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:

Number of observations are = 50
Mean  $\left(\overline{x}\right)=15$
Standard deviation  $\left(\sigma \right)=2$
Let incorrect observation is  $x_1$ & correct observation is  $\left(y_1\right)$
Given $x_1+y_1=70$ $x_1=10$
     $\Rightarrow x_1+x_2+\mathrm{.......}{x}_{50}=750$         ………… (i)
Now 
Mean of correct observation is 16 $\frac{y_1+x_2+\mathrm{......}+x_{50}}{50}=16$

   $y_1+x_2+x_3+\mathrm{.....}{x}_{50}=16\times 50$         …………….. (ii)
Eq. (i) – eq. (i) 

$\Rightarrow y_1-x_1=16\times 50-15\times 50$
$y_1-x_1=50\&x_1+y_1=70$
$y_1=60$    $x_1=10$
$\Rightarrow 4=\frac{x_1^2+x_2^2+\mathrm{.......}+x_{50}^2}{50}-{15}^2$             ………. (iii)
  $\Rightarrow {\sigma }^2=\frac{y_1^2+x_2^2+\mathrm{.....}{x}_{50}^2}{50}-{16}^2$          …….. (iv)
From (iii)
$\Rightarrow 4=\frac{{\left(10\right)}^2}{50}+\frac{x_2^2{+}_3^2+\mathrm{....}+x_{50}^2}{50}-225$
$\Rightarrow 4=2-225+\frac{\left(x_2^2+x_3^2+\mathrm{.....}+x_{50}^2\right)}{50}$
$\Rightarrow 227=\frac{\left(x_2^2+x_3^2+\mathrm{.......}+x_{50}^2\right)}{50}$
From (iv) 
${\sigma }^2=\frac{{\left(60\right)}^2}{50}+\left(\frac{x_2^2+x_3^2+\mathrm{......}+x_{50}^2}{50}\right)-{\left(16\right)}^2$

${\sigma }^2=\frac{60\times 60}{50}+227-256$
${\sigma }^2=72+227-256$
${\sigma }^2=43$