Given that the total of $16$ values is $528$ and the sum of the squares of deviations from $33$ is $9158$. The variance is:

Suppose a population $A$ has $100$ observations $101, 102, . . ., 200$ and another population $B$ has $100$ observations $151, 152, . . ., 250$. If $V_A$ and $V_B$ represent the variances of the two populations,respectively,then $V_A / V_B$ is:

If the mean of the discrete distribution $8, 9, 6, 5, x, 4, 6, 5$ is $6$,then its standard deviation (nearest to two decimal places) is

The raw data $x_1, x_2, \ldots, x_{n}$ is an $A$.$P$. with common difference $d$ and first term $0$. If $\bar{x}$ and $\sigma^2$ are the mean and variance of $x_{i}, i=1, 2, \ldots, n$,then $\sigma^2$ is:

The sum and sum of squares corresponding to length $x$ (in $cm$) and weight $y$ (in $gm$) of $50$ plant products are given below:
$\sum\limits_{i = 1}^{50} {{x_i} = 212, \sum\limits_{i = 1}^{50} {x_i^2} = 902.8, \sum\limits_{i = 1}^{50} {{y_i} = 261, \sum\limits_{i = 1}^{50} {y_i^2 = 1457.6} } }$
Which is more varying,the length or weight?

The mean and variance of a series of $5$ observations are $8$ and $24$ respectively. The mean and variance of another series of $3$ observations are $8$ and $24$ respectively. What is the variance of their combined series?