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

<p>$$\begin{aligned} &\text { Let the observations be } x_1, x_2, \ldots, x_{99}, 50\\ &\begin{aligned} & \text { Mean }=\frac{x_1+x_2+\ldots+x_9+50}{100}=40 \\ & \Rightarrow x_1+x_2+\ldots+x_{99}=4000-50 \\ & \Rightarrow x_1+x_2+\ldots+x_{99}=3950 \\ & \text { Current Mean }=\frac{3950+40}{100} \\ & \mu=\frac{399}{10}=39.9 \\ & (\text { S.D })^2=\sum_{i=1}^{99} \frac{\left(x_i\right)^2+2500}{100}-(40)^2 \\ & \sum_{i=1}^{99} x_i^2=160101 \\ & (\text { Correct S.D })^2=\frac{160101+1600}{100}-\left(\frac{399}{10}\right)^2 \\ & \sigma=5 \\ & 10(\mu+\sigma)=10(39.9+5)=449 \end{aligned} \end{aligned}$$</p>