Let $x _1, x _2, \ldots \ldots x _{10}$ be ten observations such that $\sum_{i=1}^{10}\left(x_i-2\right)=30, \sum_{i=1}^{10}\left(x_i-\beta\right)^2=98, \beta>2$ and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2\left( x _1-1\right)+4 \beta, 2\left( x _2-1\right)+$ $4 \beta, \ldots . ., 2\left(x_{10}-1\right)+4 \beta$, then $\frac{\beta \mu}{\sigma^2}$ is equal to :

If $v$ is the variance and $\sigma$ is the standard deviation, then

If mean and standard deviation of $5$ observations $x_1 ,x_2 ,x_3 ,x_4 ,x_5$ are $10$ and $3$, respectively, then the variance of $6$ observations $x_1 ,x_2 ,.....,x_3$ and $-50$ is equal to

From the data given below state which group is more variable, $A$ or $B$ ?

The mean and standard deviation of $20$ observations were calculated as $10$ and $2.5$ respectively. It was found that by mistake one data value was taken as $25$ instead of $35 .$ If $\alpha$ and $\sqrt{\beta}$ are the mean and standard deviation respectively for correct data, then $(\alpha, \beta)$ is :

Let $v_1 =$ variance of $\{13, 1 6, 1 9, . . . . . , 103\}$ and $v_2 =$ variance of $\{20, 26, 32, . . . . . , 200\}$, then $v_1 : v_2$ is