For the probability distribution given by | vedclass

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

Question

(B) The mean $(\mu)$ of the probability distribution is given by $\mu = \sum x_{i} P_{i}$.$\mu = (0 	imes \frac{25}{36}) + (1 	imes \frac{5}{18}) +

Vedclass · Accepted Answer

$\frac{1}{3} \sqrt{\frac{5}{2}}$

Vedclass · Answer

(B) The mean $(\mu)$ of the probability distribution is given by $\mu = \sum x_{i} P_{i}$.$\mu = (0 	imes \frac{25}{36}) + (1 	imes \frac{5}{18}) + (2 	imes \frac{1}{36}) = 0 + \frac{10}{36} + \frac{2}{36} = \frac{12}{36} = \frac{1}{3}$.The variance $(\sigma^2)$ is given by $\sigma^2 = \sum x_{i}^2 P_{i} - \mu^2$.First,calculate $\sum x_{i}^2 P_{i} = (0^2 	imes \frac{25}{36}) + (1^2 	imes \frac{5}{18}) + (2^2 	imes \frac{1}{36}) = 0 + \frac{10}{36} + \frac{4}{36} = \frac{14}{36} = \frac{7}{18}$.Now,$\sigma^2 = \frac{7}{18} - (\frac{1}{3})^2 = \frac{7}{18} - \frac{1}{9} = \frac{7-2}{18} = \frac{5}{18}$.Therefore,the standard deviation $\sigma = \sqrt{\frac{5}{18}} = \sqrt{\frac{5 	imes 2}{18 	imes 2}} = \sqrt{\frac{10}{36}} = \frac{\sqrt{10}}{6}$.Alternatively,$\sigma = \sqrt{\frac{5}{18}} = \frac{1}{3} \sqrt{\frac{5}{2}}$.Thus,option $B$ is correct.

$x = x_{i}$	$0$	$1$	$2$
$P_{i}$	$\frac{25}{36}$	$\frac{5}{18}$	$\frac{1}{36}$

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$F(X=x)$	$0.2$	$0.37$	$0.48$	$0.62$	$0.85$	$1$

For the probability distribution given by

$x = x_{i}$ $0$ $1$ $2$

$P_{i}$ $\frac{25}{36}$ $\frac{5}{18}$ $\frac{1}{36}$

the standard deviation $(\sigma)$ is

Similar Questions

The cumulative distribution function $F(X)$ of a discrete random variable $X$ is given by the following table:
$X$ $1$ $2$ $3$ $4$ $5$ $6$
$F(X=x)$ $0.2$ $0.37$ $0.48$ $0.62$ $0.85$ $1$

Then $P[X=4] + P[X=5] = $

Let $X$ be a random variable such that the probability function of a distribution is given by $P(X=0) = \frac{1}{2}$ and $P(X=j) = \frac{1}{3^j}$ for $j = 1, 2, 3, \ldots, \infty$. Then the mean of the distribution and $P(X \text{ is positive and even})$ respectively are:

Let $X$ be a random variable with probability distribution function,$P(X=x)=K\left(\frac{2}{5}\right)^x, x=1, 2, 3, \ldots$. Then,the value of $K$ is

Two cards are drawn simultaneously (or successively without replacement) from a well-shuffled pack of $52$ cards. Find the mean,variance,and standard deviation of the number of kings.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

For the probability distribution given by $x = x_{i}$ $0$ $1$ $2$ $P_{i}$ $\frac{25}{36}$ $\frac{5}{18}$ $\frac{1}{36}$ the standard deviation $(\sigma)$ is