If mean and standard deviation of $5$ observations $x_1 ,x_2 ,x_3 ,x_4 ,x_5$ are $10$ and $3$, respectively, then the variance of $6$ observations $x_1 ,x_2 ,…..,x_3$ and $-50$ is equal to

The variance of the data $2, 4, 6, 8, 10$ is

The mean and standard deviation of $20$ observations are found to be $10$ and $2$, respectively. On respectively, it was found that an observation by mistake was taken $8$ instead of $12$ . The correct standard deviation is

Let $n \geq 3$. A list of numbers $x_1, x, \ldots, x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers $y_1, y_2, \ldots, y_n$ is made as follows $y_1=\frac{x_1+x_2}{2}, y_2=\frac{x_1+x_2}{2}$ and $y_j=x_j$ for $j=3,4, \ldots, n$.

The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Then, which of the following is necessarily true?

If the mean and variance of the data $65,68,58,44$, $48,45,60, \alpha, \beta, 60$ where $\alpha>\beta$ are $56$ and $66.2$ respectively, then $\alpha^2+\beta^2$ is equal to