The weighted mean of the first $n$ natural numbers,where the weights are equal to the squares of the corresponding numbers,is:

The mean of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$. If an observation $x_{q}$ is replaced by $x_{q}^{\prime}$, then the new mean is:

Let $X = \{11, 12, 13, \ldots, 40, 41\}$ and $Y = \{61, 62, 63, \ldots, 90, 91\}$ be two sets of observations. If $\bar{x}$ and $\bar{y}$ are their respective means and $\sigma^2$ is the variance of all the observations in $X \cup Y$,then $|\bar{x} + \bar{y} - \sigma^2|$ is equal to $.................$.

If $\bar{x}_1$ and $\bar{x}_2$ are the means of two distributions such that $\bar{x}_1 < \bar{x}_2$ and $\bar{x}$ is the mean of the combined distribution,then

The mean of five observations is $4$ and their variance is also $4$. If three of the five observations are $1, 3, 4$,then the product of the other two is:

If the median and the range of four numbers $\{x, y, 2x + y, x - y\}$,where $0 < y < x < 2y$,are $10$ and $28$ respectively,then the mean of the numbers is