Let in a series of $2n$ observations,half of them are equal to $a$ and the remaining half are equal to $-a$. Also,by adding a constant $b$ to each of these observations,the mean and standard deviation of the new set become $5$ and $20$,respectively. Then the value of $a^{2} + b^{2}$ is equal to ....... .

$\text{Assertion (A):}$ Variance of $4x_1, 4x_2, \ldots, 4x_n$ is $16$ times the variance of $x_1, x_2, \ldots, x_n$. $\text{Reason (R):}$ If $y = ax + b$,then variance of $y$ is $a(\text{variance of } x) + b$. The correct option among the following is

Let $x_{i} (1 \leq i \leq 10)$ be ten observations of a random variable $X$. If $\sum_{i=1}^{10} (x_{i} - p) = 3$ and $\sum_{i=1}^{10} (x_{i} - p)^{2} = 9$,where $0 \neq p \in R$,then the standard deviation of these observations 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?

The variance of the observations $8, 12, 13, 15, 22$ is:

The standard deviations of two sets of observations $X=\{x_i\}$ and $Y=\{y_i\}$ $(i=1, 2, \ldots, 100)$ are respectively $5$ and $6$. If $\bar{x}, \bar{y}$ are their means and $\sum_{i=1}^{100}(x_i-\bar{x})(y_i-\bar{y})=600$,then the standard deviation of $Z=\{z_i \mid z_i=x_i-y_i\}$ is