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}} < $

The coefficients of variation of two distributions are $60$ and $70$,and their standard deviations are $21$ and $16$,respectively. What are their arithmetic means?

Let $X = \{11, 12, 13, \ldots, 40, 41\}$ and $Y = \{61, 62, 63, \ldots, 90, 91\}$ be two sets of observations. If $\bar{x}$ and $\bar{y}$ are their respective means and $\sigma^2$ is the variance of all the observations in $X \cup Y$,then $|\bar{x} + \bar{y} - \sigma^2|$ is equal to $.................$.

For $(2n+1)$ observations ${x_1}, -{x_1}, {x_2}, -{x_2}, ....., {x_n}, -{x_n}$ and $0$,where all $x_i$ are distinct,let $S.D.$ and $M.D.$ denote the standard deviation and median respectively. Which of the following is always true?

The mean of a series $x_1, x_2, \dots, x_n$ is $\bar{X}$. If $x_2$ is replaced by $\lambda$,what will be the new mean?

In a frequency distribution,if $d_i$ is the deviation of observations from $a$,and the mean is given by $\text{Mean} = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$,then what is $a$?