For given data $N=60, \sum X^2=18000$ and $\sum X=960$,then variance of data is

If the mean and variance of the data $65, 68, 58, 44, 48, 45, 60, \alpha, \beta, 60$ where $\alpha > \beta$ are $56$ and $66.2$ respectively,then $\alpha^2 + \beta^2$ is equal to

The mean and standard deviation of $10$ observations are $20$ and $2$ respectively. Later on,it was observed that one observation was recorded as $50$ instead of $40$. Then the correct variance is:

Find the mean and variance of the frequency distribution given below:
$\begin{array}{|l|l|l|l|l|} \hline x & 1 \leq x < 3 & 3 \leq x < 5 & 5 \leq x < 7 & 7 \leq x < 10 \\ \hline f & 6 & 4 & 5 & 1 \\ \hline \end{array}$

Two plants $A$ and $B$ of a factory show the following results regarding the number of workers and the wages paid to them:

\text{Parameter}	\text{Plant } $A$ \text{ and } $B$ \text{ Data}
\text{No. of workers}	$A: 500, B: 6000$
\text{Average monthly wages}	$A: Rs. 2500, B: Rs. 2500$
\text{Variance of distribution of wages}	$A: 81, B: 100$

In which plant,$A$ or $B$,is there greater variability in individual wages?

For a frequency distribution,the standard deviation is computed by: