If the variance of the frequency distribution is $3$,then $\alpha$ is ......

$X_i$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
Frequency $f_i$	$3$	$6$	$16$	$\alpha$	$9$	$5$	$6$

The mean and variance of $8$ observations are $10$ and $13.5,$ respectively. If $6$ of these observations are $5, 7, 10, 12, 14, 15,$ then the absolute difference of the remaining two observations is

Let sets $A$ and $B$ have $5$ elements each. Let the mean of the elements in sets $A$ and $B$ be $5$ and $8$ respectively,and the variance of the elements in sets $A$ and $B$ be $12$ and $20$ respectively. $A$ new set $C$ of $10$ elements is formed by subtracting $3$ from each element of $A$ and adding $2$ to each element of $B$. Then the sum of the mean and variance of the elements of $C$ is $.......$.

The mean of $100$ items is $49$. It was discovered that three items which should have been $60, 70, 80$ were wrongly read as $40, 20, 50$ respectively. The correct mean is

The mean and variance of $n$ observations are $8$ and $16$, respectively. If the sum of the first $(n - 1)$ observations is $48$ and the sum of squares of the first $(n - 1)$ observations is $496$, then the value of $n$ is:

Integers $1, 2, 3, \ldots, n$ $(n \geq 3)$ are written on a blackboard and an integer $k$ $(1 < k < n)$ is erased. The average of the remaining numbers is $16$. Then $n + k$ is