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:

Let the mean and the variance of $20$ observations $x_{1}, x_{2}, \ldots, x_{20}$ be $15$ and $9$,respectively. For $\alpha \in R$,if the mean of $(x_{1}+\alpha)^{2}, (x_{2}+\alpha)^{2}, \ldots, (x_{20}+\alpha)^{2}$ is $178$,then the square of the maximum value of $\alpha$ is equal to $...........$

For the set $A = \{x_1, x_2, x_3, x_4, x_5\}$,the variance is $4$ and the mean is $2$. For the set $B = \{y_1, y_2, y_3, y_4, y_5\}$,the variance is $5$ and the mean is $4$. Then,the variance of $A \cup B$ is

The mean and variance of $7$ observations are $8$ and $16$ respectively. If the first five observations are $2, 4, 10, 12, 14$,then the absolute difference of the remaining two observations is

The means of two groups of observations $A$ and $B$ are $\bar{x}$ and $\bar{y}$ respectively,and their standard deviations are $2$ and $3$ respectively. In order for group $A$ to be more consistent than group $B$,$\frac{\bar{y}}{\bar{x}} < $

Let $a, b, c \in N$ and $a < b < c$. Let the mean,the mean deviation about the mean,and the variance of the $5$ observations $9, 25, a, b, c$ be $18, 4$,and $\frac{136}{5}$,respectively. Then $2a + b - c$ is equal to: