If $\sum^{18}_{i=1} (x_i-8)=9$ and $\sum^{18}_{i=1} (x_i-8)^2=45$, then find standard deviation of $x_1,x_2,x_3,...,x_{18}$
Using $\sum^{18}_{i=1} (x_i-8)=9$, I got mean mean of $x_1,x_2,x_3,...,x_{18}$ as $\frac{17}{2}$ but how to use second condition to find variance so that standard deviation can be found?
\begin{align}\sum_{i=1}^{18}(x_i-8.5)^2=\sum_{i=1}^{18}(x_i-8-0.5)^2=\sum_{i=1}^{18}(x_i-8)^2-\sum_{i=1}^{18}2\cdot0.5\cdot(x_i-8)+\sum_{i=1}^{18}0.5^2\end{align}
$$\sum^{18}_{i=1} (x_i-8)=-8\times 18+\sum^{18}_{i=1}x_i=9$$ thus $$\sum^{18}_{i=1}x_i=153\implies \bar x=\frac{153}{18}=8.5$$ on the other hand $$\sum^{18}_{i=1} (x_i-8)^2=\sum^{18}_{i=1}x_i^2-16\sum^{18}_{i=1}x_i+64\times 18=45$$ therefore $$\sum^{18}_{i=1}x_i^2=1341$$ Finally we have $$\frac{\sum^{18}_{i=1}x_i^2}{18}-\bar x^2=\frac{1341}{18}-8.5^2$$