The mean and variance of $8$ observations are $10$ and $13.5,$ respectively. If $6$ of these observations are $5,7,10,12,14,15,$ then the absolute difference of the remaining two observations 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 $..........$.

One set containing five numbers has mean $8$ and variance $18$ and the second set containing $3$ numbers has mean $8$ and variance $24$. Then the variance of the combined set of numbers is

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

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.]

Let $a_1, a_2, \ldots . a_{10}$ be $10$ observations such that $\sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50$ and $\sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \cdot \mathrm{a}_{\mathrm{j}}=1100$. Then the standard deviation of $a_1, a_2, \ldots, a_{10}$ is equal to :

The frequency distribution:

$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$

 where $A$ is a positive integer, has a variance of $160 .$ Determine the value of $A$.