For the frequency distribution :

Variate $( x )$	$x _{1}$	$x _{1}$	$x _{3} \ldots \ldots x _{15}$
Frequency $(f)$	$f _{1}$	$f _{1}$	$f _{3} \ldots f _{15}$

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 

Let $v_1 =$ variance of $\{13, 1 6, 1 9, . . . . . , 103\}$ and $v_2 =$ variance of $\{20, 26, 32, . . . . . , 200\}$, then $v_1 : v_2$ is

Statement $1$ : The variance of first $n$ odd natural numbers is $\frac{{{n^2} - 1}}{3}$
Statement $2$ : The sum of first $n$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{{n\left( {4{n^2} + 1} \right)}}{3}$

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

If the mean and variance of the following data:

$6,10,7,13, a, 12, b, 12$ are 9 and $\frac{37}{4}$ respectively, then $(a-b)^{2}$ is equal to:

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 $............$.