Question

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

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

Answer 1

The correct option is C $(3,3)$

$\sum _{i=1}^{10}(x_i-5)=10$
$\Rightarrow \sum _{i=1}^{10}{x}_i-50=10$
$\Rightarrow \sum _{i=1}^{10}{x}_i=60$

$\mu =\frac{\sum _{i=1}^{10}(x_i-3)}{10}=\frac{\sum _{i=1}^{10}{x}_i-30}{10}=3$

Variance is unchanged, if a constant is added or subtracted from each observation
$\therefore \lambda =\text{Var}(x_i-3)=Var(x_i-5)$
$=\frac{\sum _{i=1}^{10}(x_i-5{)}^2}{10}-{\left(\frac{\sum _{i=1}^{10}(x_i-5)}{10}\right)}^2$
$=\frac{40}{10}-{\left(\frac{10}{10}\right)}^2$
$=3$