The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70 . If the correct mean is 16 , then the correct variance is equal to :

No. of observations= 50

mean $\left(\overline{x}\right)=15$

Standard deviation $\left(\sigma \right)=2$

Let incorrect observation is $x_1$ & correct observation is $\left(x_1^{\text{'}}\right)$

Given $x_1+x_1^{\text{'}}=70$$$\begin{gathered} \overline{x}=\frac{x_1+x_2+\dots +x_{50}}{50}=15 \left(given\right) \\ \Rightarrow x_1+x_2+\dots \dots x_{50}=750 ....\left(1\right) \end{gathered}$$

Now
Mean of correct observation is 16

$$\begin{gathered} \frac{x_1^{\text{'}}+x_2+\dots .+x_{50}}{50}=16 \\ {x}_1^{\text{'}}+x_2+x_3+\dots x_{50}=16\times 50 .....\left(2\right) \\ eq. \left(2\right) - eq. \left(1\right) \\ \Rightarrow x_1^{\text{'}}-x_1=16\times 50-15\times 50 \\ {x}_1^{\text{'}}-x_1=50\&x_1+x_1^{\text{'}}=70 \\ {x}_1^{\text{'}}=60 \\ {x}_1=10 \\ \Rightarrow 4=\frac{x_1^2+x_2^2+\dots .+x_{50}^2}{50}-{15}^2 ....\left(3\right) \\ \Rightarrow {\sigma }^2=\frac{x_1^{\text{'}2}+x_2^2+\dots x_{50}^2}{50}-{16}^2 ....\left(4\right) \\ from \left(3\right) \\ \Rightarrow 4=\frac{(10{)}^2}{50}+\frac{x_2^2+x_3^2+\dots .+x_{50}^2}{50}-225 \\ \Rightarrow 4=2-225+\frac{\left(x_2^2+x_3^2+\dots .+x_{50}^2\right)}{50} \\ \Rightarrow 227=\frac{\left(x_2^2+x_3^2+\dots .x_{50}^2\right)}{50} \\ From \left(4\right) \\ {\sigma }^2=\frac{(60{)}^2}{50}+\left(\frac{x_2^2+x_3^2+\dots .+x_{50}^2}{50}\right)-(16{)}^2 \\ {\sigma }^2=\frac{60\times 60}{50}+227-256 \\ {\sigma }^2=43 \end{gathered}$$