What is the variance of {7, 3, -1, 1, -3, 4, -2}?

Variance ${\sigma}^{2} = \frac{542}{49} = 11.0612$

Explanation:

Solve the mean $\overline{x}$ first

$\overline{x} = \frac{7 + 3 + \left(- 1\right) + 1 + \left(- 3\right) + 4 + \left(- 2\right)}{7} = \frac{9}{7}$

Solve Variance ${\sigma}^{2}$

${\sigma}^{2} = \frac{{\left(7 - \frac{9}{7}\right)}^{2} + {\left(3 - \frac{9}{7}\right)}^{2} + {\left(- 1 - \frac{9}{7}\right)}^{2} + {\left(1 - \frac{9}{7}\right)}^{2} + {\left(- 3 - \frac{9}{7}\right)}^{2} + {\left(4 - \frac{9}{7}\right)}^{2} + {\left(- 2 - \frac{9}{7}\right)}^{2}}{7}$

${\sigma}^{2} = \frac{542}{49} = 11.0612$

God bless....I hope the explanation is useful.