The mean and standard deviation of $10$ observations are $20$ and $84$ respectively. Later on, it was observed that one observation was recorded as $50$ instead of $40$. Then the correct variance is:

If the mean deviation about median for the number $3,5,7,2\,k , 12,16,21,24$ arranged in the ascending order, is $6$ then the median is

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 :

Let $X _{1}, X _{2}, \ldots, X _{18}$ be eighteen observations such that $\sum_{ i =1}^{18}\left( X _{ i }-\alpha\right)=36 \quad$ and $\sum_{i=1}^{18}\left(X_{i}-\beta\right)^{2}=90,$ where $\alpha$ and $\beta$ are distinct real numbers. If the standard deviation of these observations is $1,$ then the value of $|\alpha-\beta|$ is …… .

Let $x_1,x_2,………,x_{100}$ are $100$ observations such that  $\sum {{x_i} = 0,\,\sum\limits_{1 \leqslant i \leqslant j \leqslant 100} {\left| {{x_i}{x_j}} \right|} }  = 80000\,\& $ mean deviation from their mean is $5,$ then their standard deviation, is-