If the variance of the first $n$ natural numbers is $10$ and the variance of the first $m$ even natural numbers is $16$,then $n: m=$

The variance of the first $n$ natural numbers is

The variance of the following continuous frequency distribution is
$\begin{array}{|l|c|c|c|c|}\hline \text{Class interval} & 0-4 & 4-8 & 8-12 & 12-16 \\ \hline \text{Frequency} & 2 & 3 & 2 & 1 \\ \hline\end{array}$

Consider $10$ observations $x_1, x_2, \ldots, x_{10}$ such that $\sum_{i=1}^{10}(x_i-\alpha)=2$ and $\sum_{i=1}^{10}(x_i-\beta)^2=40$,where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The value of $\frac{\beta}{\alpha}$ is equal to :

Let $a_1, a_2, \ldots, a_{10}$ be $10$ observations such that $\sum_{k=1}^{10} a_k = 50$ and $\sum_{k < j} a_k a_j = 1100$. Then the standard deviation of $a_1, a_2, \ldots, a_{10}$ is equal to:

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: