In a series of $2n$ observations,half of them are equal to $a$ and the remaining half are equal to $-a$. If the standard deviation of the observations is $2$,then $|a|$ equals:

For a given distribution of marks,the mean is $35.16$ and its standard deviation is $19.76$. The coefficient of variation is:

Given that $\bar{x}$ is the mean and $\sigma^{2}$ is the variance of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$,prove that the mean and variance of the observations $a x_{1}, a x_{2}, \ldots, a x_{n}$ are $a \bar{x}$ and $a^{2} \sigma^{2}$ respectively,where $a \neq 0$.

The variance of the data $1001, 1003, 1006, 1007, 1009, 1010$ is:

If $\sum_{i=1}^{n}(x_{i}-a)=n$ and $\sum_{i=1}^{n}(x_{i}-a)^{2}=na$,where $n, a > 1$,then the standard deviation of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ is

The outcome of each of $30$ items was observed; $10$ items gave an outcome $\frac{1}{2} - d$ each,$10$ items gave an outcome $\frac{1}{2}$ each,and the remaining $10$ items gave an outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac{4}{3}$,then $|d|$ equals: