Let the mean and variance of $12$ observations be $\frac{9}{2}$ and $4$ respectively. Later on, it was observed that two observations were considered as $9$ and $10$ instead of $7$ and $14$ respectively. If the correct variance is $\frac{m}{n}$, where $m$ and $n$ are co-prime, then $m + n$ is equal to

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?

The mean and variance of $7$ observations are $8$ and $16,$ respectively. If five of the observations are $2,4,10,12,14 .$ Find the remaining two observations.

The mean and standard deviation of a group of $100$ observations were found to be $20$ and $3,$ respectively. Later on it was found that three observations were incorrect, which were recorded as $21,21$ and $18 .$ Find the mean and standard deviation if the incorrect observations are omitted.

The mean of the numbers $a, b, 8,5,10$ is $6$ and their variance is $6.8$. If $M$ is the mean deviation of the numbers about the mean, then $25\; M$ is equal to