Question

Sol. [Key Concept: Expand both given sums to get $\sum\ x_{i}^{2}$ and $\sum\ x_{i}$, then $\sigma^{2} = \frac{\sum\ x_{i}^{2}}{N} - (\frac{\sum\ x_{i}}{N})^{2}$.]

$$\sum_{i = 1}^{10}\mspace{2mu}\left( x_{i} + 2 \right)^{2} = 180$$

$$\begin{matrix} & & \ \sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i}^{2} + 4\sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i} + \sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} 4 = 180 \\ & \ \sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i}^{2} + 4\sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i} = 180 - 40 & \text{(1)} \end{matrix}$$

Also $\sum_{i = 1}^{10}\mspace{2mu}\left( x_{i} - 1 \right)^{2} = 90$

$$\begin{matrix} & & \ \sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i}^{2} - 2\sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i} + \sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} 1 = 90 \\ & \ \sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i}^{2} - 2\sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i} = 90 - 10 & \text{(2)} \\ & \ \sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i}^{2} + 4\sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i} = 140 & \text{(2)} \\ & \ \sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i}^{2} - 2\sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i} = 80 & \text{(2)} \\ & \ \sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i}^{2} = 100\text{~}\text{and}\text{~}\sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i} = 10 & \text{(2)} \\ & \ \sum_{i = 1}^{10}\mspace{2mu}\mspace{2mu} x_{i}^{2} & \text{(2)} \\ & \ N & \text{(2)} \\ & \ = \frac{100}{10} - \left( \frac{10}{10}x_{i} \right)^{2} & \text{(2)} \\ & \sigma^{2} = 10 - 1 = 9 & \text{(2)} \\ & \ \Rightarrow \sigma = 3 & \text{(2)} \end{matrix}$$