Variance Statistics

Question

A sample of 10 small debts from a small business were 16.39, 25.09, 16.31, 20.94, 17.58, 19.06, 17.21, 18.48, 16.88, 15.51

Can someone tell me how they are getting the variance because the xi is confusing me. Thanks

Answer 1

$x_i$ is the $i$-th value of the list $\{16.39, 25.09, 16.31, 20.94, 17.58, 19.06, 17.21, 18.48, 16.88, 15.51\}$

$x_1=16.39, x_2=25.09, \ldots , x_{10}=15.51$ and such.

$\begin{align}\overline x ~=~& \dfrac{\sum x_i}{10}\\[1ex] =~& 1.639 + 2.509 + 1.631 + 2.094 + 1.758 + 1.906 + 1.721 + 1.848 + 1.688 + 1.551\\[1ex] =~& 18.345\end{align}$

$\begin{align}s^2 ~=~& \dfrac{\sum (x_i-\overline x)^2}{9}\\[1ex] =~& \tfrac 19( (16.39-18.345)^2 + (25.09-18.345)^2 + (16.31-18.345)^2 + (20.94-18.345)^2 + (17.58-18.345)^2 + (19.06-18.345)^2 + (17.21-18.345)^2 + (18.48-18.345)^2 + (16.88-18.345)^2 + (15.51-18.345)^2)\\[1ex] \approx.~& 8.087\end{align}$

And so forth.

Most easily done with a spreadsheet application these days.

Answer 2

If $\overline{x}$ is the mean, then the variance is: $$ {1 \over 9} \left[ (x_{1} - \overline{x})^2 + (x_{2} - \overline{x})^2 + \ldots + (x_{10} - \overline{x})^2 \right] $$ The sum in $[]$'s can also be written $$ \sum_{i = 1}^{10} (x_{k} - \overline{x})^2. $$