The variance of the first $n$ natural numbers is
$\frac{{{n^2} - 1}}{{12}}$
$\frac{{{n^2} - 1}}{6}$
$\frac{{{n^2} + 1}}{6}$
$\frac{{{n^2} + 1}}{{12}}$
The mean and the standard deviation (s.d.) of $10$ observations are $20$ and $2$ resepectively. Each of these $10$ observations is multiplied by $\mathrm{p}$ and then reduced by $\mathrm{q}$, where $\mathrm{p} \neq 0$ and $\mathrm{q} \neq 0 .$ If the new mean and new s.d. become half of their original values, then $q$ is equal to
If the mean deviation about the mean of the numbers $1,2,3, \ldots ., n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to
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.
Mean and standard deviation of 100 items are 50 and $4,$ respectively. Then find the sum of all the item and the sum of the squares of the items.
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