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 :

<p>Given $\overline x = 15,\,\sigma = 2 \Rightarrow {\sigma ^2} = 4$</p> <p>$\therefore$ ${x_2} + {x_2} + \,\,.....\,\, + \,\,{x_{50}} = 15 \times 50 = 750$</p> <p>$4 = {{x_1^2 + x_2^2 + \,\,.....\,\, + \,\,x_{50}^2} \over {50}} - 225$</p> <p>$\therefore$ $x_1^2 + x_2^2 + \,\,.....\,\, + \,\,x_{50}^2 = 50 \times 229$</p> <p>Let a be the correct observation and b is the incorrect observation</p> <p>then $a + b = 70$</p> <p>and $16 = {{750 - b + a} \over {50}}$</p> <p>$\therefore$ $a - b = 50 \Rightarrow a = 60,\,b = 10$</p> <p>$\therefore$ Correct variance $= {{50 \times 229 + {{60}^2} - {{10}^2}} \over {50}} - 256 = 43$</p>