The mean and standard deviation of $15$ observations were found to be $12$ and $3$ respectively. On rechecking it was found that an observation was read as $10$ in place of $12$ . If $\mu$ and $\sigma^2$ denote the mean and variance of the correct observations respectively, then $15\left(\mu+\mu^2+\sigma^2\right)$ is equal to$...................$

The mean and variance of $20$ observations are found to be $10$ and $4,$ respectively. On rechecking, it was found that an observation $9$ was incorrect and the correct observation was $11$. Then the correct variance 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 and the variance of 6 observation $a, b$, $68,44,48,60$ be $55$ and $194 $, respectively if $a>b$, then $a+3 b$ is

Let $r$ be the range and ${S^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} $ be the $S.D.$ of a set of observations ${x_1},\,{x_2},\,.....{x_n}$, then

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