If $\mathop \sum \limits_{i = 1}^9 \left( {{x_i} – 5} \right) = 9$ and $\mathop \sum \limits_{i = 1}^9 {\left( {{x_i} – 5} \right)^2} = 45,$ then the standard deviation of the $9$ items  ${x_1},{x_2},\;.\;.\;.\;,{x_9}$ is :

Let $X=\{11,12,13, \ldots ., 40,41\}$ and $Y=\{61,62$, $63, \ldots ., 90,91\}$ be the two sets of observations. If $\bar{x}$ and $\bar{y}$ are their respective means and $\sigma^2$ is the variance of all the observations in $X \cup Y$, then $\left|\overline{ x }+\overline{ y }-\sigma^2\right|$ is equal to $……………..$.

The mean of the numbers $a, b, 8, 5, 10$ is $6$ and the variance is $6.80.$ Then which one of the following gives possible values of $a$ and $b$ $?$ 

If the data $x_1, x_2, …., x_{10}$ is such that the mean of first four of these is $11$, the mean of the remaining six is $16$ and the sum of squares of all of these is $2,000$; then the standard deviation of this data is