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Glen_b
Glen_b
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The standard deviation is the square root of the variance, as you might know. The variance is calculated by summing up the squared deviation from the mean, and dividing it by $n$.

$$\sigma^2 = \frac{\sum_{i}^{n} (x_i-\mu)^2 }{n}$$

Every difference $(x-\mu)$ is squared. When you take the square root of the variance, it is not the same as taking the square root of every $(x-\mu)^2$ and sum it up afterwards...

Because of the square term, the variance (and thus the standard deviation) gives more weight to more distant values and can't be negative, as positive and negative values get both positive when squared.

It is also wrong to calculate the standard deviation for all positive and negative values separately. The values lose their sign somehow, aswhen they get squared.

Hope this helps,

Basil

The standard deviation is the square root of the variance, as you might know. The variance is calculated by summing up the squared deviation from the mean, and dividing it by $n$.

$$\sigma^2 = \frac{\sum_{i}^{n} (x_i-\mu)^2 }{n}$$

Every difference $(x-\mu)$ is squared. When you take the square root of the variance, it is not the same as taking the square root of every $(x-\mu)^2$ and sum it up afterwards...

Because of the square term, the variance (and thus the standard deviation) gives more weight to more distant values and can't be negative, as positive and negative values get both positive when squared.

It is also wrong to calculate the standard deviation for all positive and negative values separately. The values lose their sign somehow, as they get squared.

Hope this helps,

Basil

The standard deviation is the square root of the variance, as you might know. The variance is calculated by summing up the squared deviation from the mean, and dividing it by $n$.

$$\sigma^2 = \frac{\sum_{i}^{n} (x_i-\mu)^2 }{n}$$

Every difference $(x-\mu)$ is squared. When you take the square root of the variance, it is not the same as taking the square root of every $(x-\mu)^2$ and sum it up afterwards...

Because of the square term, the variance (and thus the standard deviation) gives more weight to more distant values and can't be negative, as positive and negative values get both positive when squared.

It is also wrong to calculate the standard deviation for all positive and negative values separately. The values lose their sign when they get squared.

Hope this helps,

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bask0
bask0
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The standard deviation is the square root of the variance, as you might know. The variance is calculated by summing up the squared deviation from the mean and square it, and dividing it by $n$.

$$\sigma^2 = \frac{\sum_{i}^{n} (x_i-\mu)^2 }{n}$$

Every difference $(x-\mu)$ is squared. When you take the square root of the variance, it is not the same as taking the square root of every $(x-\mu)^2$ and sum it up afterwards...

Because of the square term, the variance (and thus the standard deviation) gives more weight to more distant values and can't be negative, as positive and negative values get both positive when squared.

It is also wrong to calculate the standard deviation for all positive and negative values separately. The values lose their sign somehow, as they get squared.

Hope this helps,

Basil

The standard deviation is the square root of the variance, as you might know. The variance is calculated by summing up the deviation from the mean and square it, and dividing it by $n$.

$$\sigma^2 = \frac{\sum_{i}^{n} (x_i-\mu)^2 }{n}$$

Every difference $(x-\mu)$ is squared. When you take the square root of the variance, it is not the same as taking the square root of every $(x-\mu)^2$ and sum it up afterwards...

Hope this helps,

Basil

The standard deviation is the square root of the variance, as you might know. The variance is calculated by summing up the squared deviation from the mean, and dividing it by $n$.

$$\sigma^2 = \frac{\sum_{i}^{n} (x_i-\mu)^2 }{n}$$

Every difference $(x-\mu)$ is squared. When you take the square root of the variance, it is not the same as taking the square root of every $(x-\mu)^2$ and sum it up afterwards...

Because of the square term, the variance (and thus the standard deviation) gives more weight to more distant values and can't be negative, as positive and negative values get both positive when squared.

It is also wrong to calculate the standard deviation for all positive and negative values separately. The values lose their sign somehow, as they get squared.

Hope this helps,

Basil

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bask0
bask0
  • 223
  • 1
  • 7

The standard deviation is the square root of the variance, as you might know. The variance is calculated by summing up the deviation from the mean and square it, and dividing it by $n$.

$$\sigma^2 = \frac{\sum_{i}^{n} (x_i-\mu)^2 }{n}$$

Every difference $(x-\mu)$ is squared. When you take the square root of the variance, it is not the same as taking the square root of every $(x-\mu)^2$ and sum it up afterwards...

Hope this helps,

Basil