Mean and variance of a set of $6$ terms is $11$ and $24$ respectively and the mean and variance of another set of $3$ terms is $14$ and $36$ respectively. Then variance of all $9$ terms is equal to
$40$
$30$
$50$
$35$
Find the standard deviation of the first n natural numbers.
Find the mean and variance of the frequency distribution given below:
$\begin{array}{|l|l|l|l|l|} \hline x & 1 \leq x<3 & 3 \leq x<5 & 5 \leq x<7 & 7 \leq x<10 \\ \hline f & 6 & 4 & 5 & 1 \\ \hline \end{array}$
Let $a_1, a_2, \ldots . a_{10}$ be $10$ observations such that $\sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50$ and $\sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \cdot \mathrm{a}_{\mathrm{j}}=1100$. Then the standard deviation of $a_1, a_2, \ldots, a_{10}$ is equal to :
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
The $S.D$. of the first $n$ natural numbers is