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 

If the standard deviation of the numbers $-1, 0, 1, k$ is $\sqrt 5$ where $k > 0,$ then $k$ is equal to

Find the variance of the following data: $6,8,10,12,14,16,18,20,22,24$

The data is obtained in tabular form as follows.

If for a distribution $\Sigma(x-5)=3, \Sigma(x-5)^{2}=43$ and the total number of item is $18,$ find the mean and standard deviation.

Let $n \geq 3$. A list of numbers $x_1, x, \ldots, x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers $y_1, y_2, \ldots, y_n$ is made as follows $y_1=\frac{x_1+x_2}{2}, y_2=\frac{x_1+x_2}{2}$ and $y_j=x_j$ for $j=3,4, \ldots, n$.

The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Then, which of the following is necessarily true?