A data consists of 20 observations $$x_1, x_2, ..., x_{20}$$. If $$\sum_{i=1}^{20}(x_i + 5)^2 = 2500$$ and $$\sum_{i=1}^{20}(x_i - 5)^2 = 100$$, then the ratio of mean to standard deviation of this data is:

$$\sum (x_i + 5)^2 = \sum (x_i^2 + 10x_i + 25) = \sum x_i^2 + 10\sum x_i + 20(25) = 2500$$

$$\sum x_i^2 + 10\sum x_i = 2500 - 500 = 2000$$  (Eq 1)

$$\sum (x_i - 5)^2 = \sum (x_i^2 - 10x_i + 25) = \sum x_i^2 - 10\sum x_i + 20(25) = 100$$
$$\sum x_i^2 - 10\sum x_i = 100 - 500 = -400$$  (Eq 2)

Subtract Equation 2 from Equation 1:

$$(\sum x_i^2 + 10\sum x_i) - (\sum x_i^2 - 10\sum x_i) = 2000 - (-400)$$
$$20\sum x_i = 2400$$
$$\sum x_i = 120$$

$$\bar{x} = \frac{\sum x_i}{20} = \frac{120}{20} = 6$$

Add Equation 1 and Equation 2:
$$2\sum x_i^2 = 2000 - 400 = 1600$$
$$\sum x_i^2 = 800$$

$$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$
$$\sigma^2 = \frac{800}{20} - (6)^2 = 40 - 36 = 4$$
$$\sigma = \sqrt{4} = 2$$

Ratio $$= \frac{\bar{x}}{\sigma} = \frac{6}{2} = 3$$