The mean of four observations is $3$. If the sum of the squares of these observations is $48$,then their standard deviation is

Let the observations $x_{i} (1 \leq i \leq 10)$ satisfy the equations $\sum_{i=1}^{10}(x_{i}-5)=10$ and $\sum_{i=1}^{10}(x_{i}-5)^{2}=40$. If $\mu$ and $\lambda$ are the mean and the variance of the observations $x_{1}-3, x_{2}-3, \dots, x_{10}-3$,then the ordered pair $(\mu, \lambda)$ is equal to:

Find the mean and variance for the following data:

$x_i$	$92$	$93$	$97$	$98$	$102$	$104$	$109$
$f_i$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

For given data $N=60, \sum X^2=18000$ and $\sum X=960$,then variance of data is

The variance of $10$ observations is $16$. If each observation is doubled,then the standard deviation of the new data will be -

Two teams $A$ and $B$ have the same mean and their coefficients of variation are $4$ and $2$,respectively. If $\sigma_A$ and $\sigma_B$ are the standard deviations of teams $A$ and $B$ respectively,then the relation between them is