The data set on the right represents square v COMPLETE Which formula should be used to calculate the variance? square COMPLETE What is the variance? square A teacher recorded all of his students' grades on the final exam as: 62,77,78,80,82,82,83,84,85,87,89,95 Consider the formulas: A s^2=((x_(1)-bar (x))^2+(x_(2)-bar (x))^2+ldots +(x_(n)-bar (x))^2)/(n-1) B s=sqrt (((x_(1)-bar (x))^2+(x_(2)-bar (x))^2+ldots +(x_(n)-bar (x))^2)/(n-1)) C. sigma ^2=((x_(1)-mu )^2+(x_(2)-mu )^2+ldots +(x_(N)-mu )^2)/(N) D sigma =sqrt (((x_(1)-mu )^2+(x_(2)-mu )^2+ldots +(x_(N)-mu )^2)/(N))

To calculate the variance, we need to use formula A:<br /><br />$s^{2}=\frac {(x_{1}-\bar {x})^{2}+(x_{2}-\bar {x})^{2}+\ldots +(x_{n}-\bar {x})^{2}}{n-1}$<br /><br />where $x_{1}, x_{2}, \ldots, x_{n}$ are the data points and $\bar{x}$ is the mean of the data.<br /><br />To find the mean, we add up all the grades and divide by the number of students:<br /><br />$\bar{x} = \frac{62+77+78+80+82+82+83+84+85+87+89+95}{12} = 80.5$<br /><br />Now we can calculate the variance using formula A:<br /><br />$s^{2}=\frac {(62-80.5)^{2}+(77-80.5)^{2}+(78-80.5)^{2}+(80-80.5)^{2}+(82-80.5)^{2}+(82-80.5)^{2}+(83-80.5)^{2}+(84-80.5)^{2}+(85-80.5)^{2}+(87-80.5)^{2}+(89-80.5)^{2}+(95-80.5)^{2}}{11}$<br /><br />Calculating each term:<br /><br />$(62-80.5)^{2} = 353.25$<br />$(77-80.5)^{2} = 12.25$<br />$(78-80.5)^{2} = 6.25$<br />$(80-80.5)^{2} = 0.25$<br />$(82-80.5)^{2} = 2.25$<br />$(82-80.5)^{2} = 2.25$<br />$(83-80.5)^{2} = 6.25$<br />$(84-80.5)^{2} = 12.25$<br />$(85-80.5)^{2} = 20.25$<br />$(87-80.5)^{2} = 42.25$<br />$(89-80.5)^{2} = 72.25$<br />$(95-80.5)^{2} = 210.25$<br /><br />Adding these up:<br /><br />$353.25 + 12.25 + 6.25 + 0.25 + 2.25 + 2.25 + 6.25 + 12.25 + 20.25 + 42.25 + 72.25 + 210.25 = 717.5$<br /><br />Dividing by 11:<br /><br />$s^{2} = \frac{717.5}{11} = 65.36$<br /><br />Therefore, the variance is approximately 65.36.