Suppose a population $A $ has $100$ observations $ 101,102, . . .,200 $ and another population $B $ has $100$ observation $151,152, . . .,250$ .If $V_A$ and $V_B$ represent the variances of the two populations , respectively then $V_A / V_B$ is
$1$
$\frac{9}{4}$
$\frac{4}{9}$
$\frac{2}{3}$
Find the mean and variance for the first $n$ natural numbers
The sum of $100$ observations and the sum of their squares are $400$ and $2475$, respectively. Later on, three observations, $3, 4$ and $5$, were found to be incorrect . If the incorrect observations are omitted, then the variance of the remaining observations is
If each of the observation $x_{1}, x_{2}, \ldots ., x_{n}$ is increased by $'a'$ where $a$ is a negative or positive number, show that the variance remains unchanged.
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
The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$
where $A$ is a positive integer, has a variance of $160 .$ Determine the value of $A$.