# Tagged Questions

In Probability and Statistics, the standard deviation of a statistical population or data set is a measure of how much variation or dispersion exists from its average value. It is defined as the square root of the variance.

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### How to calculate Standard deviation with mean 0 and Min and max value on x-axis is -1 and 1 respectively?

How to calculate Standard deviation with mean 0 and Min and max value on x-axis is -1 and 1 respectively? It is of-course a normalize distribution. I apologize in advance for stupid question.
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### Adding stochastic variables random variables where one has time component

Based on GLS regression I have identified two random variables lets call them A & B ...
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### Population confidence interval from sample SD and sample mean?

A sample has: $$\text{a sample size } n = 70,$$ $$\text{sample standard deviation } s = 184.43,$$ $$\text{and a sample mean } \bar{x} = 564.15.$$ Compute a 95% confidence interval for the population ...
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### Is the variation of the mean 0, and the variance of standard deviation just the variance?

Just trying to get some properties straight. Was looking at a problem online which reads: Let X be a random variable with mean μ and variance σ2. What is the variance of X/σ+10μ? The solution of ...
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### Relationship between median and standard deviation

I have some data and I added Gaussian noise with zero mean ($\mu=0$) and standard deviation ($SD=SD$). I was interested to see the behaviour of the noise to the data after integrations. I integrated ...
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### Why does variance divide by $n-1$? [duplicate]

The variance is: $$\dfrac{\sum_{i=1}^{n}(x_i-\bar x)^2}{n-1}$$ I read that $n-1$ is used instead of just $n$ when we are measuring the variance of a sample taken from a bigger population. I don't ...
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### Why is variance squared?

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 ...
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### question about Mean absolute deviation formula

$$\frac{\sum_{i=1}^{n}(x_i-\bar x)}{n}$$ This method will not work for calculating the mean deviation. Instead we have: $$\frac{\sum_{i=1}^{n}|x_i-\bar x|}{n}.$$ I'm not quite understanding why ...