The mean and variance of $7$ observations are $8$ and $16$ respectively. If two observations are $6$ and $8 ,$ then the variance of the remaining $5$ observations is:
$\frac{92}{5}$
$\frac{134}{5}$
$\frac{536}{25}$
$\frac{112}{5}$
Let sets $A$ and $B$ have $5$ elements each. Let the mean of the elements in sets $A$ and $B$ be $5$ and $8$ respectively and the variance of the elements in sets $A$ and $B$ be $12$ and $20$ respectively $A$ new set $C$ of $10$ elements is formed by subtracting $3$ from each element of $A$ and adding 2 to each element of B. Then the sum of the mean and variance of the elements of $C$ is $.......$.
The average marks of $10$ students in a class was $60$ with a standard deviation $4$, while the average marks of other ten students was $40$ with a standard deviation $6$. If all the $20$ students are taken together, their standard deviation will be
Determine the mean and standard deviation for the following distribution:
$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline \text { Marks } & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline \text { Frequency } & 1 & 6 & 6 & 8 & 8 & 2 & 2 & 3 & 0 & 2 & 1 & 0 & 0 & 0 & 1 \\ \hline \end{array}$
If $v$ is the variance and $\sigma$ is the standard deviation, then
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.]