Question

Lot the observations $x_{i}\left(1 \le i \le 10\right)$ the equations, $\sum\limits^{10}_{i = 1} \left(x_{i}-5\right) = 10$ and $\sum \limits ^{10}_{i = 1} \left(x_{i}-5\right)^{2} = 40$, If $\mu$ and $\lambda$ are the mean and the variance of the observations, $x_{1}-3, x_{2}- 3, \ldots\ldots, x_{10}-3$, 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)

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