Consider the statistics of two sets of observations as follows :

	Size	Mean	Variance
Observation $I$	$10$	$2$	$2$
Observation $II$	$n$	$3$	$1$

If the variance of the combined set of these two observations is $\frac{17}{9},$ then the value of $n$ is equal to ..... .

If the mean deviation about the mean of the numbers $1,2,3, \ldots ., n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to

Let the mean and variance of $12$ observations be $\frac{9}{2}$ and $4$ respectively. Later on, it was observed that two observations were considered as $9$ and $10$ instead of $7$ and $14$ respectively. If the correct variance is $\frac{m}{n}$, where $m$ and $n$ are co-prime, then $m + n$ is equal to

Let $v_1 =$ variance of $\{13, 1 6, 1 9, . . . . . , 103\}$ and $v_2 =$ variance of $\{20, 26, 32, . . . . . , 200\}$, then $v_1 : v_2$ is

The mean and variance of $7$ observations are $8$ and $16$ respectively. If one observation $14$ is omitted a and $b$ are respectively mean and variance of remaining $6$ observation, then $a+3 b-5$ is equal to $..........$.

For a given distribution of marks mean is $35.16$ and its standard deviation is $19.76$. The co-efficient of variation is..