Why we define the variance of a random variable $X$ as $\text{var}[X]=\text{E}[(X-\mu)^2]$ instead of $\text{var}[X]=\text{E}[\left|X-\mu\right|]$.  Normally we understand the standard deviation $\sigma=\sqrt{\text{var}[X]}$ as a measure of the average distance of a sample from the mean. If this is the case, isn't it more reasonable to use the absolute value instead of squaring as a measure of distance? Then this way we don't have to square afterwards to obtain $\sigma$.
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Let's take the numbers 0-10.  Their mean is 5, and the individual deviations from 5 are -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 And so the average (magnitude of) deviation from the mean is $30/11 \approx 2.72$.  However, this is not the standard deviation.  The standard deviation is $\sqrt{10} \approx 3.16$.  The first mean-deviation is a simpler and by far more intuitive definition of the "standard-deviation", so I'm sure it's the first definition statisticians worked with.  However, for some reason they decided to adopt the second definition instead.  What is the reasoning behind that decision?
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I know that the variance of the difference of two independent variables is the sum of variances, and I can prove it. I want to know where the covariance goes in the other case.
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I wish to use the  to calculate the variance of a normal-distributed function. For this, I need the expected value of $X$ as well as the one of $X^2$. Intuitively, I would have assumed that $E(X^2)$ is always equal to $E(X)^2$. In fact, I cannot imagine how they could be different.  Could you explain how this is possible, e.g. with an example?
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I have a basic question.  Say I have two random variables, $X$ and $Y$.  I can standardize them by subtracting the mean and dividing by the standard deviation, i.e., $X_{standardized} = \frac{(X - E(X))}{(SD(X))}$.  Is the correlation of $X$ and $Y$, $Cor(X, Y)$, the same as the covariance of the standardized versions of $X$ and $Y$?  That is, is $Cor(X, Y) = Cov(X_{standardized}, Y_{standardized})$?
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According to  (my emphasis),     In statistics, the standard deviation ... is a measure that is used to quantify the amount of variation or dispersion of a set of data values. A standard deviation close to 0 indicates that the data points tend to be very close to the mean (also called the expected value) of the set, while a high standard deviation indicates that the data points are spread out over a wider range of values.   The purpose of the standard deviation (SD), then, is to tell us how varied or uniform (SD → 0) the data is.  So, given a certain SD, how varied is the data? What does the size of the standard deviation mean?  Please explain the meaning of the SD by interpreting an SD = 1 (M = 0).  If you cannot interpret the size (quantity) of this SD, what other information would you need to be able to interpret it, and how would you interpret it, given that information? Please provide an example.  If, on the other hand, the quantity of the SD cannot be qualified in this manner, my argument is that it is essentially meaningless. If you disagree, please explain the meaning of the SD.  &nbsp;    The following are earlier versions to give context to the answers. Unfortunately these didn't really convey what I wanted, and my attempt to ask it elsewhere was closed. (I don't need these versions answered now):  First revision:  What does the size of the standard deviation mean?  For example, if I want to study human body size and I find that adult human body size has a standard deviation of 2 cm, I would probably infer that adult human body size is very uniform, while a 2 cm standard deviation in the size of mice would mean that mice differ surprisingly much in body size.  Obviously the meaning of the standard deviation is its relation to the mean, and a standard deviation around a tenth of the mean is unremarkable (e.g. for IQ: SD = 0.15 * M).  But what is considered "small" and what is "large", when it comes to the relation between standard deviation and mean? Are there guidelines similar to the ones that Cohen gives for correlations (a correlation of 0.5 is large, 0.3 is moderate, and 0.1 is small)?    Original question:  We always calculate and report means and standard deviations. But what does the size of the variance actually mean?  For example, assume we are observing which seat people take in an empty room. If we observe that the majority of people sit close to the window with little variance, we can assume this to mean that people generally prefer siting near the window and getting a view or enough light is the main motivating factor in choosing a seat. If on the other hand we observe that while the largest proportion sit close to the window there is a large variance with other seats taken often also (e.g. many sit close to the door, others sit close to the water dispenser or the newspapers), we might assume that while many people prefer to sit close to the window, there seem to be more factors than light or view that influence choice of seating and differing preferences in different people.  At what values can we say that the behavior we have observed is very varied (different people like to sit in different places)? And when can we infer that behavior is mostly uniform (everyone likes to sit at the window) and the little variation our data shows is mostly a result of random effects or confounding variables (dirt on one chair, the sun having moved and more shade in the back, etc.)?  Are there guidelines for assessing the magnitude of variance in data, similar to Cohen's guidelines for interpreting effect size (a correlation of 0.5 is large, 0.3 is moderate, and 0.1 is small)?  For example, if 90% (or only 30%) of observations fall within one standard deviation from the mean, is that uncommon or completely unremarkable?
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How can I prove that a random variable taking values in $[0,1]$ has variance no larger than $\frac{1}{4}$? If it matters, discrete and continuous proofs are both welcome.
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What is a standard deviation, how is it calculated and what is its use in statistics?
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How is the notation $X\sim N(\mu,\sigma^2)$ read? Is it $X$ follows a normal distribution? Or $X$ is a normal distribution? Or perhaps $X$ is approximately normal..  What if there are several variables that follow (or whatever the words is) the same distribution? How is it written?
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The mean absolute deviation is:   $$\dfrac{\sum_{i=1}^{n}|x_i-\bar x|}{n}$$   The variance is: $$\dfrac{\sum_{i=1}^{n}(x_i-\bar x)^2}{n-1}$$   So the mean deviation and the variance are measuring the same thing, yet variance requires squaring the difference. Why? Squaring always gives a non-negative value, but the absolute value is also a non-negative value. Why isn't it $|x_i-\bar x|^2$, then? Squaring just enlarges, why do we need to do this?  A similar question is , but mine is a little different. Thanks.
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I was wondering if there are any distributions besides the normal where the mean and variance are independent of each other (or in other words, where the variance is not a function of the mean).
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Equivalently, about variance?  I realize it measures the spread of a distribution, but many other metrics could do the same (e.g., the average absolute deviation). What is its deeper significance? Does it have    a particular geometric interpretation (in the sense, e.g., that the mean is the balancing point of a distribution)?  any other intuitive interpretation that differentiates it from other possible measures of spread?   What's so special about it that makes it act as a normalizing factor in all sorts of situations (for example, convert covariance to correlation)?
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I'm struggling to understand the difference between the standard error and the standard deviation. How are they different and why do you need to measure the standard error?
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I know that "variable" means "values which vary." In a simple linear regression model :  $$Y=\beta_0+\beta_1X+\epsilon$$  $X$ is variable that is the values of $X$ vary. Why is $X$ not a random variable? What is the difference between a variable and a random variable?
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My question is told in a few words: Why do you write $E[X]$ in square brackets instead of something like $E(X)$? Probably it is not a "function". How would you call it then? This question also applies for $Var[X]$.
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We have the following Random variable:  $Y=\frac{\log Z}{\log Z+\log (1-W)}$ where both $W$ and $Z$ are $U(0,1)$. The question asks to calculate the variance of the random variable $Y$.   I think the random variable $Y$ will follow the exponential distribution with parameter $2$. How I obtained this result is as follows:  First, after a little simplification, the random variable $Y$ turned out to be $\log ZW$ which can further be written as $\log Z+ \log W$. Each of which follows exponential distribution. Hence, the variance of the random variable$Y$ can be obtained as $\frac{1}{4}$. What is wrong in this approach? Because at the back, the answer is $\frac{1}{12}$ as they are saying that $Y$ will follow uniform distribution.
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My random variable $X$ is normally distributed with mean $b$ and variance $p$. I defined a new RV, $Y$, such that $Y=1/X$. Does anyone know how to find the mean of $Y$ and $Y^2$?
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To find the Variance of a Wiener Process, $Var[W(t)]$, I have to compute the integral $$ Var[W(t)]=\dots=\frac 1 {\sqrt {2 \pi t}}\int_{-\infty}^{\infty} x^2 e^{-\frac {x^2} {2t}}dx=\dots=t. $$ I've tried integration by parts to solve the integral but end up with $$ \dots=\frac 1 {\sqrt {2 \pi t}} \left(0 - \int_{-\infty}^{\infty} -\frac {x} {t} \cdot e^{-\frac {x^2} {2t}}\cdot \frac {x^3} 3 dx\right), $$ which is even worse and probably wrong.  Can anyone please help me compute the first integral and show how it becomes equal to $t$?  (I know that the Variance for a Wiener Process (Standard Brownian motion)is defined as $t$ but want to prove it with the integral above.)