Sample Variance Definition Disparity

Observe the following problem (it's fairly simple, but it's the source of my confusion):

You are given a random sample of 10 claims consisting of two claims of 400, seven claims of 800, and one claim of 1600. Determine the empirical skewness coefficient.

Here is the provided solution, word for word:

The sample mean is $0.2(400)+0.7(800)+0.1(1,600)=800$. The sample variance is $0.2(400-800)^2+0.7(800-800)^2+0.1(1,600-800)^2=96,000$. The sample third central moment is $0.2(400-800)^3+0.7(800-800)^3+0.1(1,600-800)^3=38,400,000$. The skewness coefficient is $38,400,000/96,000^{1.5}=1.29$.

The part that trips me up is the fact that the "sample variance" is equivalent to $\frac{\sum_{i=1}^n(X_i-\bar X)^2}{n}$ in this particular example... however, I have seen in countless other places that "sample variance" is instead defined as this: $S^2=\frac{\sum_{i=1}^n(X_i-\bar X)^2}{n-1}$.

So what's the difference between the two? Should one of them be called something different? Or is there some underlying assumption about the first problem I'm not realizing that's accounting for the difference? I just don't understand what the phrase "sample variance" actually refers to. What's going on here?

• They can both be referred to as "sample variance", unfortunately. $\sigma_e^2 = \frac{1}{n}\sum_i (X_i - \bar X)^2$ is the biased sample variance and $S^2 = \frac{1}{n-1} \sum_i (X_i - \bar X)^2$ is the unbiased sample variance. – bames Jan 31 '18 at 5:46
• $\sigma^2 = \frac 1 N \sum_i (X_i - \bar X)^2$ is used when the $N$ values are considered as a population. In this case $\bar X$ may be written as $\mu.$ // For samples, although $\frac 1 n \sum_i (X_j - \bar X)^2$ is biased, it has smaller Mean Square Error than the version with $n-1$ in the denominator. // MSE of estimator $T$ of parameter $\tau$ is $MSE = E[(T-\tau)^2].$ – BruceET Jan 31 '18 at 8:11