A random variable X assumes values 1, 2, 3, l | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The random variable $X$ takes values $1, 2, \ldots, n$ with probability $p_i = \frac{1}{n}$ for each $i$.The expected value $E(X) = \sum p_i x_i =

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(D) The random variable $X$ takes values $1, 2, \ldots, n$ with probability $p_i = \frac{1}{n}$ for each $i$.The expected value $E(X) = \sum p_i x_i = \frac{1}{n} \sum_{i=1}^{n} i = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$.The variance $V(X) = E(X^2) - [E(X)]^2 = \sum p_i x_i^2 - [E(X)]^2$.$E(X^2) = \frac{1}{n} \sum_{i=1}^{n} i^2 = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$.$V(X) = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 = \frac{(n+1)}{2} \left[ \frac{2n+1}{3} - \frac{n+1}{2} \right] = \frac{(n+1)}{2} \left[ \frac{4n+2-3n-3}{6} \right] = \frac{(n+1)(n-1)}{12} = \frac{n^2-1}{12}$.Given the ratio $\frac{V(X)}{E(X)} = 4$,we have $\frac{(n^2-1)/12}{(n+1)/2} = 4$.$\frac{(n-1)(n+1)}{12} \cdot \frac{2}{n+1} = 4$.$\frac{n-1}{6} = 4 \Rightarrow n-1 = 24 \Rightarrow n = 25$.

$A$ random variable $X$ assumes values $1, 2, 3, \ldots, n$ with equal probabilities. If the ratio of the variance of $X$ to the expected value of $X$ is equal to $4$,then the value of $n$ is:

Similar Questions

If $f(x) = \frac{x+2}{18}$ for $-2 < x < 4$ and $f(x) = 0$ otherwise,is the probability density function (p.d.f.) of a random variable $X$,then the value of $P(|X| < 2)$ is

If a fair coin is tossed $5$ times,the probability that heads does not occur two or more times in a row is

Two dice are thrown simultaneously. If $X$ denotes the number of sixes,find the expectation of $X$.

If $X$ is a Poisson variate such that $\alpha = P(X=1) = P(X=2)$, then $P(X=4)$ is equal to

If $f(x) = \begin{cases} 3(1 - 2x^2) & ; 0 < x < 1 \\ 0 & ; \text{otherwise} \end{cases}$ is a probability density function of $X$,then $P\left(\frac{1}{4} < x < \frac{1}{3}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module