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 correct option(s) is/are

• A. $\mu =3$
• B. $\mu =5$
• C. $\lambda =3$
• D. $\lambda =5$

Answer 1

Answer: (A), (C)

$\lambda =3$

Given $\sum _{i=1}^{10}(x_i-5)=10$
$\Rightarrow$ Mean of the observations $(x_i-5)=\frac{10}{10}=1$
Hence mean of the observations $(x_i-3)=$(old mean)$+2$

$\Rightarrow \mu =1+2=3$

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