The mean of $5$ observations is $7$. If four of these observations are $6, 7, 8, 10$ and one is missing,then the variance of all the five observations is:

Consider $10$ observations $x_1, x_2, \ldots, x_{10}$ such that $\sum_{i=1}^{10}(x_i-\alpha)=2$ and $\sum_{i=1}^{10}(x_i-\beta)^2=40$,where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The value of $\frac{\beta}{\alpha}$ is equal to :

The $S.D.$ of $5$ scores $1, 2, 3, 4, 5$ is

The mean of $100$ observations is $50$ and their standard deviation is $5$. Then,the sum of the squares of all the observations is:

Let $\sigma_1$ and $\sigma_2$ be the standard deviations of two distributions $D_1$ and $D_2$ respectively,and $D_1$ be more consistent than $D_2$. If the means of $D_1$ and $D_2$ are the same,then the percentage increase in the standard deviation of $D_2$ over the standard deviation of $D_1$ is:

Let $X = \{x \in \mathbb{N} : 1 \le x \le 19\}$ and for some $a, b \in \mathbb{R}$, $Y = \{ax + b : x \in X\}$. If the mean and variance of the elements of $Y$ are $30$ and $750$, respectively, then the sum of all possible values of $b$ is