The variance of the following continuous frequency distribution is
$\begin{array}{|l|c|c|c|c|}\hline \text{Class interval} & 0-4 & 4-8 & 8-12 & 12-16 \\ \hline \text{Frequency} & 2 & 3 & 2 & 1 \\ \hline\end{array}$

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

Let population $A$ contain $100$ observations $101, 102, \dots, 200$ and another population $B$ contain $100$ observations $151, 152, \dots, 250$. If $V_A$ and $V_B$ represent the variances of the two populations respectively,then what is $V_A / V_B$?

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:

Let the observations $x_{i} (1 \leq i \leq 10)$ satisfy the equations $\sum_{i=1}^{10}(x_{i}-5)=10$ and $\sum_{i=1}^{10}(x_{i}-5)^{2}=40$. If $\mu$ and $\lambda$ are the mean and the variance of the observations $x_{1}-3, x_{2}-3, \dots, x_{10}-3$,then the ordered pair $(\mu, \lambda)$ is equal to:

$A$ data consists of $n$ observations $x_1, x_2, ......, x_n$. If $\sum_{i=1}^n (x_i + 1)^2 = 9n$ and $\sum_{i=1}^n (x_i - 1)^2 = 5n$,then the standard deviation of this data is