1. Let the wrong data (used by the student) have:

   • number of observations $n=10$
   • mean $\bar{x}_w=15$
   • variance $\sigma_w^2=15$

2. From the wrong mean, $\sum x_w = n\bar{x}_w = 10\times 15 = 150$

3. One observation was taken as $25$ instead of the correct value $15$. So the wrong total is greater than the correct total by: $25-15=10$ Hence correct sum is: $\sum x_c = 150-10 = 140$

4. Therefore the correct mean is: $\bar{x}_c = \frac{140}{10} = 14$

5. Now use the variance formula: $\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2$ For the wrong data, $15 = \frac{\sum x_w^2}{10} - 15^2$ $15 = \frac{\sum x_w^2}{10} - 225$ $\frac{\sum x_w^2}{10} = 240$ $\sum x_w^2 = 2400$

6. Correct the sum of squares. Since $25$ was used instead of $15$, $\sum x_c^2 = 2400 - 25^2 + 15^2$ $= 2400 - 625 + 225 = 2000$

7. Now compute the correct variance: $\sigma_c^2 = \frac{2000}{10} - 14^2$ $= 200 - 196 = 4$

8. Hence the correct standard deviation is: $\sigma_c = \sqrt{4} = 2$

Therefore, the correct standard deviation is $2$.