The mean and standard deviation of $10$ observations are $20$ and $2$ respectively. Later on,it was observed that one observation was recorded as $50$ instead of $40$. Then the correct variance is:

The sum and sum of squares corresponding to length $x$ (in $cm$) and weight $y$ (in $gm$) of $50$ plant products are given below:
$\sum\limits_{i = 1}^{50} {{x_i} = 212, \sum\limits_{i = 1}^{50} {x_i^2} = 902.8, \sum\limits_{i = 1}^{50} {{y_i} = 261, \sum\limits_{i = 1}^{50} {y_i^2 = 1457.6} } }$
Which is more varying,the length or weight?

Let $v_1$ be the variance of $\{13, 16, 19, \dots, 103\}$ and $v_2$ be the variance of $\{20, 26, 32, \dots, 200\}$. Then the ratio $v_1 : v_2$ is:

Let ${x_1}, {x_2}, \ldots, {x_n}$ be $n$ observations,and let $\bar x$ be their arithmetic mean and ${\sigma ^2}$ be the variance.
Statement-$1$: The variance of $2{x_1}, 2{x_2}, \ldots, 2{x_n}$ is $4{\sigma ^2}$.
Statement-$2$: The arithmetic mean of $2{x_1}, 2{x_2}, \ldots, 2{x_n}$ is $4\bar x$.

If the variance of $x_1, x_2, \ldots, x_n$ is $\sigma_x^2$,then the variance of $\lambda x_1, \lambda x_2, \ldots, \lambda x_n$ (where $\lambda \neq 0$) is:

If $1$ is added to each of the first $10$ natural numbers,then the variance of the numbers so obtained is: