1. Given (wrongly calculated data):

   • Number of students: $n=10$
   • Mean: $\bar{x}=50$
   • Standard deviation: $\sigma=12$

2. Find the sum and sum of squares for the wrong data

   Using mean, $\sum x = n\bar{x} = 10\times 50 = 500$

   Using variance formula, $\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2$ Since $\sigma=12$, $144 = \frac{\sum x^2}{10} - 50^2$ $144 = \frac{\sum x^2}{10} - 2500$ $\frac{\sum x^2}{10} = 2644$ $\sum x^2 = 26440$

3. Correct the wrong entries

   Two marks were wrongly taken as $45$ and $50$ instead of $20$ and $25$.

   So correct sum is: $\sum x_{\text{correct}} = 500 -45 -50 +20 +25$ $= 500 - 50 = 450$

   Hence correct mean is: $\bar{x}_{\text{correct}} = \frac{450}{10} = 45$

4. Correct the sum of squares

   Wrong sum of squares: $26440$

   Remove wrong squares and add correct squares: $\sum x^2_{\text{correct}} = 26440 - 45^2 - 50^2 + 20^2 + 25^2$ $= 26440 - 2025 - 2500 + 400 + 625$ $= 26440 - 4525 + 1025$ $= 22940$

5. Find the correct variance

   $\text{Variance} = \frac{\sum x^2_{\text{correct}}}{10} - \left(\bar{x}_{\text{correct}}\right)^2$ $= \frac{22940}{10} - 45^2$ $= 2294 - 2025$ $= 269$

6. Final Answer $\boxed{269}$

The derived answer matches the stored correct answer.