The mean and variance of $7$ observations are $8$ and $16$ respectively. If two observations are $6$ and $8$,then the variance of the remaining $5$ observations is:

If both mean and variance of $50$ observations $x_1, x_2, \ldots, x_{50}$ are equal to $16$ and $256$ respectively,then the mean of $(x_1-5)^2, (x_2-5)^2, \ldots, (x_{50}-5)^2$ is

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:

In a Mathematics test,the average marks of boys is $x \%$ and the average marks of girls is $y \%$ with $x \neq y$. If the average marks of all students is $z \%$,the ratio of the number of girls to the total number of students is

If the arithmetic mean of the following frequency distribution is $50$,find the values of $f_1$ and $f_2$.

Class	Frequency
$0 - 20$	$17$
$20 - 40$	$f_1$
$40 - 60$	$32$
$60 - 80$	$f_2$
$80 - 100$	$19$
Total	$120$

The mean and variance of $20$ observations are found to be $10$ and $4,$ respectively. On rechecking,it was found that an observation $9$ was incorrect and the correct observation was $11$. Then the correct variance is