Question

In a discrete data $\frac{1}{4}^{\text {th }}$ of the observations are equal to $a$, another $\frac{1}{4}^{t h}$ of the observations are equal to $-a$. Out of the reamining, half of them are equal to $b$ and the rest are equal to $-b$. If the variance of all observations is $(a b)$, then

• A. $a^{2}=4 b^{2}$
• B. $a=-2 b$
• C. $a=b$
• D. $a=3 b$

Answer 1

Answer: (C)

Given, $\frac{1}{4}$ th observation $=a$
another $\frac{1}{4}$ th observation $=-a$
$\frac{1}{4}$ th observation $=b$
$\Rightarrow \frac{1}{4}$ th observation $=-b$
$\therefore \bar{x}=\frac{a- a+ b- b}{n}=0$
Variance $=a b$
$a b=\frac{\sum x_{i}^{2}}{n}-(\bar{x})^{2}=\frac{\sum x_{i}^{2}}{n}$
$\Rightarrow a b=\frac{\frac{n}{4}\left(a^{2}+a^{2}\right)+\frac{n}{4}\left(b^{2}+b^{2}\right)}{n}$
$\Rightarrow a b=\frac{a^{2}+b^{2}}{2}$
$\Rightarrow a^{2}+b^{2}-2 a b=0$
$\Rightarrow (a-b)^{2}=0$
$\Rightarrow a=b$