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MHT CET · Maths · Probability

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. A 35
  2. B 50
  3. C 30
  4. D 25
Verified Solution

Answer & Solution

Correct Answer

(D) 25

Step-by-step Solution

Detailed explanation

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