The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are $$\mu$$ and $$\sigma$$ respectively, then $$10(\mu + \sigma)$$ is equal to

Given: $$n = 100$$, mean $$= 40$$, standard deviation $$= 5.1$$. One observation was recorded as 50 instead of 40.

Corrected mean: $$\mu = 40 - \frac{50 - 40}{100} = 40 - 0.1 = 39.9$$

For the variance, we use $$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$$:

$$\frac{\sum x_i^2}{100} - 40^2 = (5.1)^2 = 26.01$$

$$\sum x_i^2 = 100 \times 1626.01 = 162601$$

Corrected $$\sum x_i^2 = 162601 - 50^2 + 40^2 = 162601 - 2500 + 1600 = 161701$$

Corrected variance: $$\sigma^2 = \frac{161701}{100} - (39.9)^2 = 1617.01 - 1592.01 = 25$$

Corrected standard deviation: $$\sigma = 5$$

$$10(\mu + \sigma) = 10(39.9 + 5) = 10 \times 44.9 = 449$$

Hence, the correct answer is Option D.