The variance of first $50$ even natural numbers is

In a series of $2n$ observations half of them equals $a$ and remaining half equals $-a$. If the standard deviation of observations is $2$ then $\left| a \right|$ equals

Let $ \bar x , M$ and $\sigma^2$ be respectively the mean, mode and variance of $n$ observations $x_1 , x_2,...,x_n$ and $d_i\, = - x_i - a, i\, = 1, 2, .... , n$, where $a$ is any number.
Statement $I$: Variance of $d_1, d_2,.....d_n$ is $\sigma^2$.
Statement $II$ : Mean and mode of $d_1 , d_2, .... d_n$ are $-\bar x -a$ and $- M - a$, respectively

The mean and $S.D.$ of the marks of $200$ candidates were found to be $40$ and $15$ respectively. Later, it was discovered that a score of $40$ was wrongly read as $50$. The correct mean and $S.D.$ respectively are...

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?

Consider a set of $3 n$ numbers having variance $4.$ In this set, the mean of first $2 n$ numbers is $6$ and the mean of the remaining $n$ numbers is $3.$ A new set is constructed by adding $1$ into each of first $2 n$ numbers, and subtracting $1$ from each of the remaining $n$ numbers. If the variance of the new set is $k$, then $9 k$ is equal to .... .