The arithmetic mean and standard deviation of a data set of nine numbers are $13$ and $5$ respectively. If $3$ is included as the $10^{th}$ item of the data,then the variance of the data set of ten numbers is:

If each of the observations $x_{1}, x_{2}, \ldots, x_{n}$ is increased by $a$,where $a$ is a positive or negative number,show that the variance remains unchanged.

Let $\sigma_1$ and $\sigma_2$ be the standard deviations of two distributions $D_1$ and $D_2$ respectively,and $D_1$ be more consistent than $D_2$. If the means of $D_1$ and $D_2$ are the same,then the percentage increase in the standard deviation of $D_2$ over the standard deviation of $D_1$ is:

The variance of the observations $8, 12, 13, 15, 22$ is:

$A$ scientist weighs $30$ fish. Their mean weight is $30 \text{ g}$ and the standard deviation is $2 \text{ g}$. Later,it is discovered that the weighing scale was not calibrated correctly and every fish's weight was recorded $2 \text{ g}$ less than the actual weight. What are the correct mean and standard deviation (in grams) of the fish weights,respectively?

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: