The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If $$\alpha$$ and $$\sqrt{\beta}$$ are the mean and standard deviation respectively for correct data, then $$(\alpha, \beta)$$ is:

We have $$n = 20$$ observations whose mean and standard deviation were first calculated (with the mistaken value) as $$\bar x = 10$$ and $$s = 2.5$$ respectively.

The mean formula is stated as  $$\bar x = \dfrac{\sum x_i}{n}\,.$$

So for the incorrect data the total of all observations is

$$\sum x_i = \bar x \times n = 10 \times 20 = 200.$$

One observation was taken as $$25$$ instead of the correct value $$35$$. Replacing this in the total, we get the correct sum

$$\sum x_i^{(\text{correct})}= 200 - 25 + 35 = 210.$$

Hence the correct mean is

$$\alpha = \dfrac{210}{20}=10.5.$$

For the standard deviation we first recall the relation

$$s^2 = \dfrac{\sum x_i^{2}}{n} - \bar x^{\,2}.$$

With the wrong data we have $$s = 2.5$$ and $$\bar x = 10$$, so

$$2.5^{2}= \dfrac{\sum x_i^{2}}{20} - 10^{2}.$$

This gives

$$6.25 = \dfrac{\sum x_i^{2}}{20} - 100,$$

$$\dfrac{\sum x_i^{2}}{20} = 106.25,$$

$$\sum x_i^{2}= 106.25 \times 20 = 2125.$$

The mistaken squared term $$25^{2}=625$$ must be replaced by the correct squared term $$35^{2}=1225$$. Therefore the corrected sum of squares is

$$\sum x_i^{2\,(\text{correct})}= 2125 - 625 + 1225 = 2725.$$

Substituting in the variance formula with the corrected mean $$\alpha = 10.5$$, we get

$$\sigma^{2}= \dfrac{2725}{20} - (10.5)^{2}.$$

The term $$\dfrac{2725}{20}=136.25$$ and $$(10.5)^{2}=110.25,$$ so

$$\sigma^{2}=136.25 - 110.25 = 26.$$

Thus the corrected standard deviation is $$\sigma = \sqrt{26}.$$ By the wording of the question $$\sqrt{\beta} = \sigma,$$ hence $$\beta = 26.$$

We have found $$\alpha = 10.5$$ and $$\beta = 26,$$ so the ordered pair is $$(\alpha, \beta) = (10.5, 26).$$

Hence, the correct answer is Option A.