's importance by this same distance. This means observations far away are judged more importantly. Hence squares.

I think the variance of a continuous uniform variable is the easiest to picture. Each observation can have a square drawn to it. Stacking these squares creates a pyramid. Cut the pyramid in half so half the weight is in one side and half is in the other. The face where you cut it is the variance.

• I don't know why this answer wasn't upvoted more. The point made in the second paragraph is crucial to understanding variance and differentiating it from MAD, which as correctly pointed out is what people intuitively think of when told about "measure of spread". And it's not beyond a layman to understand the idea that the weight given to a point's distance from the mean doesn't grow linearly, even if they don't understand squares mathematically. – jeremy radcliff Jul 6 '16 at 23:05
• "MAD" = en.wikipedia.org/wiki/Median_absolute_deviation for those who are wondering. I don't think such acronyms should be assumed knowledge on a question like this. – user136692 Nov 8 '16 at 3:09

Have a lot of practice giving lectures about standard deviation and variance to a novice audience.

Lets assume, one knows about average already. By average (or e.g. median) - one gets a single value from many measurements (that is how one usually uses them). But it is very import to say, that knowing some average is not enough at all. The second half of the knowledge is what is the error of the value.

Skip the next 2 paragraphs of motivation if lazy

Lets say you have some measurement device, that costed 1 000 000\$. And it gives you the answer: 42. Do you think one paid 1 000 000\$ for 42? Phooey! 1 000 000 is paid for the precision of that answer. Because Value - costs nothing without knowing its Error. You pay for the error, not the value. Here is a good live example:

Commonly, we use a ruler to measure a distance. The ruler provides a precision around one millimeter (if you use metric system). What if you have to go beyond and measure something with like 0.1mm precision? - You probably would use a caliper. Now, it is easy to check, that a cheap ruler with mm scale costs cents, while reliable caliper costs ~$10. Two orders of magnitude in price for one order of magnitude in precision. And that is very usual ratio of how much one pays for smaller errors. 1. The problem. Lets say we have a thermometer (Choose a measurement device depending on what is closer to auditory). We did N measurements of the same temperature and thermometer showed us something like 36.5, 35.9, 37.0, 36.6, ... (see the pic). But we know that the real temperature was the same all the time, and values are different because in every measurement the thermometer lies to us a bit. We can calculate the average (see red line on the picture below). Can we believe it? Even after averaging, does it have enough precision for our needs? For human health estimation for example? How can one estimate how much this little scum lies to us?  1. Max deviation - the easiest but not the best approach. We can take the farthest point, calculate the distance between it and the average (red line) and say, that this is how thermometer lies to us, because it is maximum error we see. One could guess, this estimation is too rough. If we look at the picture, most of the points are around the average, how can we decide just by one point? Actually one can practice in naming reasons why such estimation is rough and usually bad. 2. Variance. Then... lets take all distances and calculate an average distance from the average (on picture - average distance between each point and the red line)! BTW, how to calculate a distance? When you hear the "distance" it translates to "subtract" in math. Thus we start our formula with $$(x_{i} - \bar{x})$$ where $$\bar{x}$$ is the average (red line) and $$x_{i}$$ is one of the measurements (points). Then one could imagine that the formula of average distance would be summing everything and dividing by N: $$\frac{\sum(x_{i} - \bar{x})}{N}$$ But there is a problem. We can easily see, eg. that 36.4, and 36.8 are at the same distance from 36.6. but if we put the values in the formula above, we get -0.2 and +0.2, and their sum equals 0, which is not what we want. How to get rid of the sign? At this points someone usually says "Take the absolute value of each point!". Taking an absolute value is actually a way to go, but what is the other way? We can square the values! Then the formula becomes: $$\frac{\sum(x_{i} - \bar{x})^{2}}{N}$$. This formula is called "Variance" in statistics. And it fits Much better to estimate the spread of our thermometer (or whatever) values, than taking just the maximum distance. 3. Standard deviation. But still there is one more problem. Look at the variance formula. Squares make our measurement units... squared. If the thermometer measures the temperature in °C (or °F) then our error estimation is measured in $$°C^{2}$$ (or $$°F^{2}$$). How to neutralize the squares? - Use the square root! $$\sqrt{\frac{\sum(x_{i} - \bar{x})^{2}}{N}}$$ So here we come to the Standard Deviation formula which is commonly denoted as $$\sigma$$. And that is the better way to estimate our device precision. Hope it was easy to understand. From this point it should be easy go to "68–95–99.7 rule", sampling and population, standard error vs standard deviation terms Etc. P.S. @whuber pointed out a good related QA - "Why square the difference instead of taking the absolute value in standard deviation?" • History, this site, and the experience of others differ with you concerning "taking an absolute value is a little artificial." Historically, taking the absolute value was the first idea considered and it was not supplanted by taking a root mean square for at least a generation. See "Roger Boscovich and the Figure of the Earth" in S. Stigler, The History of Statistics (1986). – whuber Jan 20 at 15:09 • Good QA. There is actually no offense to absolute values here. They are used a lot and, might be the right answer and another good lecture could be about e.g. norms. But here we should come to STD. Using this line of discussion, the first thing students usually say "absolute values" then. Putting too much discussion about abs-vs-sqr/t is too much for the answer. Finally "artificial" - no offense to absolute values here. It is just rather property. It has 'if' in its definition, non diff at zero. So it is probably just my bad wording because of editing. – MajesticRa Jan 23 at 0:27 Maybe this could help. I apologise in advance that as complete amateur I may get this wrong. Imagine you ask 1000 people to correctly guess how many beans are in a jar filled with jelly beans. Now imagine that you are not necessarily interested in knowing the correct answer (which may be of some use) but you wish to get a better understanding of how people estimate the answer. Variance could be explained to a lay person as the spread of different answers (from highest to lowest). You could continue by adding that if enough people were to questioned the correct answer should lie somewhere in the middle of the spread of 'guestimates' given. I now refer to some of my more esteemed colleagues for adjudication I was sitting down trying to puzzle out variance and the thing that finally made it click into place for me was to look at it graphically. Say you draw out a number line with four points, -7, -1, 1 and 7. Now draw an imaginary Y axis with the same four points along the Y dimension, and use the XY pairs to draw out the square for each pair of points. You wind up with four separate squares consisting of 49, 1, 1, and 49 smaller squares, each. Each of them contributes to an overall sum of squares which, itself, can be represented as a large 10 x 10 square with 100 smaller squares overall. Variance is the size of the average square contributing to that larger square. 49 + 1 + 49 + 1 = 100, 100/4 = 25. So 25 would be the variance. The standard deviation would be the length of one of the sides of that average square, or 5. Obviously this analogy does not cover the full nuance of the concept of variance. There are a lot of things that need explained, such as why we often use a denominator of n-1 to estimate the population parameter, instead of simply using n. But as a basic concept to peg the rest of a detailed understanding of variance to, simply drawing it out so I could see it helped immensely. It helps understand what we mean when we say that variance is the average squared deviation from the mean. It also helps in understanding just what relationship SD has to that average. • Welcome to Cross-Validated! I like the approach, but it might be even more helpful to emphasize that the points are spread 'around' zero (i.e., they have zero mean) and you're measuring the spread relative to an "atom" located there. (+1) and I look forward to seeing more answers from you! – Matt Krause Apr 25 '16 at 2:20 I think the key phrase to use when explaining both variance and standard deviation is "measure of spread". In the most basic language, the variance and standard deviation tell us how well spread out the data is. To be a little more accurate, although still addressing the layman, they tell us how well the data is spread out around the mean. In passing, note that the mean is a "measure of location". To conclude the explanation to the layman, it ought to be highlighted that the standard deviation is expressed in the same units as the data we're working with and that it is for this reason that we take the square root of the variance. i.e. the two are linked. I think that brief explanation would do the trick. It's probably somewhat similar to an introductory textbook explanation anyway. I regard the variance of distribution as the moment of inertia with the axis that at the mean of the distribution and each mass as 1. This intuition would make the abstract concept concrete. The first moment is the mean of the distribution and the second moment is the variance. I would call it the average positive difference from the overall average. • Until you clarify the two kinds of "average" you mean (the first one is the$L^2\$ average and the second one is the arithmetic mean), it is almost certain your statement will be interpreted in ways that make it incorrect. Moreover, the term "positive difference" is strange and ambiguous: do you mean to consider only the positive residuals? Or to take the absolute values of the residuals? Or something else? – whuber Sep 21 '17 at 13:55