The mean and variance of $10$ observations were calculated as $15$ and $15$ respectively by a student who took by mistake $25$ instead of $15$ for one observation. Then, the correct standard deviation is$.....$

Consider a set of $3 n$ numbers having variance $4.$ In this set, the mean of first $2 n$ numbers is $6$ and the mean of the remaining $n$ numbers is $3.$ A new set is constructed by adding $1$ into each of first $2 n$ numbers, and subtracting $1$ from each of the remaining $n$ numbers. If the variance of the new set is $k$, then $9 k$ is equal to .... .

Let the mean and variance of $12$ observations be $\frac{9}{2}$ and $4$ respectively. Later on, it was observed that two observations were considered as $9$ and $10$ instead of $7$ and $14$ respectively. If the correct variance is $\frac{m}{n}$, where $m$ and $n$ are co-prime, then $m + n$ is equal to

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 $=13$)

If the mean and variance of the frequency distribution

are $9$ and $15.08$ respectively, then the value of $\alpha^2+\beta^2-\alpha \beta$ is $............$.

The frequency distribution:

$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$

 where $A$ is a positive integer, has a variance of $160 .$ Determine the value of $A$.