The mean of two samples of size $200$ and $300$ were found to be $25$ and $10$ respectively. Their standard deviations $(S.D.)$ are $3$ and $4$ respectively. Then,the variance of the combined sample of size $500$ is:

Consider the following frequency distribution:

$C$.$I$.	$75$-$175$	$175$-$275$	$275$-$375$	$375$-$475$	$475$-$575$	$575$-$675$	$675$-$775$
$f_i$	$3$	$2$	$1$	$0$	$1$	$2$	$3$

If the variance of this distribution is $60000$,then the coefficient of variation of the distribution is:

The diameters of circles (in mm) drawn in a design are given below:

Diameters	$33-36$	$37-40$	$41-44$	$45-48$	$49-52$
No. of circles	$15$	$17$	$21$	$22$	$25$

Calculate the standard deviation and mean diameter of the circles.
[ Hint : First make the data continuous by making the classes as $32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5-48.5, 48.5-52.5$ and then proceed.] (in $\text{ mm}$)

Find the variance of the sequence $a, a + d, a + 2d, \dots, a + 2nd$.

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 mean of $100$ observations is $50$ and their standard deviation is $5$. Then,the sum of squares of all observations is