The discrete random variable X can take all p | vedclass

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

Question

(A) The probability distribution is given by:$E(X) = \sum_{i=1}^{k} x_i p_i = \frac{1}{k} + \frac{2}{k} + \dots + \frac{k}{k} = \frac{1}{k} \sum_{i=1}

Vedclass · Accepted Answer

$\frac{k^2-1}{12}$

Vedclass · Answer

(A) The probability distribution is given by:$E(X) = \sum_{i=1}^{k} x_i p_i = \frac{1}{k} + \frac{2}{k} + \dots + \frac{k}{k} = \frac{1}{k} \sum_{i=1}^{k} i = \frac{1}{k} \cdot \frac{k(k+1)}{2} = \frac{k+1}{2}$$E(X^2) = \sum_{i=1}^{k} x_i^2 p_i = \frac{1^2 + 2^2 + \dots + k^2}{k} = \frac{1}{k} \cdot \frac{k(k+1)(2k+1)}{6} = \frac{(k+1)(2k+1)}{6}$Variance $= E(X^2) - [E(X)]^2$$= \frac{(k+1)(2k+1)}{6} - \left( \frac{k+1}{2} \right)^2$$= \frac{2k^2 + 3k + 1}{6} - \frac{k^2 + 2k + 1}{4}$$= \frac{2(2k^2 + 3k + 1) - 3(k^2 + 2k + 1)}{12}$$= \frac{4k^2 + 6k + 2 - 3k^2 - 6k - 3}{12}$$= \frac{k^2 - 1}{12}$

$X = k$	$-2$	$-1$	$0$	$1$	$2$
$P(X = k)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$\lambda$	$2\lambda$	$3\lambda$	$4\lambda$

The discrete random variable $X$ can take all possible integer values from $1$ to $k$,each with a probability $\frac{1}{k}$. Then its variance is

Similar Questions

Let a sample space be $S = \{\omega_{1}, \omega_{2}, \ldots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?

Outcome $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$

$(b)$ $1$ $0$ $0$ $0$ $0$ $0$

The variance of the random variable $X$ having the following distribution is:
$X = k$ $-2$ $-1$ $0$ $1$ $2$
$P(X = k)$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$

Find the variance of the number obtained on a throw of a fair die.

Given below is the distribution of a random variable $X$:
$X=x$ $1$ $2$ $3$ $4$
$P(X=x)$ $\lambda$ $2\lambda$ $3\lambda$ $4\lambda$

If $\alpha=P(X < 3)$ and $\beta=P(X>2)$,then $\alpha: \beta=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module