Let sets $A$ and $B$ have $5$ elements each. Let the mean of the elements in sets $A$ and $B$ be $5$ and $8$ respectively and the variance of the elements in sets $A$ and $B$ be $12$ and $20$ respectively $A$ new set $C$ of $10$ elements is formed by subtracting $3$ from each element of $A$ and adding 2 to each element of B. Then the sum of the mean and variance of the elements of $C$ is $.......$.
$32$
$38$
$40$
$36$
Let $a_1, a_2, \ldots . a_{10}$ be $10$ observations such that $\sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50$ and $\sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \cdot \mathrm{a}_{\mathrm{j}}=1100$. Then the standard deviation of $a_1, a_2, \ldots, a_{10}$ is equal to :
The mean of the numbers $a, b, 8,5,10$ is $6$ and their variance is $6.8$. If $M$ is the mean deviation of the numbers about the mean, then $25\; M$ is equal to
The variance of the data $2, 4, 6, 8, 10$ is
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 :