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

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 $A$.

Let $y_1$ , $y_2$ , $y_3$ ,..... $y_n$ be $n$ observations. Let ${w_i} = l{y_i} + k\,\,\forall \,\,i = 1,2,3.....,n,$ where $l$ , $k$ are constants. If the mean of  $y_i's$ is  is $48$ and their standard deviation is $12$ , then mean of $w_i's$ is $55$ and standard deviation of $w_i's$  is $15$ , then values of $l$ and $k$ should be

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 $\mathrm{n}$ be an odd natural number such that the variance of $1,2,3,4, \ldots, \mathrm{n}$ is $14 .$ Then $\mathrm{n}$ is equal to ..... .