A random variable $\mathrm{X}$ assumes value $1,2,3, \ldots \ldots . n$ with equal probabilities. If the…

A random variable $\mathrm{X}$ assumes value $1,2,3, \ldots \ldots . n$ with equal probabilities. If the ratio of variance of $=\sum p_i x_i^2-\left(\sum p_i x_i\right)^2$ to expected value of $\mathrm{X}$ is equal to 4 , then the value of $n$ is
  1. 35
  2. 50
  3. 30
  4. 25

Solution

$\begin{aligned} & \text { Variance }=\sum p_i X_i^2-\left(\sum p_i X_i\right)^2 \\ & =\frac{1}{n} \times \frac{n(n+1)(2 n+1)}{6}-\left\{\frac{1}{n} \times \frac{n(n+1)}{2}\right\}^2 \\ & =\frac{n(n+1)}{2 n}\left\{\frac{2 n+1}{3}-\frac{n(n+1)}{2 n}\right\} \\ & \text { Expected value }=\sum p_i x_i=\frac{1}{n} \cdot \frac{n(n+1)}{2}=\frac{n(n+1)}{2 n} \\ & \text { Ratio }=\frac{2 n+1}{3}-\frac{n(n+1)}{2 n}\end{aligned}$ $\Rightarrow n^2-25 n=0$ $\Rightarrow n(n-25)=0$ $\Rightarrow n=0$ or $n=25$ $\Rightarrow n=25$ as $n=0$ is not possible

Asked in: MHT CET 2022 (07 Aug Shift 2)

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