1. The problem is to understand why population variance and sample variance have different formulas. 2. Population variance measures the spread of an entire population and is calculated using the formula: $$\sigma^2 = \frac{1}{N} \sum_{i=1}^N (x_i - \mu)^2$$ where $N$ is the population size, $x_i$ are data points, and $\mu$ is the population mean. 3. Sample variance estimates the population variance from a sample and uses the formula: $$s^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2$$ where $n$ is the sample size, $x_i$ are sample data points, and $\bar{x}$ is the sample mean. 4. The key difference is dividing by $n-1$ instead of $n$. This is called Bessel's correction. 5. Bessel's correction corrects the bias in the estimation of the population variance from a sample because the sample mean $\bar{x}$ is itself an estimate and tends to be closer to the sample points than the true population mean. 6. Dividing by $n-1$ instead of $n$ makes the sample variance an unbiased estimator of the population variance. 7. In simple terms, sample variance uses $n-1$ to compensate for the fact that we have less information (only a sample) and to avoid underestimating the true variance. Final answer: Population variance divides by $N$ because it uses the entire population data, while sample variance divides by $n-1$ to correct bias and provide an unbiased estimate of the population variance from a sample.