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:

The mean and standard deviation of the marks of $10$ students were found to be $50$ and $12$ respectively. Later, it was observed that two marks $20$ and $25$ were wrongly read as $45$ and $50$ respectively. Then the correct variance is $............$.

The number of values of $a \in N$ such that the variance of $3,7,12 a, 43-a$ is a natural number is  (Mean $=13$)

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-

Let ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance

Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .

Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.

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