Which formula should you use to calculate the variance and which formula should you use to calculate the standard deviation of a sample of $n$ observations $x_1$, $x_2$, ..., $x_n$?

• A. Variance: $\sqrt{\frac{\sum _{i=1}{(x_i-\mathrm{̅x})}^2}{n-1}}$; Standard deviation: $\frac{\sum _{i=1}{(x_i-\mathrm{̅x})}^2}{n-1}$
• B. Variance: $\frac{\sum _{i=1}{(x_i-\mathrm{̅x})}^2}{n+1}$; Standard deviation: $\sqrt{\frac{\sum _{i=1}{(x_i-\mathrm{̅x})}^2}{n+1}}$
• C. Variance: $\frac{\sum _{i=1}{(x_i-\mathrm{̅x})}^2}{n}$; Standard deviation: $\sqrt{\frac{\sum _{i=1}{(x_i-\mathrm{̅x})}^2}{n}}$
• D. Variance: $\frac{\sum _{i=1}{(x_i-\mathrm{̅x})}^2}{n-1}$; Standard deviation: $\sqrt{\frac{\sum _{i=1}{(x_i-\mathrm{̅x})}^2}{n-1}}$

Answer: (D)

Verified step by step guidance

1. Understand that the variance of a sample measures the average squared deviation of each observation from the sample mean, and the standard deviation is the square root of the variance, providing a measure of spread in the same units as the data.
2. Recall the formula for the sample variance, which uses the sum of squared differences between each observation \(x_i\) and the sample mean \(\bar{x}\), divided by the degrees of freedom \((n - 1)\), where \(n\) is the sample size: \[ \text{Variance} = \frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n - 1} \]
3. Recognize that the sample standard deviation is the positive square root of the sample variance, so its formula is: \[ \text{Standard Deviation} = \sqrt{\frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n - 1}} \]
4. Note why we divide by \(n - 1\) instead of \(n\): this adjustment, called Bessel's correction, corrects the bias in the estimation of the population variance and standard deviation from a sample.
5. Summarize that to calculate the sample variance and standard deviation, first compute the squared deviations from the mean, sum them, divide by \(n - 1\) for variance, and then take the square root of that result for the standard deviation.