Let $n \geq 3$. A list of numbers $0 < x_1 < x_2 < \ldots < x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers is made as follows: $y_1=0, y_2=x_2, \ldots, x_{n-1}$ $=x_n-1, y_n=x_1+x_n$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?

Let ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance

Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .

Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.

The average marks of $10$ students in a class was $60$ with a standard deviation $4$ , while the average marks of other ten students was $40$ with a standard deviation $6$ . If all the $20$ students are taken together, their standard deviation will be

Let the mean and standard deviation of marks of class $A$ of $100$ students be respectively $40$ and $\alpha( > 0)$, and the mean and standard deviation of marks of class $B$ of $n$ students be respectively $55$ and $30-\alpha$. If the mean and variance of the marks of the combined class of $100+ n$ students are respectively $50$ and $350$,then the sum of variances of classes $A$ and $B$ is 

The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking it was found that, in the observations, $20$ was misread as $5$ . Then, the correct variance is equal to……