UPSC NDA I 2014 Mathematics Paper

Variance is given by $V = \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2$, where $\bar{x}$ is the mean of the given numbers.Statement 1 : Let the new numbers be $y_i = x_i + 2$.Their mean is $\bar{y} = \bar{x} + 2$.So $y_i - \bar{y} = (x_i + 2) - (\bar{x} + 2) = x_i - \bar{x}$.Hence the variance of the new set is $\frac{1}{n}\sum_{i=1}^{n}(y_i - \bar{y})^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2 = V$.Thus, adding a constant to every observation does not change the variance, so statement 1 is correct.Statement 2 : If the numbers are squared, the new set is $y_i = x_i^2$.Its variance is $\frac{1}{n}\sum_{i=1}^{n}\left(x_i^2 - \overline{x^2}\right)^2$, which is not equal to $V^2$ in general.For example, take $x_1 = 1$ and $x_2 = 3$.Then $\bar{x} = 2$ and $V = \frac{(1-2)^2 + (3-2)^2}{2} = 1$.After squaring, the numbers are $1$ and $9$, whose mean is $5$ and whose variance is $\frac{(1-5)^2 + (9-5)^2}{2} = 16$.Since $16 \neq V^2 = 1$, statement 2 is incorrect.Therefore, only statement 1 is correct, i.e. the answer is 1 only (the presently marked answer "2 only" is wrong).