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}$

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(a)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

$X$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X)$	$0.1$	$0.2$	$0.2$	$0.3$	$0.15$	$0.05$

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

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

Find the mean number of heads in three tosses of a fair coin.

For the following probability distribution,find the $Var(X)$.
$X$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X)$ $0.1$ $0.2$ $0.2$ $0.3$ $0.15$ $0.05$

(Given : $(0.25)^2 = 0.0625$,$(0.35)^2 = 0.1225$,$(0.45)^2 = 0.2025$)

$A$ random variable $X$ has the following probability distribution. For events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,what is the probability $P(E \cup F)$?

$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$

$P(X)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

The probability distribution of a random variable $X$ is given below.

$X = x$ $0$ $1$ $2$ $3$

$P(X = x)$ $\frac{1}{10}$ $\frac{2}{10}$ $\frac{3}{10}$ $\frac{4}{10}$

Then the variance of $X$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module