Q75 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics

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

  • A
    $a^2=4 b^2$
  • B
    $a=-2 b$
  • Correct answer
    C
    $a=b$
  • D
    $a=-3 b$

Solution

$\begin{aligned} & \text { Given, } \frac{1}{4} \text { th observation }=a \\ & \text { another } \frac{1}{4} \text { th observation }=-a \\ & \quad \frac{1}{4} \text { th observation }=b \Rightarrow \frac{1}{4} \text { th observation }=-b \\ & \therefore \quad \bar{x}=\frac{a-a+b-b}{n}=0 \\ & \text { Variance }=a b \\ & \quad a b=\frac{\Sigma x_i^2}{n}-(\bar{x})^2=\frac{\Sigma x_i^2}{n} \\ & \Rightarrow a b=\frac{\frac{n}{4}\left(a^2+a^2\right)+\frac{n}{4}\left(b^2+b^2\right)}{n} \Rightarrow a b=\frac{a^2+b^2}{2} \\ & \Rightarrow \quad a^2+b^2-2 a b=0 \Rightarrow(a-b)^2=0 \Rightarrow a=b\end{aligned}$

Hence, the correct answer is option: 'C'.

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