The mean and variance of $7$ observations are $8$ and $16,$ respectively. If five of the observations are $2,4,10,12,14 .$ Find the remaining two observations.
Let the remaining two observations be $x$ and $y$.
The observations are $2,4,10,12,14, x , y$
Mean, $\bar{x}=\frac{2+4+10+12+14+x+y}{7}=8$
$\Rightarrow 56=42+x+y$
$\Rightarrow x+y=14$
Varaiance $ = 16 = \frac{1}{n}\sum\limits_{i = 1}^7 {{{\left( {{x_i} - \bar x} \right)}^2}} $
$16=\frac{1}{7}[(-6)^{2}+(-4)^{2}+(2)^{2}$
$+(4)^{2}+(6)^{2}+x^{2}+y^{2}-2 \times 8(x+y)+2 \times(8)^{2}]$
$16=\frac{1}{7}\left[36+16+4+16+36+x^{2}+y^{2}-16(14)+2(64)\right]$ .......[ using $(1)$ ]
$16=\frac{1}{7}\left[108+x^{2}+y^{2}-224+128\right]$
$16=\frac{1}{7}\left[12+x^{2}+y^{2}\right]$
$\Rightarrow x^{2}+y^{2}=112-12=100$
$\Rightarrow x^{2}+y^{2}=100$ ........$(2)$
From $(1),$ we obtain
$x^{2}+y^{2}+2 x y=196$ .........$(3)$
From $(2)$ and $(3),$ we obtain
$2 x y=196-100$
$\Rightarrow 2 x y=96$ .........$(4)$
Subtracting $(4)$ from $(2),$ we obtain
$x^{2}+y^{2}-2 x y=100-96$
$\Rightarrow(x-y)^{2}=4$
$\Rightarrow x-y=\pm 2$ .........$(5)$
Therefore, from $(1)$ and $(5),$ we obtain
$x=8$ and $y=6$ when $x-y=2$
$x=6$ and $y=8$ when $x-y=-2$
Thus, the remaining observations are $6$ and $8 .$
Let $v_1 =$ variance of $\{13, 1 6, 1 9, . . . . . , 103\}$ and $v_2 =$ variance of $\{20, 26, 32, . . . . . , 200\}$, then $v_1 : v_2$ is
If $x_1, x_2,.....x_n$ are $n$ observations such that $\sum\limits_{i = 1}^n {x_i^2} = 400$ and $\sum\limits_{i = 1}^n {{x_i}} = 100$ , then possible value of $n$ among the following is
The mean and variance of a set of $15$ numbers are $12$ and $14$ respectively. The mean and variance of another set of $15$ numbers are $14$ and $\sigma^2$ respectively. If the variance of all the $30$ numbers in the two sets is $13$,then $\sigma^2$ is equal to $.........$.
The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & 2 & 3 & 4 & 5 & 6 & 7 \\ f & 4 & 9 & 16 & 14 & 11 & 6 \\ \hline \end{array}$
Find the standard deviation.
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$...................$