Consider $10$ observation $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{10}$. such that $\sum_{i=1}^{10}\left(x_i-\alpha\right)=2$ and $\sum_{i=1}^{10}\left(x_i-\beta\right)^2=40$, 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 $\frac{\beta}{\alpha}$ is equal to :

If each of the observation $x_{1}, x_{2}, \ldots ., x_{n}$ is increased by $'a'$ where $a$ is a negative or positive number, show that the variance remains unchanged.

Suppose a population $A $ has $100$ observations $ 101,102, . . .,200 $ and another population $B $ has $100$ observation $151,152, . . .,250$ .If $V_A$ and $V_B$ represent the variances of the two populations , respectively then $V_A / V_B$ is

Find the variance of the following data: $6,8,10,12,14,16,18,20,22,24$

Let $n \geq 3$. A list of numbers $0 < x_1 < x_2 < \ldots < x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers is made as follows: $y_1=0, y_2=x_2, \ldots, x_{n-1}$ $=x_n-1, y_n=x_1+x_n$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?

If the mean and variance of the data $65,68,58,44$, $48,45,60, \alpha, \beta, 60$ where $\alpha>\beta$ are $56$ and $66.2$ respectively, then $\alpha^2+\beta^2$ is equal to