The mean and $S.D.$ of $1, 2, 3, 4, 5, 6$ is

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 $b$ $?$ 

The outcome of each of $30$ items was observed; $10$ items gave an outcome $\frac{1}{2} - d$ each, $10$ items gave outcome $\frac {1}{2}$ each and the remaining $10$ items gave outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac {4}{3}$ then $\left| d \right|$ equals

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 $.................$.

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 :

If $\sum\limits_{i = 1}^{18} {({x_i} - 8) = 9} $ and $\sum\limits_{i = 1}^{18} {({x_i} - 8)^2 = 45} $ then the standard deviation of $x_1, x_2, ...... x_{18}$ is :-