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 the mean and variance of the frequency distribution

be $6$ and $6.8$ respectively. If $x_{3}$ is changed from $8$ to $7 ,$ then the mean for the new data will be:

Let the observations $\mathrm{x}_{\mathrm{i}}(1 \leq \mathrm{i} \leq 10)$ satisfy the equations, $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)=10$ and $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)^{2}=40$ If $\mu$ and $\lambda$ are the mean and the variance of the observations, $\mathrm{x}_{1}-3, \mathrm{x}_{2}-3, \ldots ., \mathrm{x}_{10}-3,$ then the ordered pair $(\mu, \lambda)$ is equal to :

Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to