1. Given wrong mean and standard deviation

For $n=20$ observations, the calculated (incorrect) mean and standard deviation are: $\bar x = 10, \qquad \sigma = 2.5$

So the incorrect sum of observations is $\sum x_i = n\bar x = 20\cdot 10 = 200$

Also, $\sigma^2 = \frac{\sum x_i^2}{n} - \bar x^2$

Thus, $2.5^2 = \frac{\sum x_i^2}{20} - 10^2$ $6.25 = \frac{\sum x_i^2}{20} - 100$ $\frac{\sum x_i^2}{20} = 106.25$ $\sum x_i^2 = 20\cdot 106.25 = 2125$

So for the incorrectly taken data: $\sum x_i = 200, \qquad \sum x_i^2 = 2125$

2. Correct the mistaken observation

One value was taken as $25$ instead of the correct value $35$.

Hence, the correct sum is: $\sum x_i(\text{correct}) = 200 - 25 + 35 = 210$

So the correct mean is $\alpha = \frac{210}{20} = 10.5$

3. Correct the sum of squares

Incorrectly, $25^2$ was used instead of $35^2$.

Therefore, $\sum x_i^2(\text{correct}) = 2125 - 25^2 + 35^2$ $= 2125 - 625 + 1225$ $= 2725$

4. Compute correct variance and standard deviation

Let the correct standard deviation be $\sqrt\beta$. Then variance is $\beta$.

Using $\beta = \frac{\sum x_i^2}{n} - \alpha^2$ we get $\beta = \frac{2725}{20} - (10.5)^2$ $= 136.25 - 110.25$ $= 26$

Hence, $\sqrt\beta = \sqrt{26}$

So, $(\alpha, \beta) = (10.5, 26)$

5. Check options

• A: $(11,26)$ ❌
• B: $(10.5,25)$ ❌
• C: $(11,25)$ ❌
• D: $(10.5,26)$ ✅

Therefore, the correct option is D.