Finding the $(X - \bar{X})^2$ of first 5 data of dataset given mean and population variance Mean and population variance of the dataset $x_{1},x_{2}..x_{10}$ are $19$ and $49$ respectively. If the value $\sum_{i=6}^{10}x_{i}^{2} = 1900$, what is the value of $\sum_{i=1}^{5}x_{i}^{2} = ?$.
I've solved it as following and it is wrong:
Population variance:
$$S^2 = \frac{\sum (x_i - \bar{x})^2}{n}\\
49 = \frac{\sum_{i=1}^{5}x_{i}^{2}}{10} + \frac{1900}{10}\\
49 = \frac{\sum_{i=1}^{5}x_{i}^{2}}{10} + 190\\
-190+49 = \frac{\sum_{i=1}^{5}x_{i}^{2}}{10}\\
\sum_{i=1}^{5}x_{i}^{2} = -141*10 = -1410$$
And this solution is wrong. How to solve this problem?
 A: You don't use the whole definition of the population variance which is
$$s^2 = \frac{\sum\limits_{i=1}^{10} (x_i - \bar{x})^2}{10}=\frac1{10}\cdot \left( \sum\limits_{i=1}^{10} x_i^2 -2 x_i\bar{x} +\overline x^2 \right)$$
$$\frac1{10}\cdot \left( \sum\limits_{i=1}^{10}x_i^2 -\sum\limits_{i=1}^{10}2 x_i\bar{x} +\sum\limits_{i=1}^{10}\overline x^2 \right)$$
$2, \overline x$ and $\overline x^2$ are constants. They can be put in front the sigma signs.
$$\frac1{10}\cdot \left( \sum\limits_{i=1}^{10}x_i^2 -2 \bar{x}\sum\limits_{i=1}^{10} x_i +\overline x^2 \sum\limits_{i=1}^{10} 1 \right)$$
We have $\frac1{10} \sum\limits_{i=1}^{10} x_i=\overline x\Rightarrow \sum\limits_{i=1}^{10} x_i=10\cdot \overline x$
$$\frac1{10}\cdot \left( \sum\limits_{i=1}^{10}x_i^2 -2 \cdot \bar{x}\cdot 10\cdot \overline x +\overline x^2 \cdot 10 \right)=\frac1{10}\cdot \left( \sum\limits_{i=1}^{10}x_i^2 -10\cdot \overline x^2  \right)$$
$$\frac1{10}\cdot \sum\limits_{i=1}^{10}x_i^2 -\overline x^2  =\frac1{10}\sum\limits_{i=1}^{10} x_i^2-\left(\frac1{10}\sum\limits_{i=1}^{10} x_i\right)^2$$
$=\frac1{10}\left(\sum\limits_{i=1}^{5} x_i^2+\sum\limits_{i=6}^{10} x_i^2\right)-\left(\underbrace{\frac1{10}\sum\limits_{i=1}^{10} x_i}_{\textrm{population mean}}\right)^2=s^2$
Now you can insert the values: $\frac1{10}\sum\limits_{i=1}^{10} x_i=19, s^2=49$ and $\sum\limits_{i=6}^{10} x_i^2=1900$. Finally solve for $\sum\limits_{i=1}^{5} x_i^2$.
