If $X$ is a random variable such that $\sigma(X) = 2.6$, then $\sigma(1 - 4X)$ is equal to:

Let $x_{i} (1 \leq i \leq 10)$ be ten observations of a random variable $X$. If $\sum_{i=1}^{10} (x_{i} - p) = 3$ and $\sum_{i=1}^{10} (x_{i} - p)^{2} = 9$,where $0 \neq p \in R$,then the standard deviation of these observations 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 $A$.

The coefficient of variation of the following distribution is

Class interval	$0-5$	$5-10$	$10-15$	$15-20$	$20-25$
Frequency	$4$	$1$	$10$	$3$	$2$

For the frequency distribution:

Variate $(x)$	$x_{1}$	$x_{2}$	$x_{3} \ldots x_{15}$
Frequency $(f)$	$f_{1}$	$f_{2}$	$f_{3} \ldots f_{15}$

where $0 < x_{1} < x_{2} < x_{3} < \ldots < x_{15} = 10$ and $\sum_{i=1}^{15} f_{i} > 0$,the standard deviation cannot be:

If $\sum_{i=1}^9(x_i-5)=9$ and $\sum_{i=1}^9(x_i-5)^2=45$,then the standard deviation of the nine observations $x_1, x_2, \ldots, x_9$ is