If $v$ is the variance and $\sigma$ is the standard deviation,then

In an experiment with $15$ observations on $x$,we have $\sum x^2 = 2830$ and $\sum x = 170$. One observation that was $20$ was found to be wrong and was replaced by the correct value $30$. The corrected variance is:

Find the mean and variance for the following data:

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

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

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:

The variance of the following frequency distribution is

Class Interval	Frequency
$0 - 6$	$10$
$6 - 12$	$8$
$12 - 18$	$6$
$18 - 24$	$4$
$24 - 30$	$2$