The variance of the variates $112, 116, 120, 125, 132$ about their $A.M.$ is

If $\sum_{i=1}^{5}(x_i-10)=5$ and $\sum_{i=1}^{5}(x_i-10)^2=5$,then the standard deviation of the observations $2x_1 + 7, 2x_2 + 7, 2x_3 + 7, 2x_4 + 7,$ and $2x_5 + 7$ is equal to-

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:

The standard deviation of $7$ observations $1, 2, 3, 4, 5, 6, 7$ is:

The sum of squares of deviations for $10$ observations taken from mean $50$ is $250$. The coefficient of variation is.....$\%$

The standard deviation of the following distribution is:

Class interval	$0-10$	$10-20$	$20-30$	$30-40$
Frequency	$1$	$3$	$4$	$2$