The mean of $5$ observations is $4.4$ and their variance is $8.24$. If three of those observations are $1, 2$ and $6$,then the other two observations are:

The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking,it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation if the wrong item is omitted.

Let the mean and variance of the frequency distribution be $6$ and $6.8$ respectively.

$x$	$2$	$6$	$8$	$9$
$f$	$4$	$4$	$\alpha$	$\beta$

If $x_{3}$ is changed from $8$ to $7$,then the mean for the new data will be:

$x_1, x_2, \ldots, x_n$ are $n$ observations with mean $\bar{x}$ and standard deviation $\sigma$. Match the items of List-$I$ with those of List-$II$:

List-$I$	List-$II$
$(a) \sum_{i=1}^n(x_i-\bar{x})$	$(i) \text{ Median}$
$(b) \text{ Variance } (\sigma^2)$	$(ii) \text{ Coefficient of variation}$
$(c) \text{ Mean deviation}$	$(iii) \text{ Zero}$
$(d) \text{ Measure used to find the homogeneity of given two series}$	$(iv) \text{ Mean of the absolute deviations from any measure of central tendency}$
	$(v) \text{ Mean of the squares of the deviations from mean}$

The mean of five observations is $5$ and their variance is $9.20$. If three of the given five observations are $1, 3$ and $8$,then the ratio of the other two observations is

Mean and standard deviation of $100$ observations were found to be $40$ and $10$,respectively. If at the time of calculation,two observations were wrongly taken as $30$ and $70$ in place of $3$ and $27$ respectively,find the correct standard deviation.