If the variance of $x_1, x_2, \ldots, x_n$ is $\sigma_x^2$,then the variance of $\lambda x_1, \lambda x_2, \ldots, \lambda x_n$ (where $\lambda \neq 0$) is:

Find the variance of the first $n$ natural numbers.

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

Let $x_1, x_2, \ldots, x_{10}$ be ten observations such that $\sum_{i=1}^{10}(x_i-2)=30$,$\sum_{i=1}^{10}(x_i-\beta)^2=98$,$\beta > 2$ and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2(x_1-1)+4\beta, 2(x_2-1)+4\beta, \ldots, 2(x_{10}-1)+4\beta$,then $\frac{\beta\mu}{\sigma^2}$ is equal to:

If the mean of the frequency distribution is $28$,then its variance is $........$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$2$	$3$	$x$	$5$	$4$