Question

Sol. [Key Concept: From $\sum(x_{i}{+ 5)}^{2}$ and $\sum(x_{i}{- 5)}^{2}$: subtract to get $\sum\ x_{i}$ (hence the mean) and add to get $\sum\ x_{i}^{2}$ (hence the variance); form the mean-to-SD ratio.]

$$\sum_{i = 1}^{20}\mspace{2mu}\left( x_{i} + 5 \right)^{2} = 2500$$

$$\begin{array}{r} \sum_{i = 1}^{20}\mspace{2mu}\mspace{2mu}\left( x_{i} - 5 \right)^{2} = 100\#(2) \end{array}$$

on (1) - (2) $\Rightarrow 20\sum_{}^{}\ x_{i} = 2400 \Rightarrow \sum_{}^{}\ x_{i} = 120$

mean, $\overline{x} = 6$

(1) + (2) $\Rightarrow 2\sum_{}^{}\ x_{i}^{2} + 40\left( 5^{2} \right) = 2600$

$${\sum x_{i}^{2} = 800 }{\sigma^{2} = \frac{\sum x_{i}^{2}}{n} - (\overline{x})^{2} = \frac{800}{20} - 36 = 4 }{\bar{x}:\sigma = 6:2 = 3:1 }$$Ans. 3 : 1