Question

Lot the observations $x_i\left(1\le i\le 10\right)$ the equations, $\sum _{i=1}^{10}\left(x_i-5\right)=10$ and $\sum _{i=1}^{10}{\left(x_i-5\right)}^2=40$, If $\mu$ and $\lambda$ are the mean and the variance of the observations, $x_1-3,x_2-3,\dots \dots ,x_{10}-3$, then the ordered pair $\left(\mu ,\lambda \right)$ is equal to :

• A. (6, 3)
• B. (3, 6)
• C. (3, 3)
• D. (6, 6)

Answer 1

Answer: (C)

$\sum _{i=1}^{10}(x_i-5)=10$
$\Rightarrow$ Mean of observation $x_i-5=\frac{1}{10}\sum _{i=1}^3(x_i-5)=1$
$\Rightarrow \mu =$ mean of observation $\left(x_i-3\right)$
$=$ (mean of observation $(x_i-5)$) $+2$
$=1+2=3$
Variance of observation
$x_i-5=\frac{1}{10}\sum _{i=1}^{10}(x_i-5{)}^2-$ (Mean of $(x_i-5)$)${}^2=3$
$\Rightarrow \lambda =$ variance of observation $\left(x_i-3\right)$
$=$ variance of observation $\left(x_i-5\right)=3$
$\therefore \left(\mu ,\lambda \right)-\left(3,3\right)$