The $S.D$ of $15$ items is $6$ and if each item is decreased or increased by $1$, then standard deviation will be
$5$
$7$
$\frac{91}{15}$
$6$
If both the means and the standard deviation of $50$ observations $x_1, x_2, ………, x_{50}$ are equal to $16$ , then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, …., (x_{50} - 4)^2$ is
Let $X=\{\mathrm{x} \in \mathrm{N}: 1 \leq \mathrm{x} \leq 17\}$ and $\mathrm{Y}=\{\mathrm{ax}+\mathrm{b}: \mathrm{x} \in \mathrm{X}$ and $\mathrm{a}, \mathrm{b} \in \mathrm{R}, \mathrm{a}>0\} .$ If mean and variance of elements of $Y$ are $17$ and $216$ respectively then $a + b$ is equal to
The variance of $10$ observations is $16$. If each observation is doubled, then standard deviation of new data will be -
The mean and variance of eight observations are $9$ and $9.25,$ respectively. If six of the observations are $6,7,10,12,12$ and $13,$ find the remaining two observations.
The variance of first $50$ even natural numbers is