Is the average of the square of the distance each value is from the means1 pointStandard DeviationVarianceMean

To clarify the concepts involved in your question:

1. Mean: This is the average of a set of values. It is calculated by summing all the values and dividing by the number of values.

2. Variance: This measures how far a set of numbers is spread out from their average value. It is calculated as the average of the squared differences from the Mean. Mathematically, it can be represented as: $\text{Variance} = \sigma^2 = \frac{\sum (x_i - \mu)^2}{N}$ where $\mu$ is the mean, $x_i$ represents each value in the dataset, and $N$ is the number of values.

3. Standard Deviation: This is the square root of the variance and provides a measure of the average distance of each data point from the mean. It can be expressed as: $\text{Standard Deviation} = \sigma = \sqrt{\sigma^2}$

So, to answer your question, yes, the average of the square of the distance each value is from the mean is indeed the variance. When you take the square root of the variance, you obtain the standard deviation, which quantifies the amount of variation or dispersion in a set of values.