Question

Can you please solve the first picture into the second picture? The more details the better, I appreciate it!

Answer 1

After solving 2nd picture we get, 1st picture

$S^2 = \frac { \sum_{i=1}^n (x_i -\bar x)^2 }{n -1}$

Where,

$\sum_{i=1}^n (x_i -\bar x)^2$

$=\sum_{i=1}^n (x_i^2 - 2x_i\bar x + \bar x^2)$

$=\sum_{i=1}^n x_i^2 - \sum_{i=1}^n 2x_i\bar x + \sum_{i=1}^n \bar x^2$

$=\sum_{i=1}^n x_i^2 - \sum_{i=1}^n 2\bar x x_i+ n \bar x^2$

$=\sum_{i=1}^n x_i^2 -2 \frac { \sum_{i=1}^n x_i} { n } \sum_{i=1}^n x_i + n( \frac { \sum_{i=1}^n x_i}{n})^2$

$=\sum_{i=1}^n x_i^2 -2 \frac {( \sum_{i=1}^n x_i)^2} { n } + n \frac {( \sum_{i=1}^n x_i)^2}{n^2}$

$=\sum_{i=1}^n x_i^2 -2 \frac {( \sum_{i=1}^n x_i)^2} { n } + \frac {( \sum_{i=1}^n x_i)^2}{n}$

$=\sum_{i=1}^n x_i^2 - \frac {( \sum_{i=1}^n x_i)^2} { n }$

$=> \sum_{i=1}^n (x_i -\bar x)^2 =\sum_{i=1}^n x_i^2 - \frac {( \sum_{i=1}^n x_i)^2} { n }$

Therefore,

$S^2 = \frac {\sum_{i=1}^n (x_i -\bar x)^2 }{n -1}$

$S^2 = \frac { \sum_{i=1}^n x_i^2 - \frac {( \sum_{i=1}^n x_i)^2} { n } }{n -1}$