One set containing five numbers has mean $8$ and variance $18$ and the second set containing $3$ numbers has mean $8$ and variance $24$. Then the variance of the combined set of numbers 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?

Let the mean of the data

be $5.$ If $m$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 \alpha}{m+\sigma^2}$ is equal to $..........$.

Mean and variance of a set of $6$ terms is $11$ and $24$ respectively and the mean and variance of another set of $3$ terms is $14$ and $36$ respectively. Then variance of all $9$ terms is equal to

The mean and variance of $10$ observations were calculated as $15$ and $15$ respectively by a student who took by mistake $25$ instead of $15$ for one observation. Then, the correct standard deviation 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