If $\mathop \sum \limits_{i = 1}^9 \left( {{x_i} - 5} \right) = 9$ and $\mathop \sum \limits_{i = 1}^9 {\left( {{x_i} - 5} \right)^2} = 45,$ then the standard deviation of the $9$ items ${x_1},{x_2},\;.\;.\;.\;,{x_9}$ is :
$4$
$2$
$3$
$9$
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
From the data given below state which group is more variable, $A$ or $B$ ?
Marks | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ |
Group $A$ | $9$ | $17$ | $32$ | $33$ | $40$ | $10$ | $9$ |
Group $B$ | $10$ | $20$ | $30$ | $25$ | $43$ | $15$ | $7$ |
Mean and standard deviation of 100 items are 50 and $4,$ respectively. Then find the sum of all the item and the sum of the squares of the items.
The first of the two samples in a group has $100$ items with mean $15$ and standard deviation $3 .$ If the whole group has $250$ items with mean $15.6$ and standard deviation $\sqrt{13.44}$, then the standard deviation of the second sample is:
If the mean and variance of the frequency distribution
$x_i$ | $2$ | $4$ | $6$ | $8$ | $10$ | $12$ | $14$ | $16$ |
$f_i$ | $4$ | $4$ | $\alpha$ | $15$ | $8$ | $\beta$ | $4$ | $5$ |
are $9$ and $15.08$ respectively, then the value of $\alpha^2+\beta^2-\alpha \beta$ is $............$.