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

$A$ $100$ mark examination was administered to a class of $50$ students. Despite only integer marks being given,the average score of the class was $47.5$. The maximum number of students who could have received marks greater than the class average is:

$\bar{x}$ and $\bar{y}$ are the arithmetic means of the runs of two batsmen $A$ and $B$ in $10$ innings respectively,and $\sigma_{A}$ and $\sigma_{B}$ are the standard deviations of their runs. If batsman $A$ is more consistent than $B$,then he is also a higher run scorer only when

Mean and standard deviation of $100$ observations were found to be $40$ and $10$,respectively. If at the time of calculation,two observations were wrongly taken as $30$ and $70$ in place of $3$ and $27$ respectively,find the correct standard deviation.

Suppose that the mean and median of the non-negative numbers $21, 8, 17, a, 51, 103, b, 13, 67$ $(a > b)$ are $40$ and $21$, respectively. If the mean deviation about the median is $26$, then $2a$ is equal to:

If $x_1, x_2, \ldots, x_n$ are $n$ observations such that $\sum_{i=1}^n x_i^2 = 400$ and $\sum_{i=1}^n x_i = 80$,then the least value of $n$ is