If $\sum_{i=1}^{5}(x_i-10)=5$ and $\sum_{i=1}^{5}(x_i-10)^2=5$ then standard deviation of observations $2x_1 + 7, 2x_2 + 7, 2x_3 + 7, 2x_4 + 7$ and $2x_5 + 7$ is equal to-

The mean and the variance of five observations are $4$ and $5.20,$ respectively. If three of the observations are $3, 4$ and $4;$ then the absolute value of the difference of the other two observations, is

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 .... .

The $S.D.$ of a variate $x$ is $\sigma$. The $S.D.$ of the variate $\frac{{ax + b}}{c}$ where $a, b, c$ are constant, is

For the frequency distribution :

where $0< x _{1}< x _{2}< x _{3}<\ldots .< x _{15}=10$ and

$\sum \limits_{i=1}^{15} f_{i}>0,$ the standard deviation cannot be