1. Given data (with the mistaken observation):

   • Number of observations: $n=100$
   • Mean: $\bar{x}=40$
   • Standard deviation: $5.1$

2. Find the incorrect total sum Since $\bar{x}=\frac{\sum x_i}{n},$ we get $\sum x_i = 100\times 40 = 4000.$

3. Correct the wrong observation in the sum One value was taken as $50$ instead of $40$. So the incorrect sum is larger by $10$.

   Hence correct sum: $\sum x_i^{(c)} = 4000 - 10 = 3990.$

   Therefore correct mean is $\mu = \frac{3990}{100} = 39.9.$

4. Use the standard deviation formula to get the incorrect sum of squares Using $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2,$ with the mistaken data, $5.1^2 = \frac{\sum x_i^2}{100} - 40^2.$

   So $26.01 = \frac{\sum x_i^2}{100} - 1600,$ $\frac{\sum x_i^2}{100} = 1626.01,$ $\sum x_i^2 = 162601.$

5. Correct the sum of squares The mistaken value $50$ should be $40$. So subtract $50^2 - 40^2 = 2500 - 1600 = 900.$

   Hence correct sum of squares: $\sum (x_i^{(c)})^2 = 162601 - 900 = 161701.$

6. Find the correct variance and standard deviation $\sigma^2 = \frac{161701}{100} - (39.9)^2.$

   Now, $\frac{161701}{100} = 1617.01,$ $(39.9)^2 = 1592.01.$

   Thus, $\sigma^2 = 1617.01 - 1592.01 = 25,$ so $\sigma = 5.$

7. Compute the required value $10(\mu+\sigma)=10(39.9+5)=10(44.9)=449.$

8. Option check

   • A: $447$
   • B: $445$
   • C: $449$ ✅
   • D: $451$

Therefore, the correct answer is Option C.